Science GR11 LB

Technical Science Grade 11 Learner's Book This book was developed with the participation of the Department of Basic Education of South Africa and funding by the Sasol Inzalo Foundation Science_GR11_LB.indb 1 2018/02/02 16:59

Technical Science Grade 11 Learner’s Book This book was developed with the participation of the Department of Basic Education (DBE) of South Africa, Magic Moments Consulting & Services and ACP Project Management & Publishing Services; and funded as an ongoing project by Sasol Inzalo Foundation. © 2017. Copyright in the work is vested in publisher. Copyright in the text remains with the contributors. © Illustrations and design ACP Project Management and Publishing Services, 2017 First published in 2017 Contributors from Magic Moments Consulting & Services Dr. Samuel Khoza, Dr. Kgadi Mathabathe, Mokhulu Matshika, Sandile Vilane Contributors from the Department of basic Education Jayanthi SK Maharaj, Morongwa Masemola, Mmaphaseka Stephen, Modiba Phosa, Mosala Leburu, Khiba Tshiu Stephen Cover design by Sasol Inzalo Foundation Cover image by Kim Stevens Illustrations by Will Alves Layout and typesetting by Damian Gibbs and Nazley Samsodien in ITC Stone Serif 9pt on 13.5pt Editing, Proofreading and Project Management by ACP Project Management and Publishing Services ISBN: 978-0-620-74313-6 Your freedom to legally copy this book This work is published under a Creative Commons Attribution-Non Commercial 4.0 Unported License (CCBY-NC 4.0) http://creativecommon.sorg/licenses/by-nc/4.0/ You are allowed to freely copy this book subject to the following criteria: You can photocopy, print and distribute it as you like. You may download it onto any electronic device, distribute it via email, and upload it to your website at no charge. You may adapt the text and illustrations. Attribution When doing any of the above, you should provide clear acknowledgement of the license/copyright holders. This attribution should include the name(s) of the original book(s) and the publisher, as well as acknowledgement of SaIF and the Department of Basic Education of South Africa. You should also provide the Creative Commons web address (http://creative commons.org/licenses/by-nc/4.0/), which explains this type of license. If you have made any changes to the material, you must indicate what changes you made. You may not in any way suggest that the licensor endorses you, or uses any of your materials or your changes to the material. Restrictions You may not make copies of this book in part or in full – in printed or electronic or audio or video or whatever form – for a profit seeking purpose. Rights of other copyright holders All reasonable efforts have been made to ensure that materials included are not already copyrighted to other entities, or in a small number of cases, to seek permission from and acknowledge copyright holders. In some cases, this may not have been possible. The publishers welcome the opportunity to redress this with any unacknowledged copyright holders. Science_GR11_LB.indb 2 2018/02/02 16:59

Contents Chapter 1: Mechanics 1 1.1 Introduction 2 1.1.1 Sign convention: The Cartesian plane 3 1.1.2 Sign convention: Compass directions 5 1.1.3 Sign convention: Bearing 5 1.1.4 Graphs and relationships between variables 7 1.2 The theorem of Pythagoras and perpendicular forces 11 1.3 Practical application of the Pythagoras’ theorem 13 1.3.1 The 3-4-5 technique for squaring corners 13 1.3.2 Using Pythagoras’ Theorem to determine immeasurable lengths 14 1.3.3 Further applications for using Pythagoras’ Theorem 15 1.4 Resultant of collinear forces 16 1.5 Resultant of forces in 2-D 18 1.5.1 Head-to-tail method 18 1.5.2 Parallelogram of forces 20 1.6 Resolving vectors 25 1.7 Friction 28 1.7.1 Importance of friction 28 1.7.2 So what is friction? 29 1.7.3 Types of frictional forces 29 Review exercises 34 Chapter summary 36 Chapter 2: Magnetism 39 2.1 Introduction 40 2.2 Magnet 41 2.2.1 Geographic North and South 41 2.2.2 Magnetic north and south 41 2.3 Magnets 42 2.3.1 What magnets are 42 2.3.2 A dipole magnet 42 2.3.3 Technological applications of magnetism 44 2.4 Magnetic domains 46 Science_GR11_LB.indb 3 2018/02/02 16:59

2.5 Inducing unmagnetised iron (Fe) 47 2.6 Magnetic field 48 2.7.1 Magnetic field between two bar magnets 48 2.7.2 Properties of magnetic field lines 51 2.7 Earth’s magnetic field 52 2.7.1 Geographic North vs Magnetic North Pole 53 2.7.2 The importance of the magnetic field to life on Earth 53 2.7.3 Aurora borealis aurorae (pronounced or-roar-ee) 54 Review exercises 56 Chapter 3: Waves and sound 59 3.1 Introduction 60 3.2 Pulses 61 3.2.1 Transverse pulse 61 3.2.2 Longitudinal pulse 62 3.3 Waves 62 3.3.1 Medium 63 3.3.2 Transverse wave 63 3.3.3 Longitudinal wave 64 3.4 Wave terminology 66 3.4.1 Frequency and period 66 3.4.2 Difference between frequency and period 67 3.4.3 Wave velocity 68 3.5 Speed of sound in different media 70 3.6 Formal experiment: Determine the speed of sound in air 70 3.6.1 Speed of light versus speed of sound 71 3.7 Superposition of waves 77 3.8 Interference 77 3.8.1 What happens when waves come together? 77 3.8.2 Constructive interference 77 3.8.3 Destructive interference 78 3.9 Sound waves 80 3.9.1 Some applications of ultrasound and infrasound 81 3.9.2 Pitch and loudness 83 3.9.3 Pitch and loudness on oscilloscope 84 3.9.4 Echo 84 Review exercises 85 Science_GR11_LB.indb 4 2018/02/02 16:59

Chapter 4: Electricity 89 4.1 Electrostatics 90 4.1.1 Introduction 90 4.1.2 Coulomb’s law 91 4.2 Electric field 95 4.2.1 Electric field lines 97 4.2.2 The electric field between two parallel plates 98 4.2.3 Applications of electrostatics in technology 100 4.3 Electric circuits 102 4.3.1 Introduction 102 4.3.2 Resistance 105 4.3.3 Ohm’s law 105 4.3.4 Series and parallel network 110 4.4 Internal resistance and emf (electromotive force) 118 Review exercises 121 Chapter 5: Heat and thermodynamics 127 5.1 Introduction 128 5.1.1 Units for measuring heat 130 5.2 Specific heat capacity 131 5.3 Heat capacity 121 5.3.1 Practical applications of heat capacity in technology 133 5.4 Law of conservation of energy 135 5.4.1 Calorimetry 137 5.5 Thermodynamics 139 5.5.1 Thermodynamics system 140 5.5.2 Thermodynamic variables 140 5.5.3 Internal energy 140 5.5.4 Heat engine 144 5.6 Efficiency of a heat engine 145 5.7 Second law of thermodynamics 146 5.8 Refrigerators 146 5.8.1 Evaporative cooling 147 Review exercises 148 Chapter summary 149 Science_GR11_LB.indb 5 2018/02/02 16:59

Chapter 6: Chemical change 151 6.1 Introduction 152 6.2 Using rules to assign oxidation numbers 153 6.2.1 Using oxidation number to identify redox reaction 154 6.2.2 Oxidation occurs when valence electrons are lost in a chemical reaction 156 6.2.3 Reduction occurs when valence electrons are gained in a chemical reaction 156 6.2.4 Oxidation and reduction are paired reactions 156 6.3 Application of redox reaction 158 6.4 Nonspontaneous reactions redox reactions occur in electrolytic cells 160 6.4.1 Electroplating 161 6.4.2 Electrolysis 162 Chapter summary 163 Selected answers 165 Science_GR11_LB.indb 6 2018/02/02 16:59

1 Mechanics Objectives By end of this chapter learners will gain a broad understanding and acquire skills relating to the following concepts: • sign convention • drawing and interpreting graphs • applying the Theorem of Pythagoras • determining the resultant force in two dimension • resolving forces into components. Science_GR11_LB.indb 1 2018/02/02 16:59

Why the study of Mechanics matters Mechanics is the study of the relationships between energy, motion and force. Mechanics is a large field and its study is essential to the understanding of physics. It also gives us an understanding of ‘how the universe works’. There are many applications of Mechanics in our daily lives. Have you ever wondered why a car would skid across a wet road? To successfully answer that question, you will need to know about force and motion; and that is part of what we will learn about in this chapter. Note: Motion is the action of changing location or position. The study of motion without regard to the forces or energies that may be involved is called kinematics. N 360 NW NE 315 45 D C W 270 90 E F2 FR 225 135 α SW SE θ = 60° 180 A B S F1 m2 m m2 θ2 1.1 Introduction In Grade 10 we learnt that physical quantities can be classified as either vector or scalar quantities. A scalar quantity has magnitude only. Speed is an example of a scalar quantity. For example, we can say that an object moves at 12 m.s–1. This quantity only tells us about the magnitude (size) of the movement. It does not tell us about the direction in which the object is moving. In Physical Sciences, speed is a scalar quantity, or physical quantity with a magnitude only. 2   Technical Science | Grade 11 Science_GR11_LB.indb 2 2018/02/02 16:59

A vector quantity is a physical quantity that has both magnitude and direction and thus can be represented using an arrow-tipped line segment known as a vector, as shown in Figure 1.1 below. Tail Head Figure 1.1 A vector If we had to, for example, say that a man is running at 10 m.s–1 in a westward direction, this would be classified as a vector quantity. 10 m.s–1 in a westward direction not only gives us the magnitude (size) of the movement, but it also gives us the direction. The magnitude of a vector quantity is represented by the length of the line segment and the direction is shown by the arrowhead. Note: Vectors can sometimes act at an angle. For example, to specify directions, pilots and air traffic controllers use bearings. Bearing is an angle measured clockwise from the North pole of the compass. The concept of bearing will be discussed later in this chapter. 1.1.1 Sign convention: The Cartesian Plane A Cartesian Plane is the set of two perpendicular axes (x- and y-axes), with the horizontal axis being the x-axis and the vertical axis being the 5 y-axis. This is the very useful tool for graphical y – axis representation of vectors. 4 The point where the two axes intersect is called 3 the origin and is indicated by the points (0; 0) 2 as shown in Figure 1.2. This is the point against which everything is measured. 1 Cartesian coordinates are numbers that indicate Origin the location of a point relative to a fixed –5 –4 –3 –2 –1 0 1 2 3 4 5 reference point (the origin), being its shortest x – axis –1 (perpendicular) distance from two fixed axes (or three planes defined by three fixed axes) that –2 intersect at right angles at the origin. –3 Cartesian coordinates are expressed as (x; y), –4 where x is the horizontal displacement from the origin (positive if to the right; negative if to –5 the left), and y is the vertical displacement from Figure 1.2 A Cartesian Plane the origin (positive if upwards; and negative if downwards). The x-coordinate is referred to as abscissa and the y-coordinate is called the ordinate. This is shown in Figure 1.3. Q (x; y) Q (4; 3) abscissa ordinate Figure 1.3 The Cartesian coordinates Chapter 1 Mechanics 3 Science_GR11_LB.indb 3 2018/02/02 16:59

Example 1.1 THE CARTESIAN PLANE Consider the Cartesian Plane shown below: y – axis 4 3 A 2 1 –4 –3 –2 –1 0 1 2 3 4 x – axis –1 –2 –3 –4 a) By how many units and in what direction has the point A been displaced along the x-axis? b) By how many units and in what direction has the point A been displaced along the y-axis? c) Write down the coordinates of point A. SOLUTION a) Trace a line that goes vertically downwards (perpendicular) to the x-axis. Read the value where your line intersects with the x-axis. In this case the values of, which is a displacement of the point A along the x-axis is 3 units, and it is directed to the left of the origin. This is a negative x direction b) If you trace a horizontal line from the point A that goes (perpendicular) to the y-axis, you will notice that the line intersects the y-axis at y = 2 units above the origin. Mathematically, this is the positive y direction. c) The coordinates of A are: A = (–3; 2). Activity 1.1 1. Thabiso decides to take his bicycle to go to the 6 shop. He rides his bicycle from his home to the 5 shop and back home. Thabiso then records the Distance (km) 4 distance he travelled from home every minute D and plots a graph of the results. 3 C Refer to the graph alongside and answer the 2 questions that follows: 1 0 1 2 3 4 5 6 7 8 9 10 11 Time (min) 4   Technical Science | Grade 11 Science_GR11_LB.indb 4 2018/02/02 16:59

Activity 1.1: Plotting a graph a) How far is the shop from Thabiso’s home? b) How far was Thabiso away from home? c) How long was Thabiso in the shop? Motivate your answer by referring to the graph. d) How far from home was Thabiso after 3 minutes? e) How far from home was Thabiso after 10 minutes? f) Explain the meaning of the flatness of the graph between the 8th and 10th minute. g) Write down the coordinates of the points C and D. Example 1.2 SIGN CONVENTIONS COMPASS DIRECTIONS The compass method is commonly used to determine the direction of a vector quantity. A compass has four cardinal points, namely, North, South, East, and West. All other directions can be derived from these cardinal points. If a vector does not point exactly in one of the cardinal points of a compass, we use an angle. For example, in Figure 1.3 below, the vector (the solid arrow), is pointing 70° North of West. COMPASS DIRECTIONS The direction of this vector can be described as: W 70° N (West 70° N North); or N 20° W (North 20° West). To determine the direction accurately it is necessary to start with the vector pointing along the West direction, then to rotate the vector towards the North point of 70° W E the compass. BEARING Depending on the context, bearing is used to refer to either the S direction of motion; the direction of a distant object relative to the current course (or the ‘change’ in course needed to get to a distant Figure 1.3 The graphical object); or the degrees away from North of a distant point relative to representation of a vector using the current point. a compass N In navigation, absolute bearing refers to the angle between the NW NE magnetic North (magnetic bearing) or true North (true bearing) and an object. For example, an object that lies to the East has an absolute bearing of 90˚. Relative bearing refers to the angle between W 135° E the object’s forward direction, and the location of another object. For example, an object’s relative bearing of 0˚ would be dead ahead; SW SE an object’s relative bearing of 180˚ would be behind. S Bearings are measured in degrees and are always written as a three- digit number, for example 275° or 080° (for 80°). Figure 1.4 Bearing as a measure of vector direction Chapter 1 Mechanics 5 Science_GR11_LB.indb 5 2018/02/02 16:59

A bearing is a direction relative to a fixed point. Given just an angle, the convention is to define the angle clockwise with respect to North. So, a vector with a direction of 135° has been rotated clockwise 135° relative to North as shown in Figure 1.4. MEASURING BEARING Bearing is an angle measured clockwise from the North pole of the compass. For example, in the morning there was a strong Easterly wind, refer to Figure 1.5a, but later it swung around to N 147˚ S (North 147˚ South), refer to Figure 1.5b. N N 90° 147° W E W E S S Figure 1.5a Compass direction Figure 1.5b Bearing DRAWING VECTORS Draw the following vectors: a) 20 km due North b) 15 m.s–1 060˚ c) 120 km.h–1 225˚ 1 ANALYSE Given: Magnitudes and directions Unknown: Graphical representation of the given information 2 SOLVE Most of these vectors cannot just be drawn without the use of a suitable conversion scale. When choosing your scale, make sure that your representation of the vector will not be too too small or too big. a) Let 1 cm : 20 km b) Let 1 cm : 3 m.s–1, so that 5 cm : 15 m.s–1 20 km of due North 0 10 20 30 40 N 50 15 m · s –1 N 60 70 60° 80 W E 90 100 1 W E 10 12 0 13 S 0 014 015 160 170 180 S 6   Technical Science | Grade 11 Science_GR11_LB.indb 6 2018/02/02 16:59

c) Let 1 cm : 24 km.h–1, so that 5 cm : 120 km.h–1 N Note: A bearing of 225˚ is more than 180˚, which is a maximum 170 180 160 150 angle a protractor can measure. Note that 225˚ = 180˚ + 45˚, 14 0 0 so rotate your protractor such that 0˚ points South, and then 13 120 measure the 45˚ that will be added to 180˚ to make 225˚. 110 100 W 90 E 80 70 60 50 40 30 20 0 10 S Activity 1.2 1. Define the following terms: 2. Draw the following vectors to scale: a) vector quantity a) 10 cm N 450˚ W b) scalar quantity b) 25 m.s–1 22˚ South of East c) bearing c) 100 km 075˚ d) 205 mm 300˚ e) 125 N 080˚ Example 1.3 GRAPHS AND RELATIONSHIPS BETWEEN VARIABLES In mathematics we learned that a function (for example, a y 5 straight line) can be represented using a graph. Different 4 graphs are used for different purposes. So, for example, Q (4:3) 3 when you are conducting an investigation that requires 2 numerical data collection, a graph may be drawn to 1 represent the information collected, and variables under investigation may show how they relate to each other. –5 –4 –3 –2 –1 0 1 2 3 4 5 x –1 Graphs can be plotted on a Cartesian plane using points. –2 The location of a point on a Cartesian plane is found by –3 indexing numerical values (coordinates) along numbered –4 grid lines. This is shown in Figure 1.6 below. –5 So, to locate point Q (4; 3) on a Cartesian Plane, we need to position the point on the horizontal line at and a Figure 1.6 Plotting points on a Cartesian vertical line (intersecting the y-axis) at 3; and drawing the Plane horizontal line (intersecting the y-axis) at 4. These are called the x-axis and y-axis respectively. Point Q will be located at the intersection of the two lines. Chapter 1 Mechanics 7 Science_GR11_LB.indb 7 2018/02/02 16:59

IDENTIFYING VARIABLES When investigating how two variables relate to one another (under controlled conditions) we need to be able to identify the variables. Generally, there is a dependent variable, independent variable and a controlled (or constant) variable. The independent variable is the one that is being manipulated and changed at constant intervals. It is plotted on the x-axis on a Cartesian plane. The dependent variable responds to the manipulated variable. It is dependent on the changes made to the independent variable and is plotted on the y-axis of the Cartesian plane. The controlled (or constant) variables are the ones that will be fixed and kept constant throughout the investigation in order to get unbiased (fair) results. GRAPHS WITH A DIRECT RELATIONSHIP Two variables are directly proportional if (and only if) they increase or decrease at the same rate. y is directly proportional to x, if (and only if) the ratio y : x is constant. Consider the table below: x y y:x 1 1 1 2 2 1 3 3 1 4 4 1 5 5 1 6 6 1 Because the ratio y : x is constant throughout, it can be y 5 concluded that y is directly proportional to x. 4 Plotting the graph of y : x will give the graph shown in 3 Figure 1.6. 2 The graph produced by variables that are directly 1 proportional is a straight line graph. If the graph is –5 –4 –3 –2 –1 0 1 2 3 4 5 x extended so that it cuts the axes, it will pass through the –1 origin. So, y is directly proportional to x, if: –2 • x and y increases by the same factor –3 • the graph of y versus x is a straight line passing through –4 the origin. –5 Mathematically this is written as: y α x Figure 1.6 Plotting a graph to predict the Whenever you have a direct proportion as stated above you relationship between variables that are can change it into an equation by using a proportionality directly proportional constant: y = kx 8   Technical Science | Grade 11 Science_GR11_LB.indb 8 2018/02/02 16:59

k in this equation is a constant (sometimes referred to as the proportionality constant). In this case it is also the slope of the graph of y versus x. The general equation we derive from this relationship is y1 y2 shown by the equation: x = x 1 2 Thus, if we graph data for two related quantities, and that graph yields a straight line through the origin, we know that the two quantities are in direct proportion. A straight line graph has a constant gradient, and the general equation y = mx + c; where m is the gradient (or slope) of the graph. To calculate the gradient of the graph, we use the following formula: m = ∆y ∆ , where c is the y-intercept of the graph. x INVERSE PROPORTIONALITY AND THE HYPERBOLA Generally, y is said to be inversely proportional to x if x increases and y decreases. There has to be the pattern at which the two variables change: y must be decreasing by the same rate at which x is increasing, such that the product of x and y is constant. Consider the table below: x y xy 1 1 1 2 0.5 1 3 0.33 0.99 ≈ 1 4 0.25 1 5 0.5 1 From the table we can see that the product, xy, is constant. Thus, y is inversely proportional to x. When we plot the points on the Cartesian Plane, the graph shown in y 5 Figure 1.7 is produced. This type of a graph is called hyperbola, and it does 4 not intersect with the axes. 3 2 It has a general equation xy = k, where k is a proportionality constant or even more general, x1y1 = x2y2: 1 0 x This relationship between x and y is written mathematically as: y α 1x , which 1 2 3 4 5 reads: y is inversely proportional to x. Figure 1.7 hyperbola graph In conclusion, we can say that two variables are inversely proportional if: • the values of y must decrease by the same factor at which the values of x are increasing • the graph of y versus x must be a hyperbola, that does not intersect with the axes. Example 1.3 DIRECT AND INVERSE PROPORTIONALITY Find the general formula and draw graphs for the relationships described below: a) a is directly proportional to F, and the proportionality constant is 5 b) the distance, D, travelled by an object is directly proportional to time, t, and the proportionality constant is 10 c) V is inversely proportional to P, and the proportionality constant is 2 d) F is inversely proportional to r2, and the proportionality constant is 6 Chapter 1 Mechanics 9 Science_GR11_LB.indb 9 2018/02/02 16:59

1 ANALYSE For two variables directly proportional to one another, the general formula has the format y = kx and the graph representing the relationship is a straight line passing through the origin. For two variables inversely proportional to one another, the general formula is xy = k, and the graph representing the relationship is a hyperbola. 2 SOLVE a) General formula: a = 5F c) The general formula: PV = 2 y a = 5F (1; 2) V α (5; 1) (2; 1) P F d) Following the same reasoning as in c) we find b) General formula: D = 10t that Fr2 = 6 D = 10t (10; 1) (1; 6) D F (2; 3) t r² Activity 1.3 1. State the condition(s) required for two variables to be directly proportional to one another. 2. State the condition(s) required for two variables to be inversely proportional to one another. 3. The following information is supplied to prepare concrete with the required strength to build a house: every one bag of cement requires four full wheel barrows of sand. a) Use the above information to complete this table: Number of cement bags Number of level-full wheel Ratio barrows with sand Sand : Cement 1 4 4 2 3 4 5 10   Technical Science | Grade 11 Science_GR11_LB.indb 10 2018/02/02 16:59

b) From the information in the table, identify the dependent and independent variable. c) Use the information in the table to plot the graph. Correctly label the axes and use a suitable scale. d) From the graph and the information in the table, determine what the relationship between the number of cement bags and the number of wheel barrows filled with sand is. e) Express the relationship between the variables as a mathematical equation. 4. Learners in a class were divided into groups and each group was asked to make a 24 cm2 rectangular chart. Each group was given one dimension of the rectangle and had to find the other dimension, such that the area of their charts equaled 24 cm2. Group Number Length (cm) Width (cm) Area (cm2) A 1 24 24 B 2 24 C 3 24 D 4 24 E 5 24 a) Complete the table: b) Use the information in your completed table to plot the graph, Width versus Length c) What is the relationship between the width and the length of rectangles with the same area? d) Use your results to two different mathematical equations that can be used to express the relationship between the length and the width for rectangles with a fixed area. e) If one group was given a length of 10 cm, use the equation obtained in d) to find the width they would have to use. 1.2 The theorem of Pythagoras and perpendicular forces The theorem of Pythagoras states that the area of a square on the hypotenuse is equal to the sum of the squares on the other two sides. It is used to calculate the dimension of one side of the right- angled triangle if two sides are given. Pythagoras’ theorem can be represented graphically, as shown in hypotenuse Figure 1.8. opposite θ In a right-angled triangle, the longest side (the one opposite the adjacent interior angle of 90˚) is called the hypotenuse. The side opposite the given angle is called the opposite and adjacent side, and are named relative to the given angle, as shown in Figure 1.8 below. When two or more forces act on an object, their resultant force can be calculated. If these forces are perpendicular to one another, Figure 1.8 A graphical Pythagoras’ Theorem becomes a very useful technique in finding representation of Pythagoras’ their resultant force. Theorem Chapter 1 Mechanics 11 Science_GR11_LB.indb 11 2018/02/02 16:59

Example 1.4 APPLICATION OF PYTHAGORAS’ THEOREM Determine the magnitude of the resultant of two forces: F⃗1 = 3 N to the right and F⃗2 = 4 N up. 1 ANALYSE Given: F⃗1 and F⃗2 Unknown: Resultant force F⃗R 2 SOLVE Start by representing the given information graphically and use Pythagoras’ Theorem to determine the magnitude of F⃗R. FR F2 F1 FR2 = F12 + F22 FR2 = (3)2 + (4)2 FR2 = 25 ∴ FR = 5 Example 1.5 APPLICATION OF PYTHAGORAS’ THEOREM Calculate the magnitude and direction of the resultant force for the forces 4 N to the left and 4 N downwards: 1 ANALYSE Given: F⃗1 and F⃗2 Unknown: Resultant force F⃗R 2 SOLVE Start by representing the given information graphically and use Pythagoras’ Theorem to determine the magnitude of F⃗R. 12   Technical Science | Grade 11 Science_GR11_LB.indb 12 2018/02/02 16:59

Fx Fy FR The magnitude of the resultant force can be calculated using Pythagoras’ Theorem: FR2 = F12 + F22 FR2 = (−4)2 + (−4)2 FR2 = 32 ∴ FR = 5,66 The direction of the resultant force can be found using the tan function: tanθ = FF2 1 tanθ = 44 = 1 θ = 45° anticlockwise from the negative x–direction F⃗R = 5,66 N at 45° anticlockwise from the negative x-direction 1.3 Practical application of Pythagoras’ Theorem 1.3.1 The 3-4-5 technique for squaring corners One of the challenges in carpentry and construction is making sure that corners are ‘square’. If they are not, any other fittings such as tiles, carpets and even roofing, will noticeably be ‘off’ from one side of the room to the other. The 3-4-5 technique can be used to get an accurate 90˚ angle. It will also ensure that all parts of the 3m 90° 4m construction fits together as planned, as shown in Figure 1.9 below. 5m This technique simply requires a builder to create a triangle using any suitable material such as wooden beams, that has sides that are 3 m, 4 m Figure 1.9 Using the 3-4-5 technique and 5 m long. Such a triangle will have a 90˚ angle opposite the 5 m side. Chapter 1 Mechanics 13 Science_GR11_LB.indb 13 2018/02/02 16:59

Note: You can use any other measurement, but they need to be in the same proportion. The beauty and simplicity of this technique is that if the builder needs to increase the accuracy on larger walls and structures, he can use any multiple of the 3-4-5 set of numbers, and it will still produce accurate results. Activity 1.4 Use Pythagoras’ Theorem to validate whether these sets of numbers will give you a right-angled triangle. 3-4-5 6-8-10 9-12-15 12-16-20 15-20-25 The above practical application is only one example of how Pythagorus’ Theorem can be used in real life situations. There are many other real life situations where using Pythagoras’ Theorem can be quite useful. Here are a few examples: 1.3.2 Using Pythagoras’ Theorem to determine ‘immeasurable length’ ‘Immeasurable length’ can be defined as the length that you cannot measure directly using a measuring device. So, for example, how would you go about using Pythagoras’ Theorem if you need to design and construct a zip line over a water pond as shown in the diagram below. A zip line is an aerial runway between two points of different heights. First you will need measure the horizontal distance of the zip line and the maximum height at the start using a measuring tape. Then you can use Pythagoras’ Theorem to determine the length of the zip line you need. 20m 50m Figure 1.10 Using Pythagoras’ Theorem to determine ‘immeasurable length’ Note: You will need to assume that the ground is perfectly horizontal. 14   Technical Science | Grade 11 Science_GR11_LB.indb 14 2018/02/02 16:59

Activity 1.5 1. Determine the length of the zip line using the horizontal and vertical lengths given in Figure 1.10. 2. What practical consideration would you need to take regarding the length of the zip line needed? Hint: think about the effect of gravity on the zip line. 1.3.3 Further applications for using Pythagoras’ Theorem Consider a solid cuboid. It is easy to measure the length, width and height of the cuboid using a measuring tape. point 1 point 2 But how can the length of the diagonal running from one corner to the opposite corner of the cuboid be measured without breaking it apart? We can use Pythagoras’ Theorem to help us. Let’s look at an example to see how it works. Activity 1.6 Application of Pythagoras’ Theorem All three sides of the cuboid form a right triangle. Using horizontal diagonal BD and the vertical edge AB, the length of diagonal AD is then found by a second application of Pythagoras’ Theorem as: BD2 = BC2 + CD2 and AD2 = AB2 + BD2 A D B C Therefore: AD2 = AB2 + BC2 + CD2 Determine the length of the diagonal AD, if the dimensions of the cuboid are as follows: AB = 4 cm; BC = 5 cm; CD = 6 cm Chapter 1 Mechanics 15 Science_GR11_LB.indb 15 2018/02/02 16:59

Activity 1.7 1. The shadow cast by a 1 m ruler held perpendicular to the ground is 8 cm long. At the same time, the shadow cast by a pine tree is 24 m long. What is the height of the pine tree? 1m 8 cm 24 m 2. Thabiso’s kite gets caught at the top of a tree as shown in the diagram below. If the string makes a 44˚ angle with the ground, and the string holder pinned to the ground is 90 m from the base of the tree, how tall is the tree? string 44° 90 m 1.4 The resultant of collinear forces Collinear forces (or vectors, in general) are forces (or vectors) that lie in the same line of action. These vectors are parallel but not necessarily pointing in the same direction. By sticking to the sign conventions discussed earlier in the chapter, collinear vectors can be added. 16   Technical Science | Grade 11 Science_GR11_LB.indb 16 2018/02/02 16:59

Example 1.7 DETERMINE THE RESULTANT FORCE Find the resultant of four forces acting on an object: F⃗1 = 2 N in the positive y-direction F⃗2 = 1,5 N in the negative y-direction F⃗3 = 2,5 N in the negative x-direction F⃗4 = 6 N in the positive x-direction 1 ANALYSE Identify collinear forces F⃗x and F⃗y assign directions conventionally. In our case there are forces acting in the x-direction and in the y-direction. 2 SOLVE Add all collinear forces. Call these resultants forces F⃗x and F⃗y: Take the direction to the right as positive, and the direction up as positive. F⃗x = +2,5 N + (–6 N) = –3,5 N This means it’s 3,5N in the negative x-direction F⃗y = +2,5 N + (–1,5 N) = 0,5 N This means 0,5N in the positive y-direction 3 NOW DRAW FR Fy Fx F2R = F2x + F2y F2R = (–3,5)2 + (0,5)2 F2R = 12,5 ∴ FR = 3,45 N The sketch represents the F⃗x and F⃗y, and calculation the resultant of F⃗x and F⃗y. This is the same as the resultant of the four forces. Chapter 1 Mechanics 17 Science_GR11_LB.indb 17 2018/02/02 16:59

Activity 1.8 1. Define the term resultant force. 2. Define the term collinear vector. 3. Calculate the resultant of the following forces: a) A force of 23,7 N in the negative x-direction acts simultaneously to a force of 9 N in the positive y-direction. b) A force of 17 N in the positive x-direction acts simultaneously to a force of 23 N in the positive y-direction. c) A force of 23,7 N in the negative x-direction acts simultaneously to a force of 9 N in the positive y-direction. 4. Four forces act simultaneously at a point, find the resultant if the forces are: F⃗1 = 2,3 N in the positive x-direction F⃗2 = 4 N in the positive y-direction F⃗3 = 3,3 N in the negative y-direction F⃗4 = 2,1 N in the negative x-direction a) Find the resultant of forces in the x-direction, F⃗x. b) Find the resultant of the forces in the y-direction, F⃗y. c) Calculate the resultant of the four forces. 5. Four forces act simultaneously at a point, find their resultant if the forces are: F⃗1 = 2,3 N in the negative x-direction F⃗2 = 4 N in the positive y-direction F⃗3 = 3,3 N in the positive x-direction F⃗4 = 2,1 N in the positive x-direction 1.5 Resultant of forces in 2-D Finding the resultant of two-dimensional forces that are not necessarily perpendicular to one another can be done by using accurate scale drawings. There are two techniques that can be used to do this. These are: • head-to-tail method • parallelogram of forces 1.5.1 Head-to-tail method Remember that a vector quantity is a physical quantity having both magnitude and direction. A vector quantity is represented by a ray, which is an arrow tipped segment, as shown in Figure 1.10 below. Tail Head Figure 1.10 A diagrammatic representation of a vector The drawing above is the general representation of a vector. The length of the line represents the magnitude of the vector, and the arrow shows the direction of the vector. 18   Technical Science | Grade 11 Science_GR11_LB.indb 18 2018/02/02 16:59

To add vectors using the head-to-tail method, move the vectors such that the head of one vector is attached to the tail of the other. The resultant vector (F⃗) is the single vector leading from the tail of the first vector to the head of the last when the vectors are placed tail to head in any order, as shown in Figure 1.11 below. R= A+B B A Figure 1.11 Graphical representation of the resultant vector (F⃗) Example 1.6 HEAD-TO-TAIL METHOD Determine the resultant of the forces F⃗1 = 30N to the right and F⃗2 = 40N upwards using the head-to- tail method. 1 ANALYSE Given: F⃗1 and F⃗2 Unknown: Resultant force F⃗ Note: Newton’s measure can’t be drawn. As a result, we should find a suitable conversion scale that will help us sketch the vector and find their resultant. 2 SOLVE Check the accuracy of the drawing. We could use Pythagoras’ Theorem to find the magnitude of the resultant force. Scale: 1 cm : 10 N 4 3 2 FR F2 = 40 N 1 0 1 2 3 4 By measuring with a ruler, we will find that the magnitude of F⃗R is 5 cm, which is an equivalent of 50 N according to the scale described above. Chapter 1 Mechanics 19 Science_GR11_LB.indb 19 2018/02/02 16:59

Activity 1.9 1. Use the head-to-tail method to find the resultant of the following forces: a) F⃗1 = 12 N to the right and F⃗2 = 5 N downwards b) F⃗1 = 16 N to the right and F⃗2 = 12 N downwards c) F⃗ = 8 N to the left and F⃗ = 6 N upwards 1 2 d) F⃗1 = 15 N to the right and F⃗2 = 20 N upwards 2. Use Pythagoras’ Theorem to verify your answers. 3. Use the head-to-tail method to determine the magnitude and direction of the resultant of the forces 12 N at 0˚ and 20 N at 35˚. 4. Use the head-to-tail method to determine the magnitude and direction of the resultant of the forces 13 N at 45° and 30 N at 30°. 1.5.2 The Parallelogram Law of Forces A parallelogram is a quadrilateral with two pairs of opposite sides equal and parallel, and two pair of interior opposite angles equal. Figure 1.12 shows an example of a parallelogram. Figure 1.12 A paralleleogram For any two simultaneous forces acting at a point and separated by an angle, we can use a technique called a parallelogram of forces to find their resultant. We call this the Parallelogram Law of Forces. This law states that if two forces acting at a point can be represented by the adjacent sides of a parallelogram both in magnitude and direction, then the diagonal from the point gives the resultant of the two forces. This is depicted in Figure 1.13 below. C D F2 FR θ F1 θ A B Figure 1.13 The Parallelogram Law of Forces 20   Technical Science | Grade 11 Science_GR11_LB.indb 20 2018/02/02 16:59

Example 1.8 THE PARALLELOGRAM LAW OF FORCES Use the Parallelogram Law of Forces to find the resultant of two forces F⃗1 = 60 N and F⃗1 = 45 N acting simultaneously at a point, and separated by the angle of 60˚. 1 ANALYSE Choose a suitable scale that you can use to convert forces to cm or mm. In this case 1 cm : 5 N is a suitable scale. Thus, 12 cm : 60 N represents F⃗1, and 9 cm : 45 N represents F⃗2. 2 SOLVE Step 1: Draw F⃗1, as converted above, represented by the horizontal-arrowed line. Step 2: Label this line AB. A F1 B Step 3: From the tail of this force, measure an angle of 60˚, and draw the second arrowed-line representing F⃗2, as converted above. Label this force AC. 11 10 9 8 7 6 490 5 80 100 70 110 12 4 60 0 13 50 0 40 14 0 3 30 2 150 2 20 160 1 170 180 10 0 0 1 2 3 4 5 6 7 8 9 10 11 12 0 –4 –2 2 4 6 8 10 12 –1 Step 4: From the head of the force AB, measure an angle of 60˚ and construct a dotted line through that angle. Label that line BD, of the same magnitude as force AC from step 3 above. Notice that BD must be parallel and equal in magnitude to AC. C D F2 60° 60° A F1 B Chapter 1 Mechanics 21 Science_GR11_LB.indb 21 2018/02/02 16:59

Step 5: Join CD with a dotted line. Note that DC will be automatically equal in magnitude and parallel to AB. C D F2 60° 60° A F1 B Step 6: Join AD. Notice that, according to the parallelogram of forces theorem, this diagonal is the resultant of the F⃗1 and F⃗2. C D F2 FR A F1 B Step 7: Use a ruler to measure the length of the diagonal. Then use the defined conversion scale to convert it to Newton and find the magnitude of the resultant force. Measure the angle made by the resultant with the horizontal, to get its direction. 8 7 90 6 5 4 3 2 1 0 –8 –7 –6 –5 –4 –3 –2 –1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Step 8: In this case, the length of the vector representing the resultant force is about 18,3 cm. Using the defined scale 1 cm : 5 N, we get that the magnitude of the resultant force is approximately: F⃗R = 91,5 N, and as shown its direction is 25˚ from the horizontal. Activity 1.10 1. State the Parallelogram Law of Forces. 2. Use the accurate scale drawing to construct the parallelogram of forces and find the resultant of the forces in each case: a) F⃗1 = 12 N and F⃗2 = 1% N separated by the angle of 120˚. b) F⃗1 = 200 N and F⃗2 = 120 N separated by the 45° 30° angle of 80˚. 3. Two forces act simultaneously on an object as m1 = 100g m2 = 200g shown alongside: 22   Technical Science | Grade 11 Science_GR11_LB.indb 22 2018/02/02 16:59

Determine the magnitude and the direction of the resultant force. 4. Use the parallelogram of forces method to determine the magnitude and direction of the resultant of forces 18 N at 45˚ and 24 N at 60˚. 5. Use the parallelogram of forces method to determine the magnitude and direction of the resultant of the forces 18 N at 125˚ and 15 N at 225˚ 6. Use the parallelogram of forces method to determine the magnitude and direction of the resultant of the force 21 N at 150˚ and 14N at 215˚. 7. Two weights of 24 N and 30 N are used to balance an object of unknown weight w⃗ on pulleys as shown in the diagram below. Determine the unknown weight w⃗ if the strings make angles of 30˚ and 40˚ with the horizontal. 30° 40° 24 N 30 N W 8. In the diagram, a mass of 5 000 kg (w = 50 kN) is hanging on the hook of the lifting cable and the arms of the cranes are stationery. The cables make the angles of 45˚ and 30˚ with the horizontal. Determine the tensions on the cables. 45° 30° 5000 kg 9. A two-legged sling and hoist chain are used to lift a car engine at constant speed as shown below. Determine the tension on each leg of the sling if the weight of the car engine is 1 500 N. 30° 20° Chapter 1 Mechanics 23 Science_GR11_LB.indb 23 2018/02/02 16:59

T⃗1 T⃗2 10 A load of 120 N is lifted by two strings connected to the same point on the load, making angles of 25˚ and 30˚ on opposite sides of the vertical. Determine 25° 30° the tensions in the string. t 11. If the threshold tension of each of the strings in the hoist below is 4 000 N, what is the maximum load that the crane can carry before the strings snap, if the angle that each string makes with the vertical is 20˚? F⃗G = 120 N 20°20° ? PRACTICAL INVESTIGATION (Informal) Determining the Resultant of two forces acting at a point Apparatus • force board • mass pieces • pulleys m2 • strings m1 • protractor m3 Procedure • Create two stands A and B on each side, and in between the stands, put a board. • On top of each stand mount a plastic pulley of at least 25 mm in diameter. • Wound a piece of string along the pulley and have masses m1 and m2 on each end of the string, that is, at the end of A and B. Put another mass of m3 at the centre of the string on the side of the board and represent F⃗1 acting at a downwards direction. • Label a point where m3 is hung as point O. Name the tension force that acts towards the left- hand side of point O towards m1, F⃗1 and name the second force that acts towards the right-hand side of point O, F⃗1. • Measure an angle that is formed between point O, F⃗1 and F⃗2. F⃗3 that acts towards m3 is now the equilibrant force because both F⃗2 and F⃗1 are in equilibrium and opposite F⃗3 above point O. When you have found F⃗3, you would have found the resultant force. • Use a parallelogram of forces diagram to make a graphical representation of your findings. 24   Technical Science | Grade 11 Science_GR11_LB.indb 24 2018/02/02 16:59

Results • Measure the magnitude of the angle between the forces F⃗2 and F⃗1. • Calculate the magnitude of the forces F⃗1 and F⃗2. This is simply the weight of the hanging masses, m1 and m2, respectively. • Use the parallelogram of forces method to find the magnitude of the resultant of these two forces • How does your answer compare to the magnitude of F⃗3? 1.6 Resolving vectors In the previous section we learned that the combined effect of two or more forces can be represented by a single force, referred to as the resultant force. In the same way, a single force that acts at an angle can be considered to have two parts – one part directed along the horizontal axis and the other along the vertical axis. The two parts (called components) of the single force are therefore perpendicular to one another – the horizontal and the vertical components of the force. Therefore, a single force acting at an angle can be resolved into a vertical and horizontal component. In Figure 1.14, the vertical component represents the influence of the force in the vertical direction, whist the horizontal component represents the influence of the force in the horizontal direction. Fverticle F = Fhorizontal Figure 1.14 resolution of vectors Any time a force is directed at an angle to the Fy horizontal, the trigonometric functions can be used to determine the components of that force. F This applies not only to forces but to all vectors. Figure 1.15 shows a force F⃗ that acts at angle θ θ from the horizontal axis. It is possible to Fx decompose force F⃗ into its components as shown in Figure 1.15 alongside. The components of a force F acting at an angle θ. From the definitions of the trigonometric ratios, Figure 1.15 Vector decompostion it is known that: opp F ⃗ sinθ = hyp = ⃗y ⇒ F⃗y = F⃗ sinθ F adj F⃗x sinθ = hyp = ⃗ ⇒ F⃗x = F⃗ sinθ F The formulae above are used to calculate the vertical (Fy) and horizontal (Fx) components of force F making an angle θ with the horizontal axis. Chapter 1 Mechanics 25 Science_GR11_LB.indb 25 2018/02/02 16:59

Example 1.9 RESOLVING A VECTOR The diagram below shows a force F⃗ = 400 N acting an angle 60˚ with the horizontal. Determine the x and y components of F⃗. 1 ANALYSE Draw a right-angled triangle and label all the sides as shown below. 60° 60° Fy F = 400 N 2 SOLVE F⃗ sin60° = 400 y F⃗ y = 400sin60° ∴ F⃗y = 346 to the right. In the positive x-direction. F⃗ cos60° = 400 x F⃗x = 400cos60° ∴ F⃗x = 200 to the right. In the negative y-direction. Example 1.10 RESOLVING FORCES USING AN ACCURATE SCALE DRAWING Determine graphically the horizontal and vertical components of a force of 800 N which acts at 20˚ below the horizontal to the left. Fy F = 400 N Let 1 cm : 100 N such that 8 cm : 800 N 20° Fx GIVEN F⃗ = 800 N STEPS 1. Draw line F⃗cosθ of any length on the horizontal and measure 20˚ anticlockwise using a protractor. Then draw line F⃗ below the horizontal. 2. From the tail of F⃗, draw a line going vertically (not measured) up and another line going horizontally to the right. 26   Technical Science | Grade 11 Science_GR11_LB.indb 26 2018/02/02 16:59

3. From the head of F⃗, draw a line going horizontally, to intersect with the vertical line drawn in step 2. Put a point where the two lines intersect. 4. From the tail of F⃗, draw a line going vertically to intersect with the horizontal line drawn in step 2. Put a point where the two lines intersect. 5. Put arrows to indicate the directions of the horizontal and vertical components of the force F⃗. 6. Measure the dimensions of the components of F⃗, and use the defined scale to get the magnitude of the forces. 7. Measuring the dimension of F⃗x and F⃗y we get: 7,5 cm and 2,75 cm. Using the scale, we get: F⃗x = 750 N in the positive x-diretion and F⃗y = 275 N in the positive y-directon. Note: Remember that the quadrant in which a force acts will determine its direction, that is, positive or negative. Activity 1.11 resolving vectors 1. Determine the horizontal and vertical components of the forces given below: a) a force of 205 N at 45˚ below the horizontal to the left b) a force of 8 N at 45˚ below the horizontal to the right c) a force of 400 N at 15˚ below the horizontal to the left d) a force of 1 500 N at 30˚ above the horizontal to the left e) a force of 90 N at 90˚ above the horizontal 2. Use the accurate scale drawing the find the components of the horizontal and vertical component of the forces described above. 3. A force of 15 N at 30˚ above the horizontal to the left and another force of 20 N at 60˚ below the horizontal to the left with the third force of 10 N at 10˚ below the horizontal to the right. a) Find the resultant horizontal force of the three forces. b) Find the resultant vertical force of the three forces. c) Calculate their resultant in magnitude and direction. HINT: Use the following formula: F⃗ ⃗ ∙F ∙ tanθ = F⃗x of θ = tan–1 ⃗x and FR2 = Fx2 + Fy2 Fy y 4. Consider the tow truck at the below. If the tensional force in the cable is 1 000 N and if the cable makes a 60˚ angle with the horizontal, then what is the vertical component of force that lifts the car off the ground? F = 400 N 60° Chapter 1 Mechanics 27 Science_GR11_LB.indb 27 2018/02/02 16:59

5. Critical thinking: One of the most important applications of vector components is in the operation of sailboats. Sailboats are propelled forward by the force that the wind exerts on the sail. It is easy to understand that if the wind blows in the direction to which the boat is moving, then the boat will move forward. But how come sailboats can move ‘into the wind’ N– towards the direction from which the wind is coming? Many people believe that this is not possible, but it does happen in real life. If it were not possible, then sailboats would not be able to reach their intended destinations because they would be tossed around by the winds in the oceans. Do your own investigation on how sailboats work and how the principle of vector components can explain the fact that sailboats can sail ‘against’ the wind. 1.7 Friction 1.7.1 The importance of friction Usually when we hear about friction, we associate it with something that needs to be reduced or eliminated. This may lead us to think that friction is always a bad thing! However, friction has many important uses in Science. Friction is a resistance force that slows down or prevents motion. It is necessary in many applications where we might want to hold items together or prevent slipping or sliding. But how does that apply to real life? Let’s look at a few examples of how having friction can be advantageous. Imagine there was no friction. You would not be able to walk properly because you need friction between your shoes and the ground. As you step forward, you push your foot backward. Friction holds your shoe to the ground, thus allowing your body to move forward. Think about when you have tried to walk on a slippery floor. Slippery surfaces usually have very little friction acting on them. How about the importance of friction when driving in a car? The car would not start moving if it wasn’t for the friction of the tires against the road. With no friction, the tires would just spin and the car would not move. Similarly, a car would not stop without the friction of the brakes and the tires. So, we need friction! Ancient people used friction to light fire. Making fire was not as easy as we know it today and it cannot be compared to lighting a match – in which case the fire lighting tool has already been created. One method of making fire involved the use of dry wooden sticks that were ground by rapidly rubbing Figure 1.17 San people in Botswana starting fire by hand them against each other or a hard surface. 28   Technical Science | Grade 11 Science_GR11_LB.indb 28 2018/02/02 16:59

This required skill, fitness and conducive environmental conditions. In this case, effort is focused on grinding dust off burnable wooden material and producing enough heat that will make the dust smoulder. The critical role played by friction in this process can be easily seen. With friction fire, effort is focused into grinding dust off of soft solid burnable material such that the dust is smouldering. Another example of the application of friction is the hand drill shown in Figure 1.16. Without friction, the hand drill would not work. It is actually suggested that the hand drill is the modern version of making fire by friction, in which case the wooden shaft that used to be spun by hand is replaced by a drill; but has the same effect of producing heat and applying pressure and grinding dust. chuck turning handle drive wheel jaw pinion drill bit main handle Figure 1.16 Manual hand drill We can think of many more examples, but just considering these examples we can deduce that friction is one of the most important phenomena in the physical world. 1.7.2 So what is friction Frictional force is the force that an object encounters when it moves or attempts to move along a surface that opposes the motion or tendency of motion. It is always directed parallel to the surface in question. When surfaces that are in contact with one another move relative to each other, the friction between the two surfaces converts kinetic energy into thermal energy (that is, it converts work to heat). This property is illustrated by the use of friction when rubbing pieces of wood together to start a fire. In this case, kinetic energy is converted to thermal energy whenever motion with friction occurs. 1.7.3 Types of frictional forces There are many different types of frictional forces in nature, but in mechanics we focus mainly on the two types. These are: • kinetic frictional force • static frictional force. Friction, like normal force, is exerted by the surface. But while normal forces is always perpendicular to the surface on which it acts, frictional force is tangent to the surface. Static frictional force, which is denoted by f⃗s, is the force that keeps an object ‘stuck’ on a surface and prevents it from moving. The direction of the static frictional force is opposite to the direction in which the object would move if there were no friction. It is given by the formula: fs = μsN, where N is the normal force Chapter 1 Mechanics 29 Science_GR11_LB.indb 29 2018/02/02 16:59

The static frictional force depends on the magnitude of the force acting on the object. Stationary objects have frictional forces between zero and a maximum value, fsmax. We define a maximum value for the static friction as fsmax = μsN. If you apply a force that is greater than fsmax, the object moves. The maximum static frictional force is also known as limiting static frictional force fsmax = msN Where fsmax = limiting static frictional force μs = the dimensionless constant called the coefficient of static friction Fn = the normal force The coefficient of friction is the ratio of frictional force to the normal force. ∴μ= f FN The coefficient of friction is dimensionless (it has no units). This can be felt when we try to accelerate a stationary object (by exerting a force on it). The difficulty we experience is mainly due to static frictional force. Once the object starts moving, accelerating it becomes easier than when it was stationary. This means that as the applied force is increased, the static frictional force is increasing until it reaches its maximum at the moment when the object is at the verge of starting to move. Once the object is in motion, it takes lesser effort to accelerate it. The object then experiences the ‘so called’ kinetic frictional force. Kinetic (dynamic) frictional force acts between the surfaces when the object is moving. Because it takes lesser effort to accelerate a moving object than to cause an fs = μ sFN object to move, the moving object experiences fs,max lesser opposition to its motion than a stationary fk = μ kFN object. Therefore, maximum static frictional force is always greater than the kinetic frictional force (as shown in Figure 1.17). O Static region Kinetic region Let’s look at each type of force using examples to Figure 1.17 Static vs kinetic frictional force illustrate them. Kinetic frictional force Kinetic frictional force is the force: • that arises when two surfaces are in contact • hinders (opposes) the mutual displacements of the two surfaces • acts along the surfaces in reciprocal contact • always acts in opposite direction to the relative motion • is tangent to the surfaces in contact. • is calculated by the formula: fk = μkFN 30   Technical Science | Grade 11 Science_GR11_LB.indb 30 2018/02/02 16:59

Example 1.11 KINETIC FRICTIONAL FORCE Temba pushes a fridge weighing 78 kg across the push horizontal floor at a constant velocity. Fnormal If the co-efficient of kinetic friction between the Fapplied floor and the fridge is 0,7; calculate the force parallel to the floor to keep the box in motion. Ffriction Note: Always use a sketch or a drawing to describe Fgravity the process friction SOLUTION Given: the mass of the fridge in kilograms = 78 kg; μk = 0,7 TO CALCULATE Note: Start by calculating the weight from the mass given F⃗N = –w⃗ = –mg F⃗N = –(78)(9,8) F⃗N = –764,4 N f⃗k = μkFN f⃗k = (0,7)(764,4) f⃗k = 535,05 N Because the fridge is moving at constant velocity, it is not accelerating (the object is in equilibrium). Therefore, the resultant of all the collinear forces acting on the fridge is zero. Therefore, the applied force required to keep the fridge moving at a constant velocity has the same magnitude as the frictional force (equilibrant force in this case) experienced by the fridge. F⃗applied = 535,08 N in the direction of motion/opposite to the frictional force Activity 1.12 1. Name four instances where friction is a disadvantage. 2. Define the co-efficient of friction. 3. The co-efficient of friction between Nomthandazo’s shoes and the floor is 0,72. If her mass is 85 kg, calculate the limiting frictional force between her shoes and the floor. 4. A wooden crate with a mass of 80 kg is dragged across a horizontal concrete floor at a constant velocity. If the co-efficient of friction between the floor and the concrete is 0,8, calculate the dynamic frictional force between the crate and the floor. Chapter 1 Mechanics 31 Science_GR11_LB.indb 31 2018/02/02 16:59

EXPERIMENT 2: FRICTION In the following experiment we will do the following: 1. Determine the relationship between the limiting frictional force and the normal force 2. Determine the coefficient of friction between the surfaces fs and FN. EXPERIMENT 2 (PART 1): RELATIONSHIP BETWEEN LIMITING FRICTION AND THE NORMAL FORMAL Apparatus • Flat table with a pulley attached to one end • String • Mass pieces • Wooden block Procedure 1. Set the experiment as indicated in the diagram below: m2 m2 2. Increase the hanging mass by adding more mass pieces until the moment the wooden block starts sliding at a constant speed. 3. Record the mass of the hanging mass m2 and the mass of the wooden m1 block 4. Increase the mass of the trolley by adding fewer mass pieces on it, until you have about five different readings. Repeat the last two steps above. EXPERIMENT 2 (PART 2) CALCULATING μ. m θ2 1. Place the block on the wooded surface. 2. Slowly lift the end of the end of the plank until the trolley start sliding from rest. 3. Fix the plank in that place and measure the inclination it makes. 4. Repeat for at least 5 values. 32   Technical Science | Grade 11 Science_GR11_LB.indb 32 2018/02/02 16:59

Results: Experiment 1 Trial # m1 Fg1 = m1g m2 Fg2 = m2g μ 1 2 3 4 5 1. The magnitude of the limiting frictional force between the block and the plank is given by: f = m2g 2. The normal force on the block is given by: FN = m1g 3. To calculate the coefficient of friction, use the formula: μ = Ff N 4. Use the results to plot the graph of f (y-axis) versus FN (x-axis) Trial # Limiting angle ms = tanθ 1 2 3 4 5 5. Calculate the average value of μs = ________________ Conclusion 1. What is the shape of the graph for Experiment 1? From the graph of Experiment 1, write a conclusion about the relationship between the limiting frictional force and the normal. 2. Calculate the average value of m for Experiment 1 from the table of results. 3. Now calculate the slope of the graph using any two points that lie on the line. 4. How do your answers to number 2 and 3 compare? Chapter 1 Mechanics 33 Science_GR11_LB.indb 33 2018/02/02 16:59

Review exercise Multiple choice questions 1. The resultant of two collinear vectors is minimum when the angle between them is: A 180˚ B 90˚ C 0˚ D Greater than 180˚ 2. Identify the vector quantities from the descriptions below: A A speed limit of 120 km·h–1 B A distance of 100 km C A car driving at 120 km·h–1 due East D An angle of 45˚ 3. The following technique cannot be used to find the resultant of two perpendicular vectors: A Head-to-tail method B Parallelogram of forces C Theorem of Pythagoras D Collinear vectors addition 4. A car accelerates from rest and reaches a certain velocity which it maintained for some time. Choose the correct statement below: A When the car was at rest, the static friction was at its maximum, that is why the car remained stationary. B When the car was moving, its static frictional force dropped to zero, and the car had kinetic frictional force. C The static frictional force increased while the car tried to start moving until it reached the maximum when it was at the verge of moving. D The car would not move unless its kinetic frictional force was equal to its static frictional force 6. The force experienced by an object parallel to the surface that opposes the motion of an object is the: A Applied force B Gravitational force C Normal force D Frictional force 34   Technical Science | Grade 11 Science_GR11_LB.indb 34 2018/02/02 16:59

Long questions 7. A Grade 12 class at Cedar High School was interested in finding the relationship between the acceleration of the trolley and the pulling force. To achieve this, they set up the experiment as shown below. They maintained the total mass, by moving some the mass pieces they placed on the trolley and adding them to the hanging mass that pulls the trolley. Each time they added the mass, they would calculate the pulling force and the acceleration. Below is the summary of the results they obtained: Trial # Handing mass (kg) Trolley mass (kg) Acceleration (m·s–2) Force (N) 1 0,005 0,525 0,094 0,049 2 0,010 0,520 0,187 0,098 3 0,015 0,515 0,27 0,147 4 0,020 0,510 0,36 0,196 5 0,030 0,500 0,54 0,294 a) From the results above, identify the variable that is: (i) a constant (ii) independent (iii) dependent b) Use the results above to plot the graph of acceleration versus pulling force. c) From the graph, and the table above, what conclusion can the learners draw about the relationship between the acceleration and the pulling force? 7. The following forces act simultaneously on an object: F⃗1 = 25 N in the positive y-direction F⃗2 = 10 N in the negative x-direction F⃗3 = 15 N in the negative x-direction F⃗4 = 10 N in the negative y-direction a) Calculate the resultant of the forces in the x-direction, F⃗x. b) Calculate the resultant of the forces in the y-direction, F⃗y. c) By using accurate scale drawings, find the magnitude of the resultant of all the forces. d) Using calculations, show the accuracy of the answer you obtained in c). above. Chapter 1 Mechanics 35 Science_GR11_LB.indb 35 2018/02/02 16:59

8. The angle between two forces F⃗1 and F⃗2, of magnitude 18 N and 15 N respectively is 60˚. Use accurate parallelogram of forces diagrams to find the resultant of the two forces. 9. A force of 100 N pulls a box at angle of 40˚ to the horizontal, as shown below: F = 100 N θ = 40° a) Calculate the horizontal component F⃗x of the force F⃗. b) Calculate the vertical component F⃗y of the force F⃗. c) If the box experiences a constant frictional force while being pulled to the right such that it moves at a constant velocity, calculate the: (i) normal force on the object (ii) coefficient of kinetic friction between the box and the surface. 10. Use accurate scale drawings to find the components of the force shown below. 200° F = 150 N Chapter summary Key terms Mechanics is a ‘branch of physics that deals with the study of the forces acting on a body and keeps it in equilibrium (objects or bodies at rest) or in motion’. Generally, mechanics can be divided into: • Kinematics – the branch of classical mechanics that describes the motion of points (alternatively ‘particles’), bodies (objects), and systems of bodies without consideration of the masses of those objects nor the forces that may have caused the motion. • Dynamics – the branch of physics (mechanics) that deals with the effect of forces on the motion of bodies. It can also be understood as the study of the forces affecting moving objects. 36   Technical Science | Grade 11 Science_GR11_LB.indb 36 2018/02/02 16:59

Sign convention Vectors have two important features: that is the magnitude and the direction. The direction of a vector can be defined in terms of Cartesian coordinates (especially for collinear vectors). Conventionally, one direction of collinear vectors is taken as positive, so that the opposite will be negative. Alternatively, the direction of a vector in 2-D can be described in terms of the bearing, and/or compass directions. A bearing is an angle that a vector makes as measured from the North pole of the compass in a clockwise direction. Graphs Certain mathematical skills are of importance in finding relationship between parameters under investigation in technical science. Two variables are said to be directly proportional to each provided that: • when the independent variable (x) is increased by the factor of a, the dependent variable (y) also increases by the same factor • the ratio is y : x is a constant (k), such that y is related to x according to the equation y = kx. The positive constant k is called the proportionality constant and is equal to slope of the graph of y = kx. y y • Since y = kx, we can generalise that: x1 = x2 1 2 the graph of y versus x is a straight line, that passes through the origin Two variables are said to be inversely (or indirectly) proportional to each other provided that: • when x increases by the factor of a, y decreases by the same factor • the product of x and y is a constant (k), such that y = kx. The positive constant k is called the proportionality constant. • the graph of y versus x is hyperbola, which does not intersect with the axes, but gets closer when x is decreased to approximately zero, or increases to a very big number. Collinear and co-planar vectors Collinear vectors are vectors that are along the same line of action. They are parallel to one other. They can be added mathematically using the conventional sign rule or graphically, by drawing the vectors along the same line of action. Co-planar vectors are vectors that lie along the same plane (for example, the x-y Cartesian plane). They are 2-D vectors. When two co-planar vectors are perpendicular to each other, either the theorem of Pythagoras or an accurate scale drawing using the head-to-tail method, may be used to calculate their resultant. Generally, the resultant of two simultaneous vectors that are separated by an angle θ can be found by using the method of the parallelogram of vectors. The Parallelogram Law of Forces (or vectors) states that if two forces acting at a point can be represented by the adjacent sides of a parallelogram both in magnitude and direction, then the diagonal from the point gives the resultant of the two forces. A vector that is acting an angle θ to the horizontal can be resolved into its horizontal and vertical components according to the formulae: Chapter 1 Mechanics 37 Science_GR11_LB.indb 37 2018/02/02 16:59

Horizontal component: Fx = Fcosθ Vertical component: Fy = Fsinθ Alternatively, the two components can be found by using accurate scale drawings Friction Frictional force is the force that an object encounters when it moves or attempts to move along a surface that opposes the motion or tendency of motion. It is always directed parallel to the surface in question. The static frictional force acts between the two surfaces when the object tends to move. When the object is accelerated by applying a force on it, the magnitude of static friction increases from zero until it reaches its maximum value (just when the object is at the verge of moving). The maximum value of static friction (called limiting static friction) is given by the formula: fsmax = μsFn When the object is moving it opposes friction in the form of kinetic frictional force. Kinetic frictional force is lesser than the static frictional. That is why it is more difficult to set a stationary object in motion, than to speed up an object already in motion. The magnitude of the kinetic frictional force is given by: fk = μsFn Frictional force is directly proportional to the applied force The coefficient of friction is the ratio of frictional force to the normal force: μs = f FN 38   Technical Science | Grade 11 Science_GR11_LB.indb 38 2018/02/02 16:59

2 Magnetism Objectives By end of this chapter learners will gain a broad understanding and acquire skills relating to the following concepts: • magnets • magnetic fields • poles of permanent magnets • direction of magnetic fields • the magnetic field of a bar magnet • the force of a magnet • the properties of magnetic field lines • Earth’s magnetic field Science_GR11_LB.indb 39 2018/02/02 16:59

Why magnetism matters Physicists’ have made many important discoveries about how we are able to use magnets in different ways. Their knowledge has also enabled the development of many technologies that we use in our everyday lives. Refrigerators, cars, elevators, and smart phones are only a few examples of gadgets that make use of magnets and magnetisms. All electric motors, generators – whether producing hydroelectric power or running bicycle lights – use magnetic fields. Even recycling facilities use magnets to separate iron from other refuse. Thus magnetism plays many important roles in our lives. S N S N S N N S 2.1 Introduction Magnets are made from minerals known as lodestones. Lodestones were discovered close to a town in the Greek Province of Magnesia, and the Magnates (people who lived in the town) found that the lodestones could attract iron objects. The lodestones became commonly known as magnets. A bar magnet has two poles: north and south. These were just labelled by convention because when the Magnesia first discovered the loadstones, they took them, magnetised a needle and placed it in a piece of cork and bucket of water. The needle pointed to the Earth’s magnetic north. Like poles of the magnet repel each other and unlike poles attract one another. A magnet generates a magnetic field around it. The magnetic field is generated by the electrons that spins around the magnet. Only certain materials, such as iron, cobalt, nickel, and gadolinium, exhibit strong magnetic effects. Such materials are called ferromagnetic, which is a combination of the Latin word for iron, ferrum, and the word magnetic. 40   Technical Science | Grade 11 Science_GR11_LB.indb 40 2018/02/02 16:59

2.2 Important definitions It is important for you to know and understand the terms we will use when we are learning about magnetism. Here are some important terms that we need to know: 2.2.1 Geographic North and South The Geographic North Pole is the northern-most point on the planet where the Earth’s axis intersects with its surface (that is, the top part of the globe). The geographic South Pole is the southern-most point of the planet where the Earth’s axis intersects with its surface (that is, the bottom part of the globe). Geographic North (the top part of the globe) Magnetic North Geographic South (the bottom part of the globe) Simply put, the Geographic North Pole and Geographic South Pole are directly opposite to each other. 2.2.2 Magnetic north and south The magnetic north is the place on Earth to which the north pole of magnets point. The magnetic south is the place on Earth to which the south pole of magnets point. Chapter 2 Magnetism 41 Science_GR11_LB.indb 41 2018/02/02 16:59

2.3 Magnets 2.3.1 What magnets are We can describe a magnet as a material that attracts magnetic objects towards it. As we have learned, all magnets have a pair of opposite poles – a north and south pole. Even if the magnet is cut into tiny pieces, each piece will still have both a north and south pole. All types of magnets attract iron-containing objects such as paper clips, and nails as shown in Figure 2.1. Figure 2.1 Paper clips attracted to the magnet A magnet can hold these objects, or even make them move. The pulling force of the magnet is called its magnetic force. The physical phenomenon, which results in attractive and repulsive forces between objects, is called magnetism. 2.3.2 A dipole magnet Look at the magnet shown in Figure 2.2. This type of magnet is known as a bar magnet. N S Figure 2.2 A bar magnet A bar magnet is a dipole. A dipole magnet is a magnet in which opposite poles (that is, the north and south poles) are on opposite sides of the magnet. The simplest example of a dipole magnet is a bar magnet. Materials containing iron have a strong attraction to the ends of such a magnet. If a bar magnet is suspended from its midpoint so that it can swing freely in a horizontal plane, it will rotate until its north pole points North and its south pole points South. Let’s look at this phenomenon. 42   Technical Science | Grade 11 Science_GR11_LB.indb 42 2018/02/02 16:59

Experiment 1: Determining the north pole of the Earth using a bar magnet Aim To determine the north pole of the Earth using a bar magnet. Material • wooden stand or something similar • bar magnet • compass Procedure 1. Tie a string in the middle of the magnet so that it can be suspended from a wooden stand as shown below. string S N bar magnet wooden stand 2. Make sure that the magnet can rotate freely. 3. Allow the magnet to come to rest. 4. Mark two points on the ground to show the position of the ends of the magnet when it comes to rest. 5. Draw a line joining the two points. This line shows the direction in which the magnet was pointing in its position of rest. 6. Now, rotate the magnet by gently pushing one end in any direction and let it come to rest. Again, mark the position of the two ends in its position of rest. Does the magnet now point in a different direction? 7. Rotate the magnet in other directions and note the final direction in which it comes to rest. Results Write down your results. Conclusion and discussion • Discuss your findings with a partner and write down your conclusion. • Did you find that the magnet always comes to rest in the same direction? • Repeat this activity with an iron bar and a plastic or a wooden scale instead of a magnet. Do not use light objects for this activity and avoid doing it where there are currents of air. • Do the other materials also always come to rest in the same direction? Chapter 2 Magnetism 43 Science_GR11_LB.indb 43 2018/02/02 16:59

2.3 Technological applications of magnetism Magnets are also used in: • meters • motors • generators • loud speakers • sound-and video-recording equipment • on computer disks. Apart from the uses of magnets discussed above and on page 40, magnets can also be used in large electromagnets that pick up heavy loads, as shown in Figure 2.4. Electromagnets are currently used to contain extremely high-temperature plasmas that are used for research. They are also used to levitate modern trains and in controlled nuclear fusion experiments. electromagnet ferromagnetic materials Figure 2.4 An electromagnet 44   Technical Science | Grade 11 Science_GR11_LB.indb 44 2018/02/02 16:59

PRACTICAL INVESTIGATION Making a simple electromagnet Scientists have found many links between electricity and magnetism, with one benefit of being an electromagnet that it can be turned on and off when needed. Let’s make our own electromagnet. You will need: • a long insulated copper wire • another shorter wire • an iron nail (with a ‘relatively soft’ core) • a cell • a switch • a box of pins or paper clips 1. Take a nail and hold it in one hand. Start winding the wire tightly around the nail. 2. Make at least 20 turns in the wire, but make sure that you leave enough wire at the ends to attach to the switch and the cell. 3. Strip the ends of the wire. Join one stripped end of the wire around the nail to the switch. Join the other end of the same wire to the negative terminal of a cell. Attach the shorter wire to positive terminal of the cell, and its end to the switch. 4. Turn on the electric current. Place the end of the nail into the box of pins or paper clips. Observe what happens. 5. Add another cell to your electromagnet. Does this make a difference? Does wrapping more wire around the nail to make more coils make a difference? 2.3.1 Like poles repel each other, and unlike poles attract each other The north pole of a magnet is attracted to the south pole of another magnet, while two north poles (or two south-poles) brought together will repel each other. The magnetic force between two magnets can be likened to the electric force between charged objects. But electric charges differ from magnetic poles because they can be isolated, whereas magnetic poles cannot. N S N S unlike poles attract eachother N S S N like poles repel eachother Figure 2.5 Like poles of the magnet repel each other and unlike poles attract one another Chapter 2 Magnetism 45 Science_GR11_LB.indb 45 2018/02/02 16:59

2.4 Magnetic domains The magnetic properties of many materials can be explained using a model. In terms of this model, an electron is said to spin much like a top does. The spinning electron represents a charge that is in motion. In substances such as wood and plastic that contain many electrons, the electrons pair up with their spin opposite, thus causing their magnetic fields to cancel each other. However, in substances such as iron, cobalt and nickel, the magnetic field produced by the electron spins do not cancel completely. Such substances are said to be ferromagnetic. In ferromagnetic substances, a strong coupling occurs between neighbouring atoms. These couplings form large groups of atoms whose net spins are aligned. The ferromagnetic substance has very small (microscopic) regions in which the magnetic moments of the atoms move in the same direction. These microscopic regions are called magnetic domains. Figure 2.6 shows that when a substance is unmagnetised, their domains are randomly orientated. Once an external magnetic field is applied, the orientation of the magnetic field of each domain may change slightly. The domains will become more closely aligned with the external magnetic field. Domains that are already aligned with an external field may also grow at the expense of other domains. This alignment enhances the applied magnetic field. Domains of unmagnetised and magnetised materials. (a) When a substance is magnetised (b) When a substance is unmagnetised its domains are more closely aligned. its domains are randomly ordered. Figure 2.6 Domains of magnetised and unmagnetised materials 46   Technical Science | Grade 11 Science_GR11_LB.indb 46 2018/02/02 16:59

2.5 Inducing unmaagnetised iron (Fe) Iron responds more strongly to magnets than all other ferromagnetics. As a result, iron can induce and magnetise all other material and make them into permanent magnets. Figure 2.7 shows how an unmagnetised piece of iron is placed between two magnets, heated, and then cooled, or simply tapped when cold. The iron then becomes a permanent magnet. Iron N S (Fe) N S heated and tapped Iron N S (Fe) N S N S N S N S Figure 2.7 Iron becomes a permanent magnet Activity 2.1 1. How many poles does a bar magnet have? 2. Describe what happens when like poles of a bar magnet get close to one other and what happens when unlike poles get close to one other. 3. Where is the force of a magnet the strongest? 4. What is the difference between permanent magnets and temporary magnets? 5. How can you turn an iron nail into an electromagnet? 6. Explain why an electromagnet is a temporary magnet. 7. How are electromagnets and bar magnets alike? How are they different? Chapter 2 Magnetism 47 Science_GR11_LB.indb 47 2018/02/02 16:59

2.6 Magnetic field Magnetic field is the region in space where another magnet or ferromagnetic material will experience a force. Magnetic field is generated by the electrons that spins around it. Just like gravitational force, magnetic force acts at a distance. We therefore define a magnetic field to represent magnetic force. The pictorial representation of magnetic field lines helps us to visualise the strength and direction of the magnetic field. Figure 2.8, shows the direction of magnetic field lines to be the direction in which the north end of a compass needle points. Magnetic field lines Compass needle N N S North Pole South Pole Compass Figure 2.8 Direction of the magnetic field lines around a bar magnet (adapted from Phet submissions) The red arrows indicate the direction of magnetic field lines. The magnetic field lines move from the north pole to the south pole outside the magnet. However, the field lines move from the south pole to the north pole inside the bar magnet. Magnetic field lines are closed – they do not have a starting point and an ending point. 2.4.1 Magnetic field between two bar magnets A simple experiment can be carried out to visualise the behaviour of the magnetic field lines around a bar magnet. When two bar magnets are placed on a flat surface, a white sheet of paper (page) is placed on top of it, iron powder is sprinkled liberally on and around the region where the bar magnet is placed, and the page is tapped gently on all sides, we will observe a magnetic field pattern. Note: You could use iron filings instead of iron powder, but iron powder gives a clearer result. This is shown in Figure 2.9(a) and (b). 48   Technical Science | Grade 11 Science_GR11_LB.indb 48 2018/02/02 16:59

N S Figure 2.9(a) Iron fillings showing magnetic field lines around two bar magnets (unlike poles) N N Figure 2.9(b) Iron filings displaying magnetic field lines around two bar magnets (like poles) Chapter 2 Magnetism 49 Science_GR11_LB.indb 49 2018/02/02 16:59

Experiment 2: Magentic field lines Aim To map the magnetic field Material • bar magnet • compass • blank A4 or A3 paper Procedure 1. Begin by putting the magnet in the middle of a large sheet of paper and outlining its form on the paper with a pencil. Label the N and S poles of the magnet on the paper. 2. Put a compass at one corner of the north pole of the magnet. On the paper, using dots, mark the positions of the N and S poles of the compass needle. 3. Move the needle a little further from the magnet until the new position of its S pole coincides with the dot marking the previous position of its N pole. Mark the new position of the N pole by a dot. 4. Again move the compass making the S pole coincide with the previous location of the N pole and mark the new position of the N pole by a dot. 5. Continue the process until the compass reaches the S pole of the magnet. This closed curve is a complete line of force. Place an arrow on it, showing its direction (away from the N and toward the S). 6. Beginning at another point on the N pole of the magnet, trace a second complete line of force. Continue with this process until you have a minimum of 8 lines of force. Results Redraw your magnetic field map. Discussion and conclusion Write down your conclusions The filed lines should be somewhat symmetric and balanced. If they are not, look for other magnets in the area. If time permits, map the field between two pole configurations assigned by your teacher. Questions 1. Are the lines more concentrated in some areas than in others? What does that tell you about the relative strength of the field in those places? 2. Lines of force never cross. Explain. You should have observed a pattern similar to that shown in Figure 2.10. 50   Technical Science | Grade 11 Science_GR11_LB.indb 50 2018/02/02 16:59

Iron filings N S Bar magnet Figure 2.10 Iron filings on thin sheets of paper over bar magnet. 3. Draw magnetic field lines for each of the following: a) around a bar magnet b) between two opposite bar magnets (i.e. north pole facing south pole) c) between two like bar magnets (i.e. north pole facing north pole of the other magnet). 2.7.2 Properties of magnetic field lines We use magnetic field lines to represent the field. Their properties can be summarised by these rules: 1. The direction of the magnetic field is tangent to the field line at any point in space. A small compass will point in the direction of the field line. 2. The strength of the field is proportional to the closeness of the lines. It is exactly proportional to the number of lines per unit area perpendicular to the lines (called the arial density). 3. Magnetic field lines can never cross, meaning that the field is unique at any point in space. 4. Magnetic field lines are continuous, forming closed loops without beginning or end. 5. Magnetic field lines move from the north pole to the south pole. Remember that the north and south poles cannot be separated. Chapter 2 Magnetism 51 Science_GR11_LB.indb 51 2018/02/02 16:59

2.8 Earth's Magnetic field Earth has a magnetic field that is similar to that of a bar magnet. The north and south pole of a small bar magnet are described as ‘north seeking’ and ‘south seeking’ poles. This description means that if a magnet is used as a compass, the south pole of the magnet will seek to, or point to, a location near the geographic North Pole of the Earth. North Pole Earth axis S N South Pole Figure 2.11 a) and b) The Earth has a magnetic north pole and south pole The Earth’s magnetic north and south pole is surrounded by a magnetic field, just as a bar magnet’s north and south pole are. But the Earth’s magnetic north pole corresponds to the south pole of a bar magnet, and the Earth’s magnetic south pole corresponds to the north pole of a bar magnet. This is true because unlike poles attract. If a compass needle is allowed to rotate both perpendicular and parallel to the surface of the Earth, the needle will be exactly parallel with respect to the Earth’s surface only near the equator. As the compass is moved northward, the needle will rotate so that it points more towards the surface of the Earth. Finally, at a point just north of Hudson Bay, in Canada, the north pole of the needle will point perpendicular to the Earth’ surface. This site is considered to be the location of magnetic South Pole of the Earth. It is approximately 1 500 km from the Earth’s geographic North Pole. Similarly, the magnetic north pole of the Earth is roughly the same distance as from geographic South Pole. The difference between true north, which is defined by the axis of rotation of the Earth, and north indicated by the compass, varies from point to point on Earth. This difference is referred to as magnetic declination. 52   Technical Science | Grade 11 Science_GR11_LB.indb 52 2018/02/02 16:59

2.8.1 Geographic North vs magnetic north pole The difference between the geographic North Pole and the magnetic north pole is that the former is a geographic pole with a stationary location at 90° North. This geographic North Pole, also known as true north, is the fixed northern-most point on the Earth from which all points lie south. The magnetic pole is not based on true north, but on the magnetosphere of the planet. The distance is about 1 500 km from true north, with its exact position constantly changing. This is shown in Figure 2.12. North Geographic Magnetic Pole North Pole 11.5˚ South Geographic South Pole Magnetic Pole Figure 2.12 The Earth is a large magnet. 2.8.2 The importance of the magnetic field to life on Earth The Earth’s magnetic field is very important for humans and other animals on Earth because it protects us from being bombarded (struck) by high energy charged particles, which are emitted by the Sun. The stream of charged particles (mainly positively charged protons and negatively charged electrons) coming from the Sun is called a solar wind. When these particles come close to the Earth, they are deflected by the Earth’s magnetic field and cannot shower down to the surface where they can harm living organisms. Astronauts in space are at risk of being irradiated by the solar winds because they are outside the zones where the charged particles are trapped. The region above the Earth’s atmosphere in which charged particles are deflected by the Earth’s magnetic field is called the magnetosphere. This is shown in Figure 2.13. Chapter 2 Magnetism 53 Science_GR11_LB.indb 53 2018/02/02 16:59

Interplanetary Magnetic Field Lines Magnetosheath Magnetopause Plasmasphere Bow Shock Magnetotail Plasma Sheet Radiation Belts Figure 2.13 A magnetosphere is that area of space, around a planet, that is controlled by the planet’s magnetic field. The shape of the Earth’s magnetosphere is the direct result of being blasted by solar wind. The solar wind compresses its sunward side to a distance of only 6-10 times the radius of the Earth. Relatively often, in addition to the usual solar winds, the Sun may eject a large bubble of material (protons and electrons) with its own magnetic field from its outer atmosphere. Sometimes these bubbles travel towards the Earth where their magnetic fields can join with Earth’s magnetic field. When this happens, a huge amount of energy is released into the Earth’s magnetosphere causing a geomagnetic storm. These storms cause rapid changes in the Earth’s magnetosphere which in turn may affect electric and magnetic systems on the Earth such as power grids, cell phone networks, and other electronic systems. 2.8.3 Aurora Borealis Aurorae (pronounced Or-roar-ee) Another effect caused by the Earth’s magnetic field is the spectacular Northern and Southern Lights, which are also called the Aurora Borealis (Figure 2.14) and the Aurora Australis respectively (Figure 2.15). 54   Technical Science | Grade 11 Science_GR11_LB.indb 54 2018/02/02 16:59

Figure 2.14 Aurora borealis photographed in Alaska Figure 2.15 Aurora australis (adapted photographed from space (adapted from NASA) from NASA) When charged particles from the solar wind reaches the Earth’s magnetosphere, they spiral along the magnetic field lines towards the North and South poles. If they collide with particles in the Earth’s atmosphere, they can cause red or green lights that stretch across a large part of the sky. This is called an aurora. As this only happens close to the North and South poles, we cannot see the aurorae from South Africa. However, people living in the high Northern latitudes in Canada, Sweden, and Finland, for example, often see the Northern lights. Chapter 2 Magnetism 55 Science_GR11_LB.indb 55 2018/02/02 17:00

Review exercise MULTIPLE CHOICE QUESTIONS 1. A bar magnet is divided in two pieces. Which of the following statements is true? A. The bar magnet is demagnetised. B. The magnetic field of each separated piece becomes stronger. C. The magnetic poles are separated. D. Two new bar magnets are created. E. The electric field is created 2. A bar magnet is divided in two pieces. Which of the following statements is true about the force between the broken pieces if they face each other with a small separation? A. There is an electric repulsive force between the broken pieces. B. There is an electric attractive force between the broken pieces. C. There is a magnetic repulsive force between the broken pieces. D. There is a magnetic attractive force between the broken pieces. E. There is no force between the broken pieces since they are demagnetised 3. Which of the following magnetic fields is correct for a single bar magnet? A B N N S S C D S S N N E N S 4. Magnetism is similar to electricity, but different from gravity, because magnetism and electricity: A. depend on charge B. can attract and repel C. depend on mass D. all three E. none of the above 56   Technical Science | Grade 11 Science_GR11_LB.indb 56 2018/02/02 17:00

5. A magnetic field is different from both an electric and a gravitational field because: A. it only acts on a few elements B. does not follow an inverse square law C. has closed field lines D. none of the above E. all of the above 6. What is true about the field lines of all electric, magnetic, and gravitational fields? They: A. form loops B. radiate from charges or masses C. show the path that a particle will follow D. have same direction E. never cross LONG QUESTIONS 7. What is a magnet? 8. Explain what is meant by the term magnetic field. 9. Explain, with the aid of a diagram, why permanent magnets have a magnetic field around them. Refer to domains in your explanation. 10. List the examples of ferromagnetic materials. 11. What happens when a permanent magnet is cut into pieces? 12. What happens when the like poles of a magnet are brought closer together? 13. What happens when the opposite poles of a magnet are brought closer together? 14. Discuss the properties of magnetic field lines. 15. Draw the magnetic field lines around the bar magnet shown below: N S 16. Explain how a compass indicates the direction of the magnetic field. 17. Compare the magnetic field of the Earth and that of the bar magnetic. Use diagrams to enhance your explanation. 18. Explain the difference(s) between the geographical North and the magnetic north pole of the Earth. 19. Give examples of phenomena that are affected by the Earth’s magnetic field. 20. Draw a diagram showing the magnetic field of the Earth. 21. Compare the geographic North Pole with the magnetic north pole. 22. Explain the difference between the geographical North pole and the magnetic North pole of the Earth. 23. Discuss qualitatively how the Earth’s magnetic field provides protection from solar winds. Chapter 2 Magnetism 57 Science_GR11_LB.indb 57 2018/02/02 17:00

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3 Waves and sound Objectives By end of this chapter learners will gain a broad understanding and acquire skills relating to the following concepts: • pulse wave and periodic wave • transverse and longitudinal wave • relationship between frequency and period • wave speed • superposition of waves • sound waves • speed of sound in different media • reflection of sound (echo) • pitch, loudness, range of sound frequencies. Science_GR11_LB.indb 59 2018/02/02 17:00

Why waves and sound matters Seeing, hearing, feeling warmth, surfing, tuning the radio, using a cell-phone – are just some of the many activities that rely on waves. Studies conducted by scientists and engineers involving vibrating strings and waves, helped them to understand what happens in atoms and molecules, electronic apparatus and antennae. Thus, learning about the vibrations in strings and the resultant waves, supports our scientific understanding. stone wavelength ripple waves Equilibrium(of the first segment) Compression Amplitude Wavelength Rare faction Rare faction/Expansion Amplitude Compression Compression Wavelength 3.1 Introduction Waves are everywhere, they transfer energy from one form to another. For example, what you listen to on the radio and watch on TV, is broadcast using waves. Talking to others on a cell phone is made possible with the use of waves. You even use waves – by using the microwave – to warm and cook some food. Thus waves are very useful to us on a daily basis. However, there are some waves that can also be harmful – even deadly – to us. 60   Technical Science | Grade 11 Science_GR11_LB.indb 60 2018/02/02 17:00

Look at Figure 3.1. Ripple waves in a river start with a disturbance at some point in the water. This disturbance causes water on the surface near that point to move, which in turn causes points farther away to move. The waves travel outward in a circular pattern away from the original disturbance. stone wavelength ripple waves Figure 3.1 Ripple waves in a river In this example, the water in the river is the medium through which the disturbance travels. Particles in the medium move up and down as the wave passes through. Remember that the medium does actually travel with the waves. So in this example, after the waves have passed, the water returns to its original position. 3.2 Pulses A pulse is a single disturbance that moves through a medium. This is shown in Figure 3.2. pulse string loose end Figure 3.2 A transverse pulse When a series of pulses are produced, in a regular, reptitive manner, a wave is formed. If the pulses are transverse pulses, then the wave formed is a transverse wave. If the pulses formed are longitudinal pulses, then the wave formed is a longitudinal wave. In this chapter we will focus on two pulses, namely, transverse and longitudinal pulses. 3.2.1 Transverse pulse A traverse pulse is a pulse where all the particles disturbed by the pulse move perpendicular (at right angles) to the direction in which the pulse is moving as shown in Figure 3.2. Chapter 3 Waves and sound 61 Science_GR11_LB.indb 61 2018/02/02 17:00

3.2.2 Longitudinal pulse A longitudinal pulse is a single disturbance in which particles of the medium move parallel to the direction in which the wave is moving (propagating). This is shown in Figure 3.3. rarefaction direction of a pulse compression wavelength Figure 3.3 A longitudinal pulse The particles move back and forth as the wave moves through. PRACTICAL INVESTIGATION Demonstrate a single vibration in different media Materials • shallow dish • water • coin or stone 1. Fill a shallow transparent Petri dish or any dish with water. The water must be still. 2. Drop a coin or small stone into the water so that it falls into the centre of the dish. Questions 1. What pattern does the coin produce in the water? 2. Where does the energy to make the pulse come from? 3. Explain your observation. 3.3 Waves What do an ocean buoy, a child in a swing, the coil inside a speaker, a guitar, atoms in a crystal, the motion of chest cavities, and the beating of hearts all have in common? They all oscillate – that is, they move back and forth between two points. Many systems oscillate, and they have certain characteristics in common. All oscillations involve force and energy. For example, you push a child in a swing to get the motion started; or you put energy into a guitar string when you pluck it. 62   Technical Science | Grade 11 Science_GR11_LB.indb 62 2018/02/02 17:00

Some oscillations create waves. So in our examples, a guitar string creates sound waves; or waves are made in a swimming pool by slapping the water with your hand. A wave is a succession of pulses. Each wave transports energy from one location (its source) to another location. During this exchange, no matter is transported, but each individual particle of the medium is temporarily displaced before returning to its original equilibrium position. This is illustrated in Figure 3.4. Figure 3.4 Succession of pulses – a wave Some electromagnetic waves such as gamma ray, x-rays, ultraviolet, visible light, infrared, microwaves and radio waves can travel through a vacuum. However, some waves require a material medium in which to travel. These waves are called mechanical waves. For example, sound waves cannot travel through vacuum, but need a medium such as air or water. 3.3.1 Medium A medium is just the substance or material through which waves move. In other words, the medium carries the wave from one place to another. The medium does not create the wave and the medium is not the wave. Air, water and rock are good examples of media (media is the plural of medium). Air is a medium for sound waves, water is a medium for water waves and rock is a medium for earthquakes (which are also a type of wave). 3.3.2 Transverse wave As we have already learned, a transverse wave is a wave in which the particles of the medium vibrate at right angles (perpendicular) to the direction of propagation of the wave. Look at Figure 3.4(a). It shows a string tied to a fix point. crest Figure 3.4(a) A string tied to a fix point. trough Chapter 3 Waves and sound 63 Science_GR11_LB.indb 63 2018/02/02 17:00

Figure 3.4(b) shows that when we shake the end of the string up and down, we will observe transverse waves propagating through the string. crest trough Figure 3.5(b) Transverse wave particles of the medium move perpendicular to the direction of propagation High points in the wave are called crests. Low points in the wave are called troughs. A series of crests and troughs form a transverse wave. The crests and the troughs propagates along the string, and the particles in the string move up and down. The distance between the successive crests or troughs is called a wavelength. This is shown in Figure 3.6. crest wavelength a rest position trough a = amplitude Figure 3.6 A transverse wave The x-axis represents the equilibrium position of the string. The curve shows the displacement of the wave at time t. 3.3.3 Longitudinal wave In a longitudinal, wave the particles move in the medium in the same dimension as the direction of movement of the wave. The wave particles vibrate back and forth along the path that the wave travels. Examples of longitudinal waves include: • sound waves • P-type earthquake waves • compression wave Longitudinal waves create areas of compression where particles are pushed together (higher density), and rarefaction where particles are pulled apart (lower density). This is shown in Figure 3.7. 64   Technical Science | Grade 11 Science_GR11_LB.indb 64 2018/02/02 17:00

Equilibrium(of the first segment) Compression Amplitude Wavelength Rarefaction Figure 3.7 A longitudinal wave PRACTICAL INVESTIGATION Generate a longitudinal wave in a slinky or spring Materials • Slinky Method 1. Hold a slinky stretched between two learners on a smooth surface such as the floor. 2. Quickly move your hand forwards and backwards. 3. Observe the wave produced. 4. Tie a piece of ribbon to one of the coils. This will represent a particle in the medium. Rarefaction/Expansion Amplitude Compression Compression Wavelength Questions 1. What is a pulse? 2. In which direction does the pulse move? 3. In which direction does the particle in the medium move? 4. How does the particle move relative to other particles in the medium? 5. What do we call a series of repeating or periodic disturbances that move through a medium from one location to another? Chapter 3 Waves and sound 65 Science_GR11_LB.indb 65 2018/02/02 17:00

3.3.4 Differences and similarities between transverse and longitudinal waves Table 3.1 compares how transverse and longitudinal waves are similar and in which ways they are different. Table 3.1 Comparison of transverse and longitudinal wave Differences Transverse wave Longitudinal waves Particle of the medium vibrate parallel to the Particle of the medium vibrate perpendicular to direction of propagation of the wave the direction of propagation of the wave The wave travels in the form of compressions and The wave travels in the form of crests and troughs rarefaction One compression and a rarefaction constitute one One crest and one trough constitute one wave wave Cannot be polarised Can be polarised Similarities They are both mechanical waves–they require medium to propagate Both a transverse and longitudinal wave can be described with a displacement time graph. 3.4 Wave terminology To master the concepts you will learn about in this Chapter, it is important for you to understand the terms used. The key terms are, frequency, period, wave speed, wavelength and wave speed. 3.4.1 Frequency and period Frequency and period are distinctly different, yet related, quantities. The quantity, frequency, is often confused with the quantity, period. Thus it is important for us to understand the difference between these two quantities. Frequency Frequency is the number of waves passing a point per second. Frequency: • refers to how often something happens • is a rate quantity • is the cycles/second. Frequency is measured in Hertz (Hz) or per second (s–1) and we use this formula to calculate it: f = number of waves time 1 =f=T 66   Technical Science | Grade 11 Science_GR11_LB.indb 66 2018/02/02 17:00

Period A period is the time taken to complete one wave. Period: • refers to the time it takes something to happen • is a time quantity • is the seconds/cycle. Period (T) is measure in seconds (s) and we use this formula to calculate it: time T = number of waves = T = 1f 1 The relationship between frequency and period is: f = T 3.4.2 Differences between frequency and period Table 3.2 lists the differences between frequency and period. Table 3.2: Differences between frequency and period Frequency Period Frequency refers to how often something happens Period refers to the time it takes for something to happen Frequency is a rate quantity (Hz or s–1) Period is a time quantity (s) Frequency is the cycles/second Period is the seconds/cycle f=1 T=1 T f Example 3.1 DETERMINE THE FREQUENCY OF MEDICAL IMAGING DEVICE A medical imaging device produces ultrasound by oscillating with a period of 0,3000 ms . What is the frequency of this oscillation? 1 ANALYSE Given: T = 0,300μs = 0,300 × 10–6 s Unknown: f = ? 2 SOLVE Use the equation relating frequency and period: f =1 T 1 f = 0,300 × 10–6 Hz ∴ f = 3,33 × 106 Hz The frequency of sound (3,33 × 106 Hz) is much higher than the highest frequency (2 000 000 Hz) that humans can hear and, therefore, is called ultrasound. Appropriate oscillations at this frequency generate ultrasound used for non-invasive medical diagnoses, such as observations of a foetus in the womb. Chapter 3 Waves and sound 67 Science_GR11_LB.indb 67 2018/02/02 17:00

Example 3.2 DETERMINE THE PERIOD OF MIDDLE C ON A MUSICAL INSTRUMENT The frequency of middle C on a typical musical instrument is 260 Hz. What is the time for one complete oscillation? 1 ANALYSE Given: f = 260 Hz Unknown: T = ? 2 SOLVE Use the equation relating frequency and period: T =1 f T = 1 260 ∴ T = 3,85 × 10–3 s = 3,85 ms The period found is the time per cycle, but this value is often quoted as simply the time in convenient units (ms or milliseconds in this case). Activity 3.1 1. A medical imaging device produces ultrasound by oscillating with a period of 0,400 ms. What is the frequency of this oscillation? 2. The frequency of middle D on a typical musical instrument is 294 Hz. What is the time for one complete oscillation? 3. Identify an event in your life (such as receiving a monthly salary) that occurs regularly. Identify both the period and frequency of this event. 3.4.2 Wave velocity Wave velocity (speed) is not the same as wave frequency. It is the speed with which a wave crest passes by a particular point in space in a particular direction. It is measured in m·s–1. So while wave frequency refers to the number of cycles occurring per second, wave velocity refers to the metres travelled per second. A wave can vibrate back and forth very frequently, yet have a small speed; or a wave can vibrate back and forth with a low frequency, yet have a high speed. We use the following formula to calculate wave velocity: wave velocity = frequency × wavelength or v = f × I. Thus, wave speed is equal to the product of frequency and wavelength. We can also derive an expression for speed of a wave in terms of its period and frequency. Let’s see how this can be done. 68   Technical Science | Grade 11 Science_GR11_LB.indb 68 2018/02/02 17:00

Table 3.3 Derivation of the wave speed formula We know that speed is equal to displacement divided by the time it takes to undergo v = ∆x ∆ that displacement. t For waves, a displacement of one wavelength (λ) occurs in a time interval equal to v = λT one period of vibration (T). But frequency and period are inversely proportional ∙that is, T = 1f ∙. Substituting this λ ∴ v = 1 = fλ frequency relationship in v = TI . f v = fλ Let’s look at an example to help us understand wave velocity. Example 3.3 MEASURING THE WAVE VELOCITY OF A TRANSVERSE WAVE Measure the wave velocity of the transverse wave below: –2 y(m) 0.5 0 0.2 0.4 0.6 0.8 1 t(s) –0.5 –1 0,30 cm 1 ANALYSE 1. Count the number of crest(s) passing a point in one second. 2. In the figure above there are 3,25 waves passing a point in one second. 3. The wavelength is given as 0,30 cm. 2 SOLVE 1. Calculate the frequency of the source that generated the waves. Since there are 3,25 waves passing a point in one second, the frequency is 1 = 0,31 Hz. 3,25 2. To calculate the speed of a wave, use the formula v = f × λ. Note Remember to make sure that the quantities are expressed in SI units. Since the wavelength is expressed in centimetres, convert it to metres: 0,30 cm = 0,30 cm = 1 m = 0,0030 m 1 100 cm Therefore, the speed of a wave will be 0,31 × 0,0030 = 0,00093 m·s–1 Chapter 3 Waves and sound 69 Science_GR11_LB.indb 69 2018/02/02 17:00

3.5 Speed of sound in different media The speed of a sound wave depends on what type of medium it is traveling through, and if that medium is a gas or liquid. A significant change in temperature can also affect the wave speed. Table 3.4 shows the speed of sound in different media. For example, the speed of sound in water at 25˚ is 1 493 m·s–1. Table 3.4 Speed of sound in different media Medium Velocity (m·s–1) Air (20 ˚C) 343 Air (0 ˚C) 331 Water (25 ˚C) 1 493 Sea water 1 533 Diamond 12 000 Iron 5 130 Copper 3 560 Glass 5 640 3.6 Formal experiment: Determine the speed of sound in air The average speed of sound through air is about 344 metres per second at room temperature. However, changes in temperature and humidity will affect this speed. Apparatus • stop watch • two blocks of wood, or other items that make a loud, sharp sound when struck together or a toy pistol • a friend to help with the experiment • a tape measure or long line of known length Method 1. Find a large empty area, such as a field or large court (at least more than 100 m). 2. Choose two spots on opposite ends of the area where each person will stand. 3. Measure the distance between the two spots using a tape measure. Alternatively, you can count off measured steps between the two spots. 4. Have your friend take the blocks and stand at one spot, holding them up high. 5. Take the stopwatch and stand at the other spot. Make sure you have a clear view of the blocks. 6. Signal your friend to bang the two blocks together hard. 7. Start the stopwatch as soon as you see the blocks hit each other. 8. Press stop as soon as you hear the sound from the blocks. 70   Technical Science | Grade 11 Science_GR11_LB.indb 70 2018/02/02 17:00

9. Calculate the speed of the sound by dividing the distance between you and your friend by the elapsed time. To get a more accurate measurement, repeat the above steps a few times and then take an average of the results. 10. By measuring the length of the area – and so the length of the distance between you – and dividing that by the number of seconds you measured as a delay, you’ll get the approximate speed of sound. 11. Repeat the experiment on different times of the day, different humidity conditions and different temperatures. 3.5.1 Speed of light versus speed of sound Why do we first see lightning flash before we hear the thunder during a thunderstorm? To answer this question, we need to compare the speed of sound to that of light. From Table 3.4 above, we can see that the speed of sound through air is about 343 m·s-1 whilst light travels through air at an enormous speed of 300 000 000 m·s-1. This is about a million times faster than sound. When lightning occurs in the clouds, the flash and the sound associated with it travel away from where they are produced. For a person standing 5 km away from where the lightning occurred, the light will reach him almost instantaneously but the sound will only reach him a few seconds later because of the enormous speed at which light travels. [Source: Jan Smit: Quarks, Crickets, Catfish . . . Through the eye of a physicist] Example 3.4 DETERMINE THE WAVELENGTH OF PIANO STRING A piano string tuned to middle C vibrates with a frequency of 262 Hz. Assuming that the speed of sound in air is 343 m·s–1, find the wavelength produced by the string. 1 ANALYSE Given: v = 343 m·s–1; f = 262 Hz Unknown: λ = ? 2 SOLVE Use the equation relating speed, wavelength, and frequency of a wave: v = fλ 343 = 262 × λ ∴ λ = 1,31 m Chapter 3 Waves and sound 71 Science_GR11_LB.indb 71 2018/02/02 17:00

Example 3.5 CALCULATE THE FREQUENCY OF WAVE PROPAGATION A transverse wave with a wavelength of 300 mm has a speed of 9 m·s–1. Calculate the frequency of the wave. 1 ANALYSE Given: λ = 300 mm = 300 = 0,3 1 000 v = 9 m·s–1 Unknown: f=? 2 SOLVE Use the equation relating speed, wavelength, and frequency of a wave: v = fλ 9 = f × 0,9 ∴f= 9 0,3 ∴ f = 30 Hz Activity 3.2 Relationship between speed, frequency, period and wavelength 1. Define the following terms: a) wave speed b) frequency c) period d) wave velocity 2. The Note A above middle C on a piano emits a sound wave with a wavelength of 0.77 metres. a) What is the frequency of the wave? Use 340 m·s–1 as the speed of the sound wave. b) Calculate the period of the sound wave. 72   Technical Science | Grade 11 Science_GR11_LB.indb 72 2018/02/02 17:00

PRACTICAL ACTAIVITY (PROJECT) How to make a string telephone (project) Besides travelling through air, sounds can also travel through solid objects. Materials • 2 paper or plastic cups (empty yogurt containers work well) or empty tins • 1 piece of string about 2 metre long • scissors • 1 thick needle • 2 paperclips Method 1. Poke a small hole in the bottom of each cup with the needle. 2. Thread the string through the holes and tie a paperclip to each end of the string to ensure that it is secured inside the cup. 3. Give one cup to a friend, and take the cups far apart, so that the string is tight. Talk and listen to each other through the string telephone. Questions 1. What happens if the string hangs loose? 2. What happens if you hold the string? 3. What happens if you connect three or four string telephones together? FORMAL EXPERIMENT Determining the difference between pitch and loudness using an oscilloscope Background The amplitude and the frequency of longitudinal waves cannot be measured directly. Instead we use an oscilloscope. An oscilloscope detects the compressions and rarefactions of the longitudinal waves and translate them into an electric signal to produce a graph of distance against time. The distance read on the vertical axis represents an amplitude of the longitudinal waves. Do not confuse the graph you observe on the screen with a transversal wave. The oscilloscope detects the change in pressure as the sound waves strike the detecting device. When the oscilloscope Chapter 3 Waves and sound 73 Science_GR11_LB.indb 73 2018/02/02 17:00

detects high pressure, it corresponds to the compressions; and when it detects low pressure, it corresponds to rarefactions. Apparatus • oscilloscope • speaker • microphone • power supply Method 1. Setup the apparatus as shown below: 2. Switch on the power supply and bring the microphone closer to the speaker. 3. Write down what you observed from the oscilloscope screen. Now increase the frequency to 200 Hz. a) Describe what you hear? b) Draw the pattern you see on the screen. 4. Increase the frequency to 2 000 Hz. a) Describe how this sound differs from the one you heard. b) Draw the pattern you see on the screen. 5. Now increase the volume and describe the sound you heard and draw the graph you see on the screen. 6. Write down your conclusion. Activity 3.3 Transversal waves 1. If a string is continuously shaken from side 1.5 B C D to side, it generates a series of transverse 1 pulses, which forms a transverse wave train 0.5 as shown alongside: It takes 0,3 s to complete A E 0 the pattern from A to E. –0.5 Study the wave pattern and answer the –1 questions that follow: F G H –1.5 120 cm 74   Technical Science | Grade 11 Science_GR11_LB.indb 74 2018/02/02 17:00

Note All measurements are in centimetres. a) Define the transverse pulse? b) Explain what is meant by points in phase. c) Name two points that are in phase. d) Give three sets of points, which represent one wavelength. e) What is the name given to the line AE? f) Determine the amplitude of the wave? g) Calculate the period of the wave? h) Calculate the frequency of the wave? i) Calculate the wavelength and the speed of the wave. PRACTICAL DEMONSTRATION Make a pulse on a string and on a rope Part A 1. Put the slinky on a smooth floor. Keep it a bit stretched. Place your foot down as shown in the figure. hand B slinky hand A foot floor 2. Pull the slinky sideward, that is, parallel to the floor (hand B). Release hand B. A pulse will travel along the slinky. By putting the foot and hand B in different positions, pulses of different shapes can be made. 3. Put the rope on a smooth floor and keep it tight or pulled straight. From the rest position (shown by the arrow in the figure), quickly move rope your hand sideward (parallel to the floor) and back again to the rest position. floor Chapter 3 Waves and sound 75 Science_GR11_LB.indb 75 2018/02/02 17:00

4. By moving your hand faster or slower, and by varying the distance that you move away from the rest position, you can make pulses with different shapes (this is not as easily done as with the slinky.) Part B 1. What is the direction the coils of the slinky move? 2. Tie a short piece of coloured string (wool) to one of the coils. 3. This represents a particle in the medium. (Or use the rope – Tie a short piece of coloured string (wool) somewhere to the rope.) Coloured string Questions 1. In what direction does the coloured string move as the pulse passes? 2. Write down your observation. Activity 3.4 Calculations involving wave equation 1. Use the general wave equation: v = fλ to answer the questions that follow: a) What is the ƒ if v = 345 m·s–1 and λ = 0,25 m? b) What is the λ if v = 345 m·s–1 and ƒ = 790 Hz? c) What is the v if λ = 3 m and ƒ = 10 Hz? d) What is the λ if v = 100 m·s–1 and ƒ = 3 Hz? e) What is the v if λ = 1 m and ƒ = 345 Hz? 2. How long is the wavelength of metro radio whose broadcast frequency is 97,1 MHz? (97,1 MHz = 97,100,000 Hz and v = 300 000 000 m·s–1) 3. Joe, the whistle maker, knows that the maximum volume for a whistle will occur if the length of the whistle is exactly 14 of the wavelength. If Joe must make a whistle that plays at a pitch of 320 Hz, how long will the whistle be? 4. Using the velocity of sound at 343 m·s–1 and given the frequencies of a piano scale, compute the wavelengths of that scale. Note Frequency wavelength Note Frequency Wavelength C4 261,6 G4 392 D4 293,6 A4 440 E4 323,6 B4 493,9 F4 349,2 C5 523,2 76   Technical Science | Grade 11 Science_GR11_LB.indb 76 2018/02/02 17:00

3.8 Superposition of waves When two waves travel through a medium, they interact. Fortunately, we can often assume that the interaction is linear, that is, the amplitude of the sum is the sum of the amplitudes. This relationship is known as the principle of superposition. Superposition of waves Superposition of waves is an algebraic sum of the amplitudes of the waves that meet at the same point simultaneously. 3.9 Interference Interference is what happens when two or more waves come together. Depending on how the crest and troughs of the waves are matched up, the waves might add together or they can partially or even completely cancel each other. Note that in this chapter we only discuss interference as it applies to transversal and longitudinal waves, but it applies to other waves as well. 3.9.1 What happens when waves come together? When waves come together, those waves are superimposed: they add together, with the amplitude at any point being the addition of the amplitudes of the individual waves at that point. 3.9.2 Constructive interference Constructive interference occurs whenever waves come together A+B A B so that they are in phase with each other. This is shown in Figure 3.12. B A+B For two waves of equal amplitude A interfering constructively, the B A resulting amplitude is twice as large as the amplitude of an individual wave. This Figure 3.12 Wave A and B meet Figure 3.13 Constructive and combine constructively interference of two waves phenomenon is shown in Figure 3.13. The diagram shows two pulses coming together, interfering constructively, and then continuing on together as if they had never encountered one another. Although the waves interfere with each other when they meet, they continue travelling as if they had never encountered each other. When the waves move away from the point where they came together, their form and motion is the same as it was before they came together. Chapter 3 Waves and sound 77 Science_GR11_LB.indb 77 2018/02/02 17:00

3.9.3 Destructive interference Destructive interference occurs when waves come together in A such a way that they completely B A cancel each other out. This is A+B=0 shown in Figure 3.14. A+B A When two waves interfere B B destructively, they must have the same amplitude and propagate in opposite directions. Figure 3.15 Figure 3.14 wave A and B meet Figure 3.15 Wave A and B meet depicts two pulses with the same and combine destructively and combine destructively amplitude moving in opposite directions. The pulses meet and combine destructively, and cancel each other. After interference the pulses continue to move in their respective directions. Figure 3.15 shows two pulses interfering destructively. Again, the pulses moves away from the point where they combine as if they never met each other. PRACTICAL ACTIVITY Interference of pulses Material • slinky • rope Method 1. Create two identical pulses with a string or rope, one on each side of the rope. 2. Release both pulses at the same time. 3. Create the pulses, one on the negative and one on the positive side of the rope and then both on the positive side. 4. Observe what happens at the point where the two pulses meet. A B B A Figure (a) Figure (b) 5. Fill in the missing words and delete the incorrect words. 78   Technical Science | Grade 11 Science_GR11_LB.indb 78 2018/02/02 17:00

a) In Figure (a), the displacements of the two pulses are in _______ directions. When the two pulses meet, there is a large / small / no displacement of the medium. It seems that when the two pulses meet we can find the total displacement by ______ (fill in: the correct mathematical operation) the displacements of the individual pulses. b) In Figure (b), the displacements of the two pulses are in the ______ direction. When the two pulses meet, there is a large / small / no displacement. It seems that when the two pulses meet, we can find the total displacement by _____ (fill in: the correct mathematical operation) the displacements of the individual pulses. c) In both situations we observe that a little while after the two pulses meet, both pulses appear again. The pulse travelling to the right continues to travel to the ______ and its ______ remains the same. The pulse travelling to the left continues travelling to the _______ and its ________ remains the same. 6. Formulate the Principle of Superposition. Activity 3.5 The ripple tank experiment 1. Set up a ripple tank with the two vibrator beads to produce just a single pulse as shown in Light source Figure 3.16. Power supply 2. Place a clean sheet of paper Source of pulse Shallow tank of water under the ripple tank. 3. Fill the ripple tank with water. 4. Switch the light source above the ripple tank on. 5. Set the vibrator of the ripple tank to produce single vibrations. The two vibrating beads must produce single pulses. 6. The vibrating beads produce a radiating outward single pulse. Pulse pattern on white sheet of paper Questions Figure 3.16 Ripple tank experiment setup – adapted from DBE 1. Observe what happens if the waves and sound guide distance between the two beads change. 2. How does the distance between the sources of the pulses influence the pattern formed on the sheet of paper? 3. Why are there regions that appear dark? Which type of interference cause these dark bands? 4. When the crest of one pulse coincides with a trough of another pulse, it results in the ______ of the pulse as they move over A series of pulses produce this interference pattern. Chapter 3 Waves and sound 79 Science_GR11_LB.indb 79 2018/02/02 17:00

each other. If the pulses are of equal amplitude, the result will be _____ at that point. We call this ________ interference. 5. When the trough of one pulse coincides with the trough of another, it results in a _________ trough. This is visible on the paper as a region of lighter bands. This is known as ________ interference Activity 3.5 Superposition of waves 1. In each set of waves below, the two waves at the left represent two waves traveling at the same time. Combine the two waves and show the results at the right. a) + b) + c) In the top example, (constructive, destructive) _________interference produces a (louder, softer) __________ sound. d) In the bottom example, (constructive, destructive) ___________ interference produces a (louder, softer) __________ sound 3.10 Sound waves A sound wave is the pattern of disturbance caused by the movement of energy propagating through a medium (such as air, water, or any other liquid, or solid matter). Sound waves do not travel in a vacuum. The source causes a vibration, such as a ringing telephone, or a person’s vocal chords. The vibration disturbs the particles in the surrounding medium; then those particles disturb those next to them, and so on. The pattern of the disturbance creates outward movement in a wave pattern, like waves of seawater on the ocean. The wave carries the sound energy through the medium, usually in all directions and less intensely as it moves farther from the source. The frequency of an audible sound wave determines how high or low we perceive the sound to be. This is known as a pitch. As the frequency of the sound wave increases, the pitch rises. The 80   Technical Science | Grade 11 Science_GR11_LB.indb 80 2018/02/02 17:00

frequency of a wave is an objective quantity that can be measured, while pitch refers to how different frequencies are perceived by a human ear. An individual’s hearing depends on variety of factors, including age and experiences with sound noises. The frequencies above and below the audible range are referred to as ultrasound and infrasound respectively. Ultrasound waves have a frequency above the normal range of human hearing. They can be used to scan for birth defects in unborn babies (OR monitor the heartbeat of a foetus) and for defects in manufactured equipment. 3.10.1 Some applications of ultrasound and infrasound Sound waves with frequencies lower that 20 Hz are called infrasonic waves and those above 20 000 Hz are called ultrasonic waves. Table 3.4 compares these two types of waves. Table 3.4 Comparing ultrasound and infrasound Ultra sound Infrasound Ultrasound – used as a Sonar Infrasound- used in pipelines as a leak detector Ships on the ocean make use of the reflecting Infrasound is used to forecast natural disasters like properties of sound waves to determine the depth earthquakes, volcanic eruptions and tsunamis. of the ocean. A sound wave is transmitted and bounces off the seabed. transmitter reciever Infrasound can be used to track animals and monitor seismic activity. seabed Because the speed of sound is known and the time lapse between sending and receiving the sound can be measured, the distance from the ship to the bottom of the ocean can be determined. Did you know? Elephants use infrasonic sound waves to communicate with one another. Their large ears enable them to detect these low-frequency waves, which have relatively long wavelengths. Elephants can effectively communicate in this way, even if they are separated by many kilometres. Chapter 3 Waves and sound 81 Science_GR11_LB.indb 81 2018/02/02 17:00

BELL JAR EXPERIMENT (INFORMAL EXPERIMENT) Aim To investigate whether sound travels in a vacuum Material required • bell jar • electric bell • power source • connecting wires • vacuum pump Figure 3.17 Bell jar experiment setup Procedure 1. Setup the apparatus as shown in Figure 3.17. 2. Connect the electric bell and the vacuum pump to the air tight glass bell jar. 3. Close the switch to complete the circuit. 4. Reduce the air pressure inside the chamber by pumping out the air through the vacuum pump. Observations As the air in the jar is being pumped out, the sound Hose from the tap. becomes fainter, although it is the same current that is passing through the bell. On reducing the air pressure further, the intensity of the sound reduces and stops. Hose from the bell jar. Conclusion Sound waves do not travel in a vacuum. Note In a case where the school does not have a vacuum pump. A t-connector can be purchased (it is very cheap and available in all hardware stores). It can be connected to the short pieces of used garden pipes. The water tap Hose to the drain. must be opened to create a vacuum. 82   Technical Science | Grade 11 Science_GR11_LB.indb 82 2018/02/02 17:00

INVESTIGATION: APPLICATION OF SOUND Humans can only hear sounds with frequencies of between 20 Hz and 20 000 Hz. Each individual’s hearing however depends on a variety of factors. This includes age and experiences with sound. Dogs, however, can hear sounds of frequencies of up to 45 000 Hz. Have you observed that dogs generally respond excitedly to whistling sounds? This is because high frequency sounds are in the comfortable range of hearing for dogs. Try using a thick low pitch voice when communicating with a dog and compare its response to the ringing of a high pitch bell for instance. For you to do: Do your own investigation on the use of ultrasounds as repellant for insects and rodents in homes, gardens and factories. Think of any other possible application of this property of sound. You can also read more about some of the applications below. Did you know? Infrasound is used in monitoring earthquakes, raw signal charting rock and petroleum formations below the earth, and also in ballistocardiography and seismocardiography to study the mechanics of time the heart. Did you know? Ultrasound technology is used in water treatment to suppress algal growth and biofilm force sensor formation. 3.10.2 Pitch and loudness Sound is vibrations. These vibrations can make solids, liquids or gases vibrate. These vibrations move in all directions from the source. They in turn make different parts of our ear vibrate so we can hear and interpret those vibrations. Loudness is the description of how loud or soft a sound is perceived to be. Loudness is determined by the amplitude of the sound. The higher the amplitude, the louder sound. Two rules for objects that make sound: a) The shorter the column of air or the string being vibrated, the higher the pitch of the sound will be. b) The harder an object is hit, blown or plucked, the louder the sound will be. Chapter 3 Waves and sound 83 Science_GR11_LB.indb 83 2018/02/02 17:00

INVESTIGATION: INVESTIGATION (PITCH OF WAVES) 1. Place varying amounts of water into cool drink bottles. 2. Blow across each bottle and write down your observations. 3. Now tap each bottle with a spoon and write down your observations. 4. Write your conclusions. 3.10.3 Pitch and loudness on oscilloscope The amplitude vs time graphs for pitch and loudness are shown in Figure 3.18. From the graphs we can deduce that the higher the frequency, the higher the pitch; and the higher the amplitude the louder the sound. Figure 3.18 Loudness vs pitch on oscilloscope amplitude amplitude 0s 0.01s 0.02s 0.03s 0.04s 0.05s 0s 0.01s 0.02s 0.03s 0.04s 0.05s quiet sound louder sound amplitude amplitude 0s 0.01s 0.02s 0.03s 0.04s 0.05s 0s 0.01s 0.02s 0.03s 0.04s 0.05s low pitch high pitch 84   Technical Science | Grade 11 Science_GR11_LB.indb 84 2018/02/02 17:00

3.10.4 Echo An echo is a sound wave that is reflected back (bounced back) from a surface. Sound waves can bounce off smooth, hard objects in the same way that a rubber ball bounces off the ground. Although the direction of the sound changes, the echo sounds the same as the original sound. Echoes can be heard in small places with hard walls, like wells; or where there are lots of hard surfaces around, for example, in the school hall or at a mountain range. However, sound waves are not always reflected. If sound waves meet a soft surface, such as cushion, they will be absorbed and will not bounce back. Activity 3.6 1. Define an echo and mention its applications. 2. A SONAR receives a strong echo of ultrasonic signal it sent downwards into the sea 3,6 seconds after the transmission. Calculate the depth of the sea-bed if the speed of the ultrasonic waves in sea-water is 1 250 m·s–1. 3. Name the units used to measure: a) pitch b) frequency c) loudness 4. You hook up a stereo system. When you test the system, you notice that in one corner of the room, the sounds seem dull. In another area, the sounds seem excessively loud. Describe how the sound moving about the room could have resulted in this. 6. You observe two musical instruments that you cannot identify. One plays high-pitched sounds and the other plays low-pitched sounds. How could you determine which is which without hearing either of them play? 7. Imagine that you observe two fireworks displays. You hear the explosion of one as soon as you see it. However, you see the other display several milliseconds before you hear the explosion. Explain why this is so. Review exercises Multiple choice questions 1. What is the highest point on a wave? A. trough B. wave C. crest D. bones 2. A student holds two vibrating objects. Object A is vibrating faster than Object B. Which sentence below is TRUE? A. A is LOUDER than B. B. B is LOUDER than A. Chapter 3 Waves and sound 85 Science_GR11_LB.indb 85 2018/02/02 17:00

C. A will have a higher PITCH than B. D. B will have a higher PITCH than A. 3. What is the distance between two consecutive crests or troughs called? A. wavelength B. amplitude C. speed D. frequency 4. A boy thought that the sound level in his room was too high. He turned down the music in his room. What physical property of the sound changed due to this and how did it change? A. The frequency increased B. The amplitude increased C. The amplitude decreased D. The frequency decreased 5. What does the highness or lowness of a sound represent? A. volume B. pitch C. wavelength D. amplitude 6. If the energy in a longitudinal wave travels from south to north, the particles of the medium ____. A. move from north to south, only. B. vibrate both north and south. C. move from east to west, only. D. vibrate both east and west. 7. When the particles of a medium are vibrating at right angles to the direction of energy transport, the type of wave is described as a _____ wave. A. longitudinal B. sound C. standing D. transverse 8. The main factor which effects the speed of a sound wave is the ____. A. amplitude of the sound wave B. pitch of the sound wave C. loudness of the sound wave D. properties of the medium 86   Technical Science | Grade 11 Science_GR11_LB.indb 86 2018/02/02 17:00

9. A transverse wave is travelling through a medium as illustrated in the diagram below. The particles of the medium are moving. A B E F G C D H I J A. parallel to the line joining AD. B. along the line joining CI. C. perpendicular to the line joining AD. D. at various angles to the line CI. 10. Many wave properties are dependent upon other wave properties. Yet, one wave property is independent of all other wave properties. Which one of the following properties of a wave is independent of all the others? A. wavelength B. frequency C. period D. velocity Long questions 11. Why doesn’t sound travel in a vacuum? 12. What factors does the speed of sound depend upon? What are some factors that it does not depend upon? 13. A cat can hear sound frequencies up to 70 000 Hz. Bats send and receive ultra-frequency squeaks up to 120 000 Hz. Which hears sound of shorter wavelengths, cats or bats? Explain your answer. 14. Why is it a safe prediction that you will have a significantly greater loss of hearing in your later years than your grandparents experienced? 15. If the frequency of a sound is doubled, what change will occur in its speed? What change happens to its wavelength? 16. The average wavelength in a series of ocean waves is 15,0 meters. A wave crest arrives at the shore on an average of every 10,0 seconds, so the frequency is 0,100 Hz. What is the average speed of the waves? 17. An FM radio station broadcasts electromagnetic waves at a frequency of 94,5 MHz (equal to 94 500 000 Hz). These radio waves have a wavelength of 3,17 meters. What is the speed of the waves? 18. Sound waves travel in air with a speed of 343 m·s–1. The lowest frequency sound that the average person can hear is 20,0 Hz; the highest frequency is 20 kHz. Find the wavelengths corresponding to these frequencies. Chapter 3 Waves and sound 87 Science_GR11_LB.indb 87 2018/02/02 17:00

88   Technical Science | Grade 11 Science_GR11_LB.indb 88 2018/02/02 17:00

4 Electricity Objectives By end of this chapter learners will gain a broad understanding and acquire skills relating to the following concepts: • Coulomb’s law • electric fields • electric field lines • electric field between parallel plates • Ohm’s Law • ohmic and non-ohmic conductors • electric circuit calculations • emf • internal resistance. Science_GR11_LB.indb 89 2018/02/02 17:00

Why electrostatics matters Electrostatics has various benefits in our everyday life. Even flowers use electric fields to communicate with insects! Some examples of where electrostatics is used is in the q+ operation of photocopiers, spray painting, electrostatic + precipitators, and so on. An electrostatic copier works by arranging positive charges in a pattern to be copied on the surface of a non-conducting drum, and then gently sprinkling negatively charged dry toner particles onto the drum. In spray painting, particles of paint are given a positive charge A as they leave the nozzle of a spray gun. 10Ω R1 R2 25Ω – – – – – – – – – – – negative 24V + d 35Ω R3 R4 20Ω E + + + + + + + + + + + positive 4.1 Electrostatics 4.1.1 Introduction Electrostatics is a branch of physics that deals with the phenomena and properties of stationary or slow-moving electric charges. One example of how electrostatics is used, is in the operation of electrostatic sprayers. One such example is shown in Figure 4.1 below. Did you know? At the tip of the sprayer there is a small electrode. This electrode creates high voltage electricity. As the spray particles passes through the nozzle at a slow speed, they become negatively charged. The positively charged surface is the crop surface. It has a magnetic attraction to the negatively charged Figure 4.1 An electrostatic sprayer particles from the sprayer. As you spray the liquid, the negatively charged spray particles will stick to the positively charged crop surface. In Grade 10 we learned that there are two types of charges. These are the positive charges called protons, and negative charges called electrons. A neutral object has an equal number of protons and 90   Technical Science | Grade 11 Science_GR11_LB.indb 90 2018/02/02 17:00

electrons, resulting in a zero net charge. A positively charged object has a deficit of electrons, while a negatively charged object has an excess of electrons. We also learned that like charges repel each other and that unlike charges attract each other (as shown in Figure 4.2). This is evidence that two charges exert a force on each other. This force is called an electrostatic force. Q1 Q2 Q1 Q2 F21 F12 F21 + + F12 + – Like charges repel eachother Unike charges attract eachother Figure 4.2 unlike and like charges The charge of an electron (an elementary charge) is qe = –1,6 × 10–19 C; whereas a proton has a charge of qe = 1,6 × 10–19 C. This is the smallest value a charge can have. Mathematically this can be expressed as Q = nqe. Here η represent the number of electrons in excess (for a negative charge), or deficit (for a positive charge). We learned further that the total charge of an isolated system will remain constant during any physical process. This statement is referred to as the law of conservation of charge. 4.1.2 Coulomb’s law Coulomb’s law describes the relationship between force exerted by the charges, the charges themselves and the distance separating the charges. Coulomb’s law states that: The force (F) of attraction or repulsion between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance (r) between the charges. Mathematically this relation can be expressed as: F α Q1Q2 and 1 F α r2 Combining the two yields: Q1Q2 F α r2 This can be written as a mathematical equation by introducing the proportionality constant k, such that: (kQ1Q2) F= r2 Where: Q1 = the magnitude of the point charge in Coulombs (C) r = the distance between the stationary point charges in metres (m) k = proportionality constant and has the value k = 9 × 109 N.m2·C–2 F = the electrostatic force between Q1 and Q2 in Newton (N) Chapter 4 Electricity 91 Science_GR11_LB.indb 91 2018/02/02 17:00

Conditions for Coulomb’s Law validity There are certain conditions that must be kept in order for the Coulomb’s Law to hold true. These conditions are: • the charges under investigation must be point charges • the charges must be stationary relative to each other Calculations using the Coulomb’s law: Before doing any calculations, let’s look at how to do unit conversions, especially for the electric charges. Table 4.1 below summarises typical submultiples of SI units. Table 4.1 Typical submultiples of SI units Micro μ × 10–6 Nano n × 10–9 Pico ρ × 10–12 Femto f × 10–15 Using the submultiples above, a charge of 6μC, will be 6 × 10–6 C in SI units. Note: When using Coulomb’s Law equation, we only consider the magnitude, and not the sign, of the charges. Example 4.1 CALCULATIONS FOR ELECTROSTATIC FORCE Two identical spheres A and B carrying charges of QA = +8nC and QB = –12nC, respectively, are placed at a distance 10 mm. Calculate the electrostatic force the charges exert on one another. 1. ANALYSE Given: QA = +8nC and QB = –12nC; r = 10 mm Convert all the given variables to their main SI units. In our case charges must be in C and not in n, and the distance be in m not mm QA = +8 C × 10–9 C and QB = –12 C × 10–9 C; r = 10 × 10–3 m Unknown: F = ? 2. SOLVE kQ1Q2 Use Coulomb’s Law: F= r2 F = (9 × 10 )(8 × 10 )(12 × 10 ) 9 –9 –9 (10 × 10-3) F = 8,64 × 10–3 N Because the particles are oppositely charged, they exert an attractive force between them. 92   Technical Science | Grade 11 Science_GR11_LB.indb 92 2018/02/02 17:00

Example 4.2 CALCULATIONS FOR ELECTROSTATIC FORCE USING RATIOS Two identical spheres carrying charges QA and QB are placed a distance r apart, and exert a force F on each other. If the distance separating the charges is halved, what will be the magnitude of the new force, with respect to F, the charges exert on each other? 1. ANALYSE Given: The force that the charges initially exert on each other is given by: kQ1Q2 F= r2 ∙ 1 Halving the distance will result in new distance rnew = 2r ∙ Unknown: Fnew = ? 2. SOLVE kQ1Q2 Use Coulomb’s Law: Fnew = 1r 2 ∙2 ∙ kQ1Q2 Fnew = 1r 2 4 4kQ1Q2 Fnew = 2 r ∴ Fnew = 4 F Halving the distance between the charges increases the force they exert on each other by the factor of 4. Activity 4.1: Application of Coulomb’s law 1. State Coulomb’s law in words. 2. Use your understanding of relationship between two variables to draw the generic graphs of: a) F versus Q1Q2 b) F versus r2 c) F versus 12 r 3. Two charged spheres carrying charges QA and QB are a distance r apart, and exert a force F⃗ on one another. a) If the distance between the charges is doubled, and the charge on one of the two spheres is doubled, what will the new force that the charges will exert on each other be, relative to F⃗? b) By what factor will the force the two charges exert on each other change, if the charges are both doubled? c) If the distance between the charges is decreased by the factor of 3, by what factor will the force the charges exert on each other increase? Chapter 4 Electricity 93 Science_GR11_LB.indb 93 2018/02/02 17:00

4. Two charged objects A and B are placed 5 cm apart, as shown in the sketch. The charge on A, QA = +2,6μC and charge on B, QB = +3,2μC. QA QB + 5 cm + a) Draw the force diagram (not to scale) of the force of A on B (F⃗A on B) and the force of B on A (F⃗B on A). b) Will the two objects attract or repel each other? c) Will (F⃗A on B) be equal to (F⃗A on B)? Give a reason for your answer d) Calculate the magnitude of the force between objects A and B. 5. Two charged objects M and N are placed 4,5 cm apart. a) Calculate the magnitude of the force between the objects if the charge on M is QM = +2,5μC and on N is QN = +3,2μC. b) Calculate the force between the objects in (a) above when the distance between them is doubled. c) Compare your answers to (a) and (b) above and determine the factor by which the force is reduced when the distance between the charges is doubled. Is this outcome in line with your initial expectation? Explain. 6. Two identical spheres carrying equal charges are placed a distance of 4 cm apart, and exert a force of 3,6 × 10–4 N. Calculate the magnitude of the charge on each sphere. 7. Three charged objects A, B and C lie on a straight line as shown in the diagram. The distance between A and B is 12 cm and between B and C is 7 cm. QA QB QC – 12 cm + 7 cm – a) Given that the magnitude of the charge on A is QA = +3μC , and on B is QB = +5μC, and on C is QC = +2μC: (i) how many forces are exerted on charge A? (ii) how many forces are exerted on charge B? (iii) how many forces are exerted on charge C? b) Calculate the forces on charge A, and hence determine the net force on charge A. (Hint: draw force diagrams showing all the forces acting on A.) c) Calculate the forces on charge B. Determine the net force on charge B. 8. Two identical spheres carrying charges QA = –3,2μC and QB = +4,8μC are placed a distance of 50 mm apart. The spheres are then brought in contact with one another and then placed back to their original positions. a) What will the new charge on each sphere be after they have been separated? b) Calculate the force that the charges on the spheres exert on one another after they have been separated 94   Technical Science | Grade 11 Science_GR11_LB.indb 94 2018/02/02 17:00

4.2 Electric field The previous section described the interaction between charges in terms of the force between them, the distance between them and their magnitudes. According to this description, the force exerted on the charges acts over a distance between charges, with no intervening medium. The strength of the electrostatic force between two charges decreases as the distance separating the charges increases. Thus, at a certain point, the force the charges exert on each other will be significantly small such that it can be approximated to zero. This means that there must be a region within which two charged particles will experience the force due to the charges they carry. In 1838, Michael Faraday introduced another type of interaction between charges. According to Faraday, there is ‘something’ that exists around charges. This ‘something’ he called an electric field. A charge gives rise to an electric field that permeates the space around it. This in turn exerts a force on other charges. Thus, a charge A in the vicinity of another charge B; will experience a force exerted on it by the electric field of charge B. Electric fields act in the same direction as the force on charges, and are therefore vector quantities (they have definite direction). Note Even though electric fields cannot be seen, they do exist in reality. They exhibit physical qualities whose effects can be measured. The electric field strength at a point is defined as the force per unit positive test charge. That is: E = Fq Where: E = the electric field strength in Newton per Coulomb (N·C–1) F = the electrostatic force experienced by the test charge, q, measured in Newton (N) q = the test charge in Coulombs (C) To describe the electric field quantitatively, we use the term electric field strength. The electric field strength is the magnitude of the electric field at a point. The electric field at a point is a vector quantity with magnitude as well as direction. The electric field at a point is defined as: The electric field at a point ∙E⃗ ∙ is the electrostatic force ∙F⃗ ∙ experienced by the test charge (q) placed at that point divided by the charge itself. That is: ⃗ E⃗ = Fq Chapter 4 Electricity 95 Science_GR11_LB.indb 95 2018/02/02 17:00

Example 4.3 CALCULATIONS FOR ELECTRIC FIELD STRENGTH A QA = +9μC point charge experiences a force of 2,7 × 10–5 N. Calculate the electric field strength at the point where a point charge Q. 1. ANALYSE Given: F = 2,7 × 10–5 N Q = +9μC = 9 × 10–9 C Unknown: E = ? 2. SOLVE Use the formula for electric field strength. Thus: F E=Q 2,7 × 10-5 E = 9 × 10-9 E = 3 000 N·C–1 The electric field strength on a charge Q experiencing a force E = 3 000 N·C–1. Example 4.4 CALCULATIONS FOR ELECTRIC FIELD STRENGTH If the electric field of magnitude 2 500 N·C–1 causes a charge Q to experience a force of 1,25 × 10–5 N, calculate the charge Q. 1. ANALYSE Given: E = 2 500 N·C–1 F = 1,25 × 10–5 N Unknown: Q = ? 2. SOLVE Use the formula for electric field strength. Thus: F E=Q 1,25 × 10–5 2 500 = Q 1,25 × 10–5 ∴ Q = 2 500 Q = 5 × 10–9 C Q = 5nC 96   Technical Science | Grade 11 Science_GR11_LB.indb 96 2018/02/02 17:00

Activity 4.2: Calculating the electric field strength 1. Define the following terms: a) electric field b) electric field strength 2. A charge of –5nC experiences an upward force of 4 × 10–6 N at a point in an electric field. At that point calculate: a) the electric field strength b) the force that an electron will experience. 3. Given that there is the electric field of 300 N·C–1, calculate the magnitude of the force that the field exert on an Alpha particle. An alpha particle is a helium nucleus with a charge of two protons. (Charge of a proton =+1,6 × 10–19 C. 4.2.1 Electric field lines Electric field lines are a way of representing an electric field. Arrows on the field lines indicate the direction of the electric field. Figure 4.3 show electric field lines around: a) a positive charge b) a negative charge c) two positive point charges d) two negative point charges; and e) the two opposite point charges. Figure 4.3 electric field lines around the charge(s) a) Positive charge b) Negative charge + _ Electric field of a positive point charge Electric field of a positive point charge Chapter 4 Electricity 97 Science_GR11_LB.indb 97 2018/02/02 17:00

c) Two positive point charges d) Two negative point charges +Q +Q –Q –Q e) Positive and a negative point charges: +Q –Q Hints on drawing the electric field lines • The electric field lines are perpendicular to the surface of the charge carrier. • The do not cross or intersect with one another • They indicate the strength of the field. The closer they are to one another, the stronger the field strength. • The lines are merely a representation, they do not really exist. The electric field spreads everywhere around the charge, not only 2-dimensional as we have drawn above. 4.2.2 The electric field between two parallel plates The strength of an electric field of a point charge depends on the distance from the point to where the charge is placed. However, it is independent of the charge generating the field. This electric field is said to be non-uniform, but an electric field could be also uniform. Figure 4.4 shows an example of a uniform electric field between two parallel plates. + + + + + + + + + + + – – – – – – – – – – – Figure 4.4 Electric field lines between two oppositely parallel plates Characteristics of the electric field lines between two opposite parallel plates • The electric field lines between parallel plates is uniform. • The electric field lines are parallel and equally spaced between the parallel plates. 98   Technical Science | Grade 11 Science_GR11_LB.indb 98 2018/02/02 17:00

• The electric field strength between the parallel plates is calculated using the formula: E = V d • The electric field strength between two parallel plates is inversely proportional to the distance between the plates ∙E α 1∙. d Example 4.5 CALCULATIONS FOR ELECTRIC FIELD STRENGTH The electric field strength between the parallel plates is the same at any point between the plates. a) Calculate the electric field strength between two charged parallel plates that have a potential difference of 3 V and are 3 cm apart. b) Calculate the force that the field exert will exert on the proton between the plates. 1. ANALYSE Given: V=3V d = 3 cm = 3 × 10–2 m Unknown: F = ? Ensure all the variables are in their SI units 2. SOLVE V E=d 3 E = 0,03 E = 100 N·C-1 The force that the proton will experience because of this field can be calculated using E = F Q Here the charge is the charge of the proton. Thus: Q = +1,6 × 10-19 C F E =Q F 100 = 1,6 × 10-19 ∴ F = 1,6 × 10-17 N Activity 4.3: Application of electric field strength 1. Draw an electric field diagram: a) surrounding a negative point charge b) surrounding two positive point charges c) surrounding two oppositely charged point charges Chapter 4 Electricity 99 Science_GR11_LB.indb 99 2018/02/02 17:00

2. Two parallel plates are 2 cm apart and have a potential difference of 15 V. a) Draw the electric field diagram of the two plates. b) Calculate the electric field between the two plates. c) If the distance between the plates is doubled, what will be the new field between the plates? (Do not use any formulae to answer this question.) 3. A distance between parallel plates is varied at constant intervals, while keeping the potential difference constant. Draw a graph showing the relationship between V and E. 4. The horizontal parallel plates A and B are separated by a distance of 32 mm. B is at a potential of +1 000 V. plate A S 32 mm V + 1000 V Q plate B a) Draw a sketch to show the electric field lines between the plates A and B. b) Calculate the magnitude of the electric field strength between the plates. c) A tiny charged particle is stationary at S, 8 mm below plate A, which is at zero electrical potential. It has a charge of 3,2 × 10–12 C. (i) State whether the charge on this particle is positive or negative. (ii) Calculate the force due to the electric field on the charge. 4.2.3 Applications of electrostatics in technology Electrostatics in an inkjet printer An inkjet printer uses electrostatic charge to direct the tiny ink droplets to the correct place on the page. Coloured ink is passed through a very small hole called a nozzle. The nozzle separates the ink into many tiny droplets, which are given an electrostatic charge. The direction in which the charged ink droplets move can be controlled by electrically charged metal plates. A voltage on the plates means that the charged ink droplets will be attracted to one plate and repelled by the other. Figure 4.6 How is Electrostatic Charge used The ink droplets have a positive charge. The ink droplets in an Inkjet Printer? are attracted to the negative plate and repelled by the positive plate. By controlling the voltage on the plates, a particular ink drop can be precisely positioned on the paper. There are many nozzles, and the final picture is made up of a very large number of coloured ink drops, each in exactly the right place for the image. 100   Technical Science | Grade 11 Science_GR11_LB.indb 100 2018/02/02 17:00

Deflection plates Paper Charging electrode Ink nozzle Gutter Ink reservoir [Source: Adapted from http://www.gcsescience.com/pse9-electrostatic-charge-inkjet-printer-nozzle.htm] Electrostatics in spray painting Millions of cars are made each year and the steel car bodies must all be painted to prevent them from rusting. The paint is sprayed onto the car bodies and the process is made more efficient by using electrostatic charges. The spray paint goes past high voltage positive needles as it leaves the spray gun and the tiny droplets of paint pick up a positive charge. They do this by losing negative electrons. The car body is then given a high voltage negative charge which attracts the positively charged paint droplets. The paint droplets spread out more as they leave the gun because Figure 4.7 How is Electrostatic Charge they all get the same positive charge and so they all repel used in Paint Spraying? each other. It covers a wider area more evenly. The paint [Source: Adapted from http://www.gcsescience.com/pse8-electrostatic- charge-paint-spray.htm] droplets are attracted to the negative metal car body, and so less paint is wasted. Electrostatics in industrial chimneys Pollution from industrial chimneys as well as the waste gases from burning coal (carbon dioxide and sulfur dioxide), can be reduced by using electrostatic charge. The chimney contains many small particles of unburnt fuel (mainly carbon). The chimney has a high voltage negative grid across it and this gives the small particles a negative charge as they escape causing them to gain electrons. Further up, the chimney has positively charged plates. These plates attract the negatively charged particles. The particles of pollution build up on the plates until they are heavy enough to fall down into containers. In this way, much of the smoky Figure 4.8 Application of electrostatics in pollution is removed from the chimney before it can industrial chimneys get out into the atmosphere. [Source: Adapted from: http://www.gcsescience.com/pse11-electrostatic-chimney- pollution.htm] Chapter 4 Electricity 101 Science_GR11_LB.indb 101 2018/02/02 17:00

4.3 Electric circuits 4.3.1 Introduction We rely on electric forces, electricity and electrical appliances to make many things possible in our daily lives. To understand the technology that uses electricity in the real-world, the study of electrical circuits is essential. In this section we will learn about electric circuits, series and parallel combinations, and the application of Ohm’s law. Did you know? During load shedding in our country we could not use the kettle to boil water for tea or coffee; could not use electric stoves or ovens to cook dinner; couldn’t charge our cell phones nor watch a television. We were in the dark! According to Eskom, load shedding is a measure of last resort to prevent the collapse of the power system country-wide. When there is insufficient power station capacity to supply the Figure 4.9 Load shedding demand (load) from all the customers, the electricity system becomes unbalanced. This can cause a country-wide blackout), which could take days to restore. When power is insufficient, Eskom can thus either increase supply or reduce demand to bring the system back into balance. As the difference between supply and demand becomes small, we refer to the system becoming ‘tight’. This implies that action has to be taken to prevent the system from becoming unstable. Eskom normally takes a sequence of steps to keep the system stable and to avoid load shedding. The steps include first asking large customers to reduce the load voluntarily. However, if several power station units trip suddenly and unexpectedly, we may have to skip those steps and go straight to load shedding to prevent the system from becoming unstable. Scheduled load shedding is a controlled way of sharing the available electricity amongst all its customers. By switching off parts of the network in a planned and controlled manner, the system remains stable throughout the day, and the impact is spread over a wider base of customers. In Grade 10 we learned the following about resistors connected in series and connected in parallel: 102   Technical Science | Grade 11 Science_GR11_LB.indb 102 2018/02/02 17:00

Resistance, current and potential difference Resistance, current and potential difference in series circuit in parallel circuit I A VT AT V1 I1 R1 I2 R2 I3 A1 A1 A2 A3 R1 R2 A2 V1 V2 V2 V Effective/(equivalent) resistance: Effective/(Equivalent) Resistance: R = R1 + R2 1 = 1 + 1 + 1. Rp R1 R2 R3 The equivalent resistance in a series circuit is The equivalent resistance is always smaller than the worked out by calculating the sum of all the smallest resistance in the circuit. resistors in the circuit. For two resistors connected in parallel the following equation can also be used: product R1 × R2 Rp = sum = R1 + R2 Current: Current: I1 = I2 = I3. I = I1 + I2 + I3 + … The current is the same anywhere in a series circuit. Resistors in parallel are current dividers. Potential Difference: Potential Difference: VT = V1 + V2 + V3 V = V1 = V2 = V3 Resistors connected in series are voltage dividers. The potential difference across all resistors connected in parallel will always be the same. It will also be the same the potential difference across the whole parallel combination. Figure 4.10 Series and parallel connections Activity 4.4: Electric circuits 1. Use the keywords to complete the paragraph below. negative loop  positive  path  charges  closed  energy A circuit will only work if it is ____________________. This means that it must make a complete _______________, otherwise the electrical _______________cannot be transferred. A circuit is a ______________that allows ________________to pass through it. The conventional direction of the electric current is always travels from the _______________ terminal to the__________________________terminal. Chapter 4 Electricity 103 Science_GR11_LB.indb 103 2018/02/02 17:00

2. In your own word, describe what is happening in the diagram. Make reference to the following conventional direction of the current, direction of electrons and energy conversion. PRACTICAL INVESTIGATION (Informal) Relationship between potential difference and current Aim: Investigate the relationship between the potential difference (V) across a resistor and the current through it. To investigate the relationship between the potential difference (V) across a resistor and the current through it, we are going to apply the scientific method. Investigative question What is the relationship between the potential difference (V) across a resistor and the current through it? Variables • Independent: Potential difference • Dependent: Current • Controlled (fixed): Temperature of the resistor/ resistance of the resistor Hypothesis When current through a resistor increases, the potential difference across the resistor increases. Apparatus • 4 cells (1,5 V each) • variable resistor • resistor • switch • voltmeter • connecting wires • ammeter 104   Technical Science | Grade 11 Science_GR11_LB.indb 104 2018/02/02 17:00

Procedure Step 1: Set up the circuit according to the following circuit diagram. A Step 2: Close the switch and measure the voltage across the resistor using the voltmeter, and the current in the circuit using the ammeter. Step 3: Add one more 1,5 V cell to the circuit and repeat your measurements. V Step 4: Repeat until you have four cells. Circuit diagram1 Step 5: Record your result in the following table. Number of cells Voltage, V (V) Current, I (A) 0 1 2 3 4 Step 6: Use the information in your table to draw a graph illustrating the relationship between the potential difference (V ) and the current (I ). Plot (I ) on the y-axis, and (V ) on the x-axis. Step 7: Calculate the slope (gradient) of the graph. Step 8: What physical quantity is represented by the slope of the graph? Step 9: Analyse your results and graph and write down a conclusion for your investigation. 4.3.2 Resistance In Grade 10 we defined resistance as the opposition to the flow of electric current. We also learned that the unit measure of the resistance is ohms (Ω). 1Ω = 1 V.A–1. From the investigation we found that the gradient of the graph (the slope) represents the reciprocal of the resistance, which is called conductance. ∆y ∆I gradient = ∆x = ∆V 4.3.3 Ohm’s law Below is the summary of the results obtained in the experiment performed in the previous practical activity: Number of cells Voltage, V (V) Current, I (A) 0 0 0 1 1,5 0.0015 2 3 0.003 3 4,5 0.0045 4 6 0.006 Chapter 4 Electricity 105 Science_GR11_LB.indb 105 2018/02/02 17:00

Graph of current versus potential 0.007 deference across the resistor 0.006 From the results obtained above, one 0.005 Current (A) can therefore conclude that: 0.004 • the current flowing through the 0.003 conductor is directly proportional 0.002 to the potential difference across 0.001 the resistor (I α V). 0 • from the graph we get the 0 1 2 3 4 5 6 7 following linear equation Potential difference (V) y = mx + c. Making: y=I x=V m=1 R c=0 ∆I • we get 1 = , which is the constant of proportionality (slope/gradient of the graph). ∆V R Therefore, R = V. I However, for the statement above to be true we must ascertain that the resistance of the conductor remains constant. To ensure this, we must also ascertain that the: • length of the conductor remains constant • thickness of the conductor remains constant • material of which the conductor is made of remains unchanged • temperature of the conductor. The first three factors are almost always guaranteed. The one factor that needs to be closely monitored is the temperature of the conductor. Therefore, we can modify the initial drawn statement from the experimental results to be: • The current flowing through the conductor is directly proportional to the potential difference across the resistor, provided the temperature of the conductor remains constant. The statement above was named after the German Physicist and Mathematician, George Ohm. Using equipment of his own creation, Ohm found that there is a direct proportionality between the potential difference (voltage) applied across a conductor (resistor) and the current through it, provided that the temperature remains constant. This relationship is known as Ohm’s law. 106   Technical Science | Grade 11 Science_GR11_LB.indb 106 2018/02/02 17:00

Example 4.6 CALCULATIONS USING OHM’S LAW Calculate the resistance of a resistor if the potential difference across the terminals of the resistor is 6 V and current passing through is 1,5 A. 1. ANALYSE Given: I = 1,5 A V=6V Unknown: R = ? 2. SOLVE Using Ohm’s Law, we know that: R=V I ∴R= 6 1,5 R = 4Ω Therefore, the resistance of such a resistor is 4Ω. Example 4.7 CALCULATIONS USING OHM’S LAW What is the current flowing through the resistor, if the potential difference across that resistor is 4 V and the resistance of the resistor is 10Ω? 1. ANALYSE Given: R = 10Ω V=4V Unknown: I = ? 2. SOLVE Using Ohm’s Law: R=V I 10 = 4 I ∴ I = 0,4 A The current passing through such a resistor is 0,4 A. Chapter 4 Electricity 107 Science_GR11_LB.indb 107 2018/02/02 17:00

EXPERIMENT 10 (formal) Determine the resistance of an unknown resistor. Background Ohm’s law states that: The current flowing through the conductor is directly proportional to the potential difference across the resistor, provided the temperature of the conductor remains constant. Mathematically I=V R Aim To determine the resistance of an unknown resistor applying Ohm’s law. Apparatus • battery/DC power supply • resistance wire (resistor) • rheostat (variable resistor) • ammeter • voltmeter • connecting wires Precautions • Do not leave the switch closed at all times. The battery will run flat. • Close the switch only for taking the readings on the voltmeter and ammeter. • Take your readings accurately. • Do not conduct the experiment with wet hands. Method/Procedure 1. Set up a circuit according to the circuit diagram shown below, connecting the unknown resistor between points A and B. A A B V Circuit diagram 2 2. Close the switch and adjust the rheostat to give a low voltage across the unknown resistor. Measure the voltage across the resistor using the voltmeter, and the current in the circuit using the ammeter. 3. Close the switch and change the resistance in the rheostat and take the ammeter and voltmeter reading. 108   Technical Science | Grade 11 Science_GR11_LB.indb 108 2018/02/02 17:00

4. Repeat the readings for different values of the resistance of the rheostat and record your results in a table below (Open the switch while recording the readings in order to spare the batteries’ life span). Trial No. Ammeter reading (A) Voltmeter reading (V) 1 2 3 4 5 5. Use your reading to plot a graph of potential difference against current. 6. Do the analysis and interpretation of the result. a) What is the shape of the graph? b) What does the gradient of the graph represent? 7. Use the graph to calculate the resistance of the unknown resistor. 8. Write down a conclusion for the experiment. Example 4.8 OHM’S LAW 1. The learners were investigating the relationship between the potential difference across the resistor and the current passing through it. They added cells in series, starting with one cell. Before adding a new cell, a switch was closed and the voltmeter and ammeter reading were taken. Use the results below to answer the following questions: Number of cells V (V) I (A) 1 3.0 0.375 2 6.0 0.75 3 9.0 1.125 4 12.0 1.50 a) Plot the graph of V vs. I. b) Explain how you would use the graph to determine the resistance of the resistor. c) Determine the resistance of the resistor used. d) From the graph, what conclusion can be drawn about the relationship between V and I? 2. Calculate the resistance of the resistor, if the potential difference across the resistor’s terminals is 3 V, and the current passing through it is 0,4 A. 3. The current flowing through a 10Ω resistor is 0,21 A. a) What is the potential difference across the 10Ω resistor? b) If the switch was on for 10 s, while taking the measurements, what is the energy delivered to the resistor. ∙Hint: recall that V = W ∙ Q Chapter 4 Electricity 109 Science_GR11_LB.indb 109 2018/02/02 17:00

4. Below is a table of results for which Ohm’s law is valid. Calculate the missing quantities. Questions Potential difference (V ) Current (I ) Resistance (R) 5.1 4,5 V (a) 15 Ω 5.2 (b) 3A 12 Ω 5.3 9V 0,5 A (c) 5.4 (d) 2,5 A 3Ω 5.5 220 V (e) 80 Ω 4.3.4 Series and parallel network In Grade 10 we looked at how to calculate effective resistances of series and parallel circuits. At the beginning of this chapter we summarised those results as a reminder of what we did. The following examples will still require us to apply such principles to which parallel and series circuits adhere. Example 4.9 ELECTRIC CIRCUITS CALCULATIONS Study the circuit diagram below and answer the questions 12V S that follow: a) Calculate the effective resistance of the circuit. b) Determine the reading on the ammeter A1. A1 c) What is the reading on the voltmeter V? A2 R1 =12Ω d) Determine the reading on the ammeter A2 e) If R2 burned out what would happen to the reading on R2 =12Ω the ammeter A1? V Solution By tracing the path followed by the conventional current, we can see where the current divides. This tells us which resistors are in parallel. Below is the presentation of the flow of current in the circuit above: IT IT IT IT A1 I1 A2 R1 IT IT I2 IT R2 V We clearly see that at the junction of R1 and R2 the current divides. 110   Technical Science | Grade 11 Science_GR11_LB.indb 110 2018/02/02 17:00

We can thus deduce that R1 and R2 are parallel. To simplify the problem, we can replace the resistors with one resistor R12, which has the same effect as the two resistors. IT IT IT IT A1 IT IT IT R12 V We can see that the current flowing through R12 is the total current. Since this is now the only resistor in the circuit, the potential difference measured by the voltmeter is the total external potential difference. a) Since we only have two resistors connected in parallel, the total resistance is: RT = R R+1R3 R 1 2 RT = 44 × 12 + 12 RT = 3Ω b) From the sketches, it is clear that A1 measures the total current flowing in the circuit. Since we know the total voltage and the total resistance, we can easily calculate the total current: IT = VRT T IT = 12 3 IT = 4A c) As discussed above, the voltmeter V measures the total voltage of the circuit, since R12 = RT. Therefore, the reading on the voltmeter is 12 V. d) We already know that the resistors in parallel have the same voltage. Therefore, potential difference across R1 is also 12 V. Using Ohm’s law we can easily calculate the current through R1, which will be reading on the ammeter A1. I1 = VR1 1 I1 = 12 R4 I1 = 3 A e) If R2 burns, the total resistance of the circuit will increase. As the total potential difference is constant, the total current, according Ohm’s law, will decrease. Therefore, the reading on A1 will decrease. Chapter 4 Electricity 111 Science_GR11_LB.indb 111 2018/02/02 17:00

Example 4.10 ELECTRIC CIRCUITS CALCULATIONS 24V Study the diagram alongside, then answer the questions that follow. 2A Calculate: a) the resistance of resistor, R1. A b) the reading on A. V R2 = 15Ω R1 Solution: R4 = 3Ω Follow the same approach we used in the previous example. Trace the path of the current, thereby identifying R3 = 10Ω the resistors that are parallel. IT 2A IT A IT I1 V R2 = 15Ω R1 R4 = 3Ω I1 IT I2 R3 = 10Ω IT From the diagram above, we can see that current dives at the R2 and R3 junction. Therefore, the resistors are parallel to each other. This allows us to replace the two resistors with one resistor (R23), which has the same effect as the two resistors. 24V IT 2A IT IT IT R4 = 3Ω R1 V IT IT R 23 Now the sketch is simplified, and we can answer the questions. 112   Technical Science | Grade 11 Science_GR11_LB.indb 112 2018/02/02 17:00

a) We know the total current of the circuit, as well as the total voltage. We are therefore able to calculate total resistance from which we will be able to calculate R1. RT = VI T T RT = 24 2 RT = 12Ω But the three resistors, R1, R23 and R4 are in series. So if we can calculate the effective resistance of R23, then we can easily calculate R1. R23 = R R1 R4 1 × R4 15 × 10 R23 = 15 + 10 R23 = 6Ω Now for resistors in series we have: RT = R1 + R23 + R4 12 = R1 + 6+ 3 ∴ R1 = 3Ω b) Ammeter A, measures the current through R2. If we can find the potential difference across effective resistance R23, then we ca solve the problem. For resistors in series current is the same as the total current. Therefore, we have: V1 + V23 + V4 = VT I1R1 + V23 + I4R4 = ITRT 2 × 2 + V23 + 2 × 3 = ITRT ∴ 2 × 2 + V23 + 2 × 3 = 24 ∴ V23 = 12 V V23 ∴ I2 = R1 12 ∴ I2 = 15 ∴ I2 = 0,8 A In the next section we will explore how the results for good conductors such as copper compare with that of semiconductors. Activity 4.5: Series and Parallel Combination 1. In circuit diagram below, a 12 V battery is connected 12V to a combination of resistors. R1 = 3Ω, R2 = 6Ω and R3 = 2Ω. V2 R3 A1 Calculate: R2 a) The total resistance of the circuit. b) The reading on the ammeter A1. R1 c) The reading on the voltmeter V2. V1 d) The current flowing through the resistor R2. Chapter 4 Electricity 113 Science_GR11_LB.indb 113 2018/02/02 17:00

e) What will happen to the reading on ammeter A1 if the resistor R1 burned out? (Write only INCREASE, DECREASE, or STAY THE SAME). Explain how you reached your conclusion. 2. Calculate the missing information in the table for the following circuit. Show all your calculations. Note that the order in which the calculation must be done does not matter. R2 R5 R4 R3 R1 R6 V I R Total (m) 2A (n) R1 (a) (b) 5Ω R2 3,5 V (c) (d) R3 (e) 1,5 A (f) R4 4V (g) (h) R5 (i) 1A (j) R6 (k) (l) 2Ω EXPERIMENT OHMIC AND NON-OHMIC CONDUCTORS Aim: To investigate the voltage-current relationship for a copper wire (resistor) and a diode (semiconductor) to determine which one obeys Ohm’s law. Materials • battery/DC power supply • resistance wire (resistor) • rheostat (variable resistor) • ammeter • voltmeter • connecting wires • diode/led (semiconductor) Precautions Do not leave the switch closed at all times. The battery will run flat. Close the switch only for taking the readings on the voltmeter and the ammeter. Take your readings accurately. Do not conduct the experiment with wet hands. 114   Technical Science | Grade 11 Science_GR11_LB.indb 114 2018/02/02 17:00

Method 1. Set up a circuit according to the circuit diagram shown below, connecting the copper wire (resistor) between points A and B. A V R 2. Close the switch and adjust the rheostat to give a low voltage across the unknown resistor. Measure the voltage across the resistor using the voltmeter, and the current in the circuit using the ammeter. 3. Close the switch and change the resistance in the rheostat and take the ammeter and voltmeter reading. 4. Repeat the readings for different values of the resistance of the rheostat and record your results in a table below. (Open the switch while recording the readings in order to spare the batteries’ life span.) Ammeter reading (A) Voltmeter reading (V) 1 2 3 4 5 5. Replace the copper wire (resistor) with the diode/LED (semiconductor). 6. Close the switch and adjust the rheostat to give a low voltage across the diode/LED (semiconductor). Measure the voltage across the diode/LED (semiconductor) using the voltmeter, and the current in the circuit using the ammeter. 7. Repeat the readings for different values of the resistance of the rheostat and record your results in a table below (Open the switch while recording the readings in order to spare the batteries’ life span). Trial No. Ammeter reading (A) Voltmeter reading (V) 1 2 3 4 5 8. Use your reading to plot a graph of potential difference against current for both the copper wire and the diode/led (semiconductor. Chapter 4 Electricity 115 Science_GR11_LB.indb 115 2018/02/02 17:00

9. Do the analysis and interpretation of the result. What is the shape of the graph for each the copper wire and the semiconductor? 10. Write down a conclusion for the experiment. a) Which one of these components obey Ohm’s law? b) Which factor must remain constant for a component to obey Ohm’s law? From the experiment we can deduce the following: • Resistance is defined by a simple relationship: resistance is the ratio of potential difference across a circuit component and the current passing through it. The actual resistance of materials does not depend on the potential difference across it, nor on the current through it. Resistance is an intrinsic property of material that only depends on the cross-sectional area and the length of the material. • Ohm’s law is only valid for metallic conductors if the temperature of the conductors remains constant. When metallic conductors are heated, their resistance increases. However, resistance (resistivity) decreases with increasing temperature in semi-conductors and insulators. • The correct causal relationship between potential difference and current in an electric circuit is that potential difference gives rise to current. Current cannot flow through a conductor if potential difference does not exist between the ends of the conductor. Ohm’s law is also valid for the colour-coded ceramic resistors used in electronic equipment. Conductors that obey Ohm’s Law are referred to as Ohmic conductors and those that don’t are referred to as non-ohmic conductors. Examples of ohmic conductors and non-ohmic conductors are given in Table 4.2. Table 4.2 Ohmic vs non-ohmic conductors Ohmic conductors Non-Ohmic conductors Copper Wire Acid battery Nichrome Transistor 116   Technical Science | Grade 11 Science_GR11_LB.indb 116 2018/02/02 17:00

Platinum Ionized gas tubes Carbon/graphite conductors Light Bulbs Light Emitting Diodes (LED) Activity 4.6: Ohmic and non-Ohmic conductors 1. Define the following terms a) Ohm’s law b) Ohmic conductor c) Potential difference d) Current e) Resistance 2. List two examples of ohmic and non-ohmic conductors Chapter 4 Electricity 117 Science_GR11_LB.indb 117 2018/02/02 17:00

4.4 Internal resistance and emf (electromotive force) To produce a steady flow of charge in an electric circuit we need a device to do work on the charge carriers. Such a device is called an emf-device. This device is said to provide an emf (ε). The emf- device has a positive terminal (at a higher potential) and a negative terminal at lower potential). The terminals are called electrodes. Until now we have only worked with ideal batteries. These batteries are not affected by the circuit or current in any way, and provide a precise voltage until they go flat. Thus, if you were to measure the potential difference across the terminals of a battery on its own, you will get a different value to what you measure when it is connected in a complete circuit. The value will be less when the battery is included in a complete circuit. The emf of an emf-device is the maximum electric potential difference (in volts) that exists between the terminals of the emf-device such as a battery. Thus we can say that: the emf of an emf-device is the work (energy transferred) per unit charge that the device does in moving charge from its low- potential terminal to its high-potential terminal. Note: Sometimes the difference is called the lost volts. Nothing has actually been lost but energy has been transferred. R Real batteries are made from materials that have resistance. This means that real batteries are not just sources of potential difference (voltage), but they also possess internal resistance. If the total potential difference source is referred to as the emf, ε, then a real battery can be represented as an emf ε connected in series with a resistor r. r Internal resistance is the resistance to flow of charges through a voltage source V such as a battery that arises from the resistance of the materials from which the source is made. In a real battery, the internal resistance is due to chemicals within the battery. In a generator, the internal resistance is the resistance of the wires and the other components within the generator. The internal resistance of the battery is represented by the symbol r. The circuit diagram alongside shows a cell with an emf ε and its ε internal resistance r, an external resistance R, an ammeter and a voltmeter: A In the circuit diagram, the reading of the ammeter will be: ε R I= R+r This is sometimes called Ohm’s Law for a whole circuit. V Now making ε the subject of the previous equation we have: ε = I(R + r) OR ε = IR + Ir IR = (Vload) [It is called terminal potential difference (potential difference V across the external circuit)] Ir = the drop of potential in the internal resistance (Vinternal resistance) IR = ε – Ir OR Vload = ε – Vint OR Vext = ε – Vint 118   Technical Science | Grade 11 Science_GR11_LB.indb 118 2018/02/02 17:00

EXPERIMENT Determine the internal resistance of a battery (informal) Aim: Determining the internal resistance of a battery Apparatus • 1 voltmeter • 1 ammeter • 1 rheostat • 1 switch • a battery (or cells • connecting wires Procedure: 1. Connect a circuit consisting of a rheostat, switch, ammeter and V battery connected in series as illustrated in the diagram. Ensure that, for the purpose of this experiment, you use new cells (or batteries). 2. Connect the voltmeter across the battery as illustrated. A 3. Observe the reading on the voltmeter and the ammeter when the switch is open. Record these values in the table. 4. Close the switch and observe the readings on the voltmeter and the ammeter. Record these values in the table. 5. Repeat steps 3 and 4 two or three times. 6. Calculate the average potential difference (V) and current (I) for both open and closed switch. SWITCH OPEN SWITCH CLOSED Trial # V (V) I (A) V (V) I (A) 1 2 3 Average 7. Calculate the average drop in potential difference (∆V = Vopen – Vclosed) 8. Calculate the increase in current (∆I = Iclosed – Iopen) 9. ∙ ∆l ∙ Calculate the internal resistance (r) of one cell r = ∆v 10. Write down a conclusion for the experiment. Chapter 4 Electricity 119 Science_GR11_LB.indb 119 2018/02/02 17:00

Discussion When the switch in a simple electric circuit is open, no current is flowing. If there is a voltmeter connected across the battery (or a cell), it will give a certain reading. The reading on the voltmeter connected across a battery when the circuit is open (that is when there is there is no current) flowing is called the emf. – 3 000 V 0.00 When the switch is closed, the reading on voltmeter across the battery changes (it decreases). This is due to the resistance to the current as it flows through the battery. The resistance inside a cell when there is current flowing through it is referred to as the internal resistance (denoted by the small letter r). The potential difference across the battery when the current is flowing in the circuit is called the terminal voltage. – X 2.941 V 0.294 A Screenshot from Electroplus English software: The voltmeter reading across the battery decreases when the switch is closed and there is current flowing in the circuit. 120   Technical Science | Grade 11 Science_GR11_LB.indb 120 2018/02/02 17:00

Review exercise Multiple choice questions Choose one correct answer from the options below: 1. Two small identical metal spheres, on insulated stands, carry charges –q and +3q respectively. When the centres of the spheres are separated by a distance d the one exerts an electrostatic force of magnitude F on the other. The spheres are now made to touch each other and are then brought back to the same distance d apart. –q +3q d What will be the magnitude of the electrostatic force which one sphere now exerts on the other? A. 1F 4 B. 1F 3 C. 1F 2 D. 3F 2. A positive charge Q is released from rest at the centre of a uniform electric field. positive plate +Q negative plate How does Q move immediately after it is released? A. It accelerates uniformly. B. It moves with an increasing acceleration. C. It moves with constant speed. B D. It remains at rest in its initial position. 3. The sketch below shows two sets of parallel A plates which are connected together. A potential difference of 200 V is applied 20 mm across both sets of plates. The distances 40 mm P 200 V between the two sets of plates are 20 mm and 40 mm respectively. When a charged particle Q is placed at point C R, it experiences a force of magnitude F. R D Chapter 4 Electricity 121 Science_GR11_LB.indb 121 2018/02/02 17:00

Q is now moved to point P, halfway between plates AB and CD. Q now experiences a force of magnitude: A. 1F 2 B. F C. 2F D. 4F Long questions 4. Two charges of +3nC and –5nC are separated by a distance of 40 cm. Calculate the magnitude of the electrostatic force that the point charges exert on each other. 5. Two insulated metal spheres carrying charges of +6nC and –10nC are separated by a distance of 20 mm. a) State Coulomb’s law in words. b) Draw the electric field diagram for the two charges c) What is the electrostatic force between the spheres? d) The two spheres are touched and then separated by a distance of 60 mm. What are the new charges on the spheres? e) What is new electrostatic force between the spheres at this distance? 6. Calculate the potential difference between two parallel plates if it takes 5 000 J of energy to move 2,5nC of charge between the plates. a) If the plates are 5 mm apart, what is the electric field intensity between the two plates? b) In this uniform field region, what will be the force experienced by an electron? 7. Calculate the electric field between the plates of a capacitor if the plates are 20 mm apart and the potential difference between the plates is 300 V. A 8. Consider the circuit below and answer the questions that follow. The battery used has the negligible resistance. R2 10Ω R1 25Ω a) Calculate the equivalent resistance of the four resistors. 24V b) What is the total current that flows in the circuit? 35Ω R3 R4 20Ω c) Calculate the potential difference across and the current that flows through (i) The 10Ω resistor (ii) The 35Ω resistor (iii) The 25Ω resistor (iv) The 20Ω resistor A1 9. In the circuit diagram below, the resistors R1 = 5Ω, R2 = 10Ω, R3 = 8Ω. The battery has a negligible resistance. When the switch 20V V1 R1 is closed, calculate the following: a) the total resistance of the circuit. R2 A2 b) the reading on the ammeter A1 V2 R3 V2 c) the reading on V1 A3 d) the reading on V2 122   Technical Science | Grade 11 Science_GR11_LB.indb 122 2018/02/02 17:00

e) the reading on A3 f) the reading on V3 g) the reading on A2 10. Rank each of the following bulbs in each combination below in terms of their increasing brightness strength (that is, from the dimmest to the brightest). Assume that all the bulbs are identical. A B C A a) c) B C A B A b) d) C B 10. State Ohm’s law in words. 11. Draw the graph that shows the relationship between the current in the circuit and the potential difference. 12. Explain the difference(s) between the Ohmic and the non-Ohmic conductors. 13. List two examples of non-Ohmic resistors 14. List the factors that affect the resistance of a substance 15. A group of learners conducted an experiment in which they wanted by adding to determine the unknown resistance of a resistor. To achieve this, they set up an experiment add cells in series, while recording the voltmeter and ammeter readings. Below is the table summarising the results they obtained: Number of Cells V (V) I (A) R = V(Ω) I 1 1,44 0,14 2 2,88 0,29 3 4,33 0,43 a) Complete the table below by calculating R for each number of cells. b) Calculate the average resistance. Chapter 4 Electricity 123 Science_GR11_LB.indb 123 2018/02/02 17:00

c) Plot the graph of V versus I. d) Calculate the slope of the graph, and use the slope the determine the resistance of the resistor. e) How does your answer to number b) compare to that of d) above? 16. Study the circuit diagram below, and answer the questions that follow. The battery has an emf of 12 V, and a negligible resistance. R2 = 6.8Ω R1= 2.2Ω R4 = 1Ω R3 = 4.7Ω 12V Do the appropriate calculations and complete the table below: (Hint: study the path of the conventional electric current) Quantity R1 R2 R3 R4 Total Voltage 12 V Current Resistance 2,2Ω 6,8Ω 4,7Ω 1Ω 17. Study the circuit diagram below, and answer the questions that follow. The battery has an emf of 9 V, and a negligible resistance. R1= 2.2Ω R2= 4.7Ω R4 = 1Ω R3 = 4.7Ω R3= 3.3Ω 9V Do the calculations, and complete the table below: Quantity R1 R2 R3 R4 R5 Total Voltage 9V Current Resistance 2,2Ω 4,7Ω 4,7Ω 1Ω 3,3Ω 18. Define the following terms: a) emf b) internal resistance 124   Technical Science | Grade 11 Science_GR11_LB.indb 124 2018/02/02 17:00

c) resistance d) potential difference e) electric current 19. In a series circuit, certain general rules may be stated with regard to quantities of voltage, current and resistance. Express these rules using your own words: a) In a series circuit, voltage… b) In a series circuit, current… c) In a series circuit, resistance… 20. In a parallel circuit, certain general rules may be stated with regard to quantities of voltage, current and resistance. Express these rules using your own words: a) In a parallel circuit, voltage… b) In a parallel circuit, current… c) In a parallel circuit, resistance… Chapter 4 Electricity 125 Science_GR11_LB.indb 125 2018/02/02 17:00

126   Technical Science | Grade 11 Science_GR11_LB.indb 126 2018/02/02 17:00

5 Heat and thermodynamics Objectives By end of this chapter learners will gain a broad understanding and acquire skills relating to the following concepts: • define the specific heat capacity (c) of a substance • define the heat capacity (C) of a substance • use the equation for specific heat capacity to perform simple calculations • discuss practical applications of heat capacity in technology. • state the law of conservation of energy • experimentally determine the heat capacity of a solid. • define terms used in the study of thermodynamics • list thermodynamic variables • define internal energy of a thermodynamic system. • state the first and second laws of thermodynamics • describe how a heat engine operates • describe how refrigerators operate Science_GR11_LB.indb 127 2018/02/02 17:00

Why heat and thermodynamics matter Most of the machines and appliances used in industry or in our homes, cars and even airplanes are driven by some kind of engine. The mechanism that drives most machines and engines can be explained through the laws of thermodynamics. Over the year scientists have found ways to harness the heat generated by different fuels and use it to the benefit of mankind. This chapter aims to introduce you to thermal physics and its applications in everyday life. 5.1 Introduction In matter and materials you learnt that all matter is made up of constantly vibrating molecular particles, a branch of science known as thermodynamics. The observable states of matter are due to the nature of vibrations experienced by molecules that make up matter. In solids molecules vibrate but not at a fast pace. In liquids the molecules vibrate at a faster pace which allows them to slide over each other resulting in the fluid nature of liquids. Sometimes the molecules move so fast that the intermolecular forces disconnect which Figure 5.1 A cup of hot coffee 128   Technical Science | Grade 11 Science_GR11_LB.indb 128 2018/02/02 17:00

allows them to break free resulting in the gaseous state. Therefore, the difference between a cup of hot tea and a cup of cool tea lies in the movement of molecules that make up the two substances as shown in Figure 5.1. The molecular particles in a cup of hot tea vibrate faster that the molecules in a cup of cool tea. When objects are in motion we say that they have kinetic energy and when they are still they are said to possess potential energy. Remember these concept as you will come across them again in the next sections. Therefore, we can say that the temperature of solids, liquids and gases depends on the average kinetic energy of their particles as shown in Figure 5.2. So temperature of a gas refers to how fast the gas particles bounce back and forth. For a liquid, temperature is Figure 5.2 Arrangement of molecules in solids, liquids and gases determined by how fast the liquid particles slide past one another and for a solid temperature refers to how fast the particles vibrate in place. In Grade 10 you were introduced to the concepts of temperature, heat and how it can be transferred from one object to another. So what is temperature? What is heat? In Grade 10 you learnt that: Temperature is a measure of how hot or cold an object is. Temperature is a state variable that quantifies the hotness or coldness of a system (object). Temperature is a measure of the average kinetic energy of molecular particles that make up a substance. However, as temperature is one of the seven SI quantities, temperature is measured on a Kelvin scale. A unit often used to measure temperature is degrees Celsius (˚C). But in this chapter we do not use degrees Celsius to report temperature. Instead we use Kelvins. For example, 270 K is read (270 Kelvins) Thermal energy is transferred in the direction from a hotter system (object) to a cooler system (object). Each system consisting of molecular particles has thermal energy. Internal energy (U) Internal energy (U) is the sum of various kind of energy that the atom or molecules of a substance possesses. This include translational and rotational kinetic energy, vibrational energy and potential energy, chemical, and so on. Chapter 5 Heat and thermodynamics 129 Science_GR11_LB.indb 129 2018/02/02 17:00

Thus, the internal energy (U) of a system (substance) is the sum of the kinetic energy (energy due to motion) and the potential energy (energy due to the interaction of particles) of all the particles constituting the system. The faster the particles move, the higher the kinetic energy. So the temperature of an object depends on the average kinetic energy (movement) of the particles that make up the substance and not the amount of substance. If you pour half of the water contained in a cup onto the floor the temperature of the remaining water does not change. This is because the particles remaining in the cup still have the same average kinetic energy. When you touch a hot stove we say that thermal energy is transferred from the stove to your hand because the stove is warmer. So heat is the thermal energy in transfer or the thermal energy that is transferred from one object to another due to a temperature difference, commonly measured in calories or joules. In other words: Heat is the energy transferred between a system and its environment because of a temperature difference that exist between them. OR Heat is the energy that flows from a higher temperature object to a lower temperature object because of the difference in temperatures. Before scientists realised that heat is transferred energy, heat was measured in terms of the ability to raise the temperature of water. Thus the calorie was defined as the amount of heat that would raise the temperature of 1 g of water from 14,5 ˚C to 15,5 ˚C. In 1948 the scientific community decided that since heat (like work) is transferred energy, the SI unit of heat is joule (J). Note: A substance (system) does not contain heat, it contains thermal energy. Heat is thermal energy in transit/transferred energy. Once thermal energy has been transferred it is no longer referred to as heat. 5.1.1 Units for measuring heat In science, heat is commonly measured in joules (J), but in America, energy is measured Nutrition facts Serving size 1 oz (28g About 15 chips) in Calories. Amount per serving Calories 160 Calories from Fat 90 % Daily Value Did you know? Total Fat 16% Saturated Fat 1.5g 8% 1 calorie = 4,19 J Trans Fat 0g Cholesterol 0mg 0% Food labels usually show the energy rating Sodium 170mg 7% Potassium 350mg 10% of food in joules or calories because food is Total carbohydrate 15g 5% considered as fuel burnt by our bodies to Dietary Fiber 1g Sugars less than 1g release energy. For example, this packet of Protein 2g chips can give you 670 joules or 160 calories Vitamin A 0% Vitamin C 10% Calcium 0% Iron 2% of energy when digested. Vitamin E 6% Thiamin 4% Niacin 6% Viamin B Figure 5.4 Nutrition facts on the back of a packet of potato chips 130   Technical Science | Grade 11 Science_GR11_LB.indb 130 2018/02/02 17:00

5.2 Specific heat capacity After warming a pie in the microwave oven, the filling could be too hot to eat even though the crust is not. This is because different substances have different thermal capacities for storing energy. Equal amounts of different materials require different quantities of heat to change their temperatures by a specific number of degrees. One gram of water requires one calorie (4,19 J) of energy to raise its temperature by 1 ˚C. However, it takes less energy to raise the temperature of one gram of iron by 1 ˚C. As a result, we can say water has a higher specific heat capacity than iron. Specific heat capacity is denoted by the small letter ‘c ’. It can also be regarded as the resistance of a substance to a change in temperature or resistance to releasing thermal energy. We saw in the previous example that it takes more energy input for one gram of water to increase its initial temperature by 1 ˚C compared to the same mass of iron. 5.3 Heat capacity Heat capacity Denoted by the capital letter ‘C ’, refers to the amount of heat required to raise the temperature of the whole substance by 1 K or 1 ˚C. In addition, the heat capacity (C) of an object is the proportionality constant between the heat (Q) that the object absorbs or loses and the resulting temperature change (∆T). Q = C∆T OR Q = C(Tf – Ti) Heat capacity C has the unit of energy per degree (J∙˚C–1) or degree per kelvin (J∙K–1) The energy needed to produce a certain temperature rise depends on the mass of the substance and what substance is being heated. Knowing the specific heat capacity of substances enables us to calculate the amount of energy required to raise their temperatures to specific amounts. Table 5.1 shows a list of specific heat capacities of some common substances. Table 5.1 specific heat capacities of some common substances Substance Specific heat capacity (J∙kg∙K–1) Water 4 200 Ethanol 2 500 Ice 2 100 Aluminium 900 Concrete 800 Steel 500 Copper 380 Mercury 150 Chapter 5 Heat and thermodynamics 131 Science_GR11_LB.indb 131 2018/02/02 17:00

Table 5.1 shows that the specific heat capacity of water is 4 200 J/kg∙K (4 200 J∙kg–1∙K–1). This means: to raise the temperature of 1 kg of water by 1 K, you need 4 200 J of energy Did you know? Water has a higher specific heat capacity for storing thermal energy than almost any other substance. This means that it takes longer to warm up and longer to cool down. This property allows it to preserve aquatic life during cold winter and hot summer days. Knowing the specific heat capacity of water means that we can calculate how much energy is required for any mass of water and any temperature rise. For example: To raise the temperature of 1 kg of water by 5 K you need (5 × 4 200 J) 21 000 J of heat To raise the temperature of 3 kg of water by 1 K you need (3 × 4 200 J) 12 600 J of heat Two objects of the same material will have heat capacities proportional to their masses. It is therefore convenient to define heat capacity per unit mass: C = c hence C = cm where m is the mass of the substance and c is the specific heat capacity. m A general equation deduced from this reasoning can be written in words as follows: Energy transferred = specific heat capacity × mass × temperature change. Or in symbols as follows: Q = cm∆T Where: Q is the heat c is the specific heat capacity of the substance m is the mass of the substance in kg ∆T (Tfinal – Tinitial) is the corresponding change in temperature of the substance in Kelvin (K) Let us use the above equation to perform a simple calculation. DETERMINE THE AMOUNT OF ENERGY REQUIRED TO INCREASE Example 5.1 THE TEMPERATURE OF A SUBSTANCE Jabu uses an aluminium saucepan to boil 0,5 kg of water. The saucepan has a mass of 0,4 kg. The saucepan and the water start at a temperature of 20 ˚C. How much heat is needed to: (a) boil the water, (b) increase the temperature of the saucepan to 100 ˚C? 1 ANALYSE Given: msaucepan = 0,4 kg mwater = 0,5 kg 132   Technical Science | Grade 11 Science_GR11_LB.indb 132 2018/02/02 17:00

Tinitial = 20 ˚C The specific heat capacities of water and Aluminum are given in Table 1. Unknown: Qwater = ? Qsaucepan = ? 2 SOLVE For both water and the saucepan, the temperature change is from 20 ˚C to 100 ˚C (at standard atmospheric pressure). So ∆T = Tfinal – Tinitial = 100 °C – 20 ˚C = 80 ˚C or 80 K We use the equation: Q = cm∆T For the water: Q = 4 200 J × 0,5 kg × 80 K kg,K = 168 000 J For the saucepan: Q = 900 J × 0,4 kg × 80 K kg,K = 28 800 J 168 000 J of heat is needed to boil the water and 28 000 J of heat is needed to increase the temperature of the saucepan to 100 °C. This is assuming that the whole saucepan reaches 100 °C, which may not be quite true in reality. Activity 5.1 Calculations for the amount of heat Show that the quantity of heat needed to raise the temperature of a 10 kg piece of steel from 0 ˚C to 100 ˚C is 450 000 J. The specific heat capacity of steel is 450 J∙kg–1∙˚C–1 or J∙kg–1∙K–1). 5.3.1 Practical applications of heat capacity in technology When the temperature of a substance rises, its molecules vibrate faster and move farther apart. The increase in the distance between molecules results in thermal expansion. The opposite is true when the temperature drops. Most substances expand when heated and contract when cooled. Even though this happens at the molecular level these changes are sometimes noticeable. Telephone wires are longer and they droop on a hot day compared to a cold day as shown in Figure 5.6. Figure 5.6 Sagging telephone lines on a hot summer’s day Chapter 5 Heat and thermodynamics 133 Science_GR11_LB.indb 133 2018/02/02 17:00

Metal lids on glass bottles can often be loosened by holding the lid under a hot tap water as shown in Figure 5.7. The hot water makes both the glass and the metal lid expand but because different materials expand with different amounts of heat, the metal lid expands more than the glass bottle and also expands much faster because of direct exposure. Figure 5.7 Opening a jar using hot water Technicians and Engineers must take thermal expansion into account when building structures and devices of all kinds. Engineers use reinforcing steel with the same expansion rate as concrete so that both materials can expand and contract the same under different temperature conditions as shown in Figure 5.8. Using reinforcement with a lower heat capacity than concrete may result in cracks in the concrete over time. Figure 5.8 Concrete railway line of the Gautrain Have you noticed that it gets bumpy while driving over a bridge? The bumps and gaps on the bridge are called expansion joints, they allow the bridge to expand and contract as shown in Figure 5.9. Even concrete roads and sidewalks have to be intersected with gaps which are sometimes filled with tar, to allow for expansion in summer and contraction in winter. Figure 5.9 Expansion joints on a bridge 134   Technical Science | Grade 11 Science_GR11_LB.indb 134 2018/02/02 17:00

The fact that different substances expand and contract at different temperatures is taken advantage of in the making of bimetallic strips. A bimetallic strip is made out of two strips of different metals, welded together, one made out of brass and the other of iron. When heated the brass part of the strip expands more and bends the strip as shown in Figure 5.10. The resulting bend may be used to a turn a pointer, regulate a valve or close a switch in an electric circuit. This concept is used in thermostats. contact rivet wire wire bimetallic strip bimetallic base strip Figure 5.10 Bimetallic strip temperature A perfect example is the thermostat inside control an electric iron shown in Figure 5.11. When it gets too cold, the bimetallic strip in a thermostat bends towards the brass side and activates an electrical switch and turns the iron on. When the electric iron gets too warm, the coil bends towards the iron metal which breaks the electrical circuit and turns the electrical iron off. bimetallic strip fixed Thermal expansion of liquids is also the reason contact why a petrol tank should not be filled to the brim Figure 5.11 Bimetallic strip in a thermostat of an on a hot summer’s day. electric iron 5.4 Law of conservation of energy We already know that heat is the energy transferred between a system and the environment as a consequence of a temperature difference between them. Q lost Q gained hotter system colder system On hot days people frequently put ice cubes into a hot soft drink to cool it quickly. Figure 5.12 Two systems interacting thermally Figure 5.12 shows two systems interacting thermally, that is, heat is transferred from the hotter system to the colder system. Energy cannot be created or destroyed, energy conservation indicates that energy leaving the hotter system must enter the colder system. That is the system exchange energy with no net loss or gain (Qnet = 0 J) Chapter 5 Heat and thermodynamics 135 Science_GR11_LB.indb 135 2018/02/02 17:00

To express this idea mathematically, we must take into consideration the signs of heat gained and heat lost by the hotter system. If we consider Qgained as positive and Qlost as negative, then the law of conservation of energy can be expressed mathematically as follows in terms of heat: Qgained = –Qlost When there is no exchange of energy with the environment (isolated system), then Qnet = Qgained + Qlost = 0 If there are more than two systems, then systems Q 1 + Q 2 + Q 3 + … = 0 The same equation (Q = cm∆T) discussed in section 5.3 above applies to both heating and cooling. As a substance cools, its temperature falls and it loses thermal energy. The energy released can be represented as follows: Energy released = specific heat capacity × mass × temperature drop Therefore: (c1m2∆T2)gained = (c2m2∆T2)lost Where ∆T is always Tfinal – Tinitial Law of conservation of energy (thermal energy) The amount of thermal energy lost equals the amount of thermal energy gained, when no thermal energy is transferred to the environment. OR The total thermal energy of an isolated system is constant. In symbols, the law of conservation of thermal energy is represented as follows: Thermal energy gained = Thermal energy lost ETH gained = ETH lost In terms of heat we can write: Qgained = –Qlost Let's use the above equation to perform a simple calculation. Example 5.2 DETERMINE THE HEAT GAINED Lebogang mixes 100 g of 25 ˚C water with 75 g of 40 ˚C water. Show that the final temperature of the mixture is 31.4 ˚C. 1 ANALYSE Given: Beaker one: Tintitial = 25 ˚C, m = 100 g 136   Technical Science | Grade 11 Science_GR11_LB.indb 136 2018/02/02 17:00

Beaker two: Tintitial = 40 ˚C, m = 75 g Prove: Tfinal = 31,4 ˚C 2 SOLVE Because we have two different masses of water mixed together we must equate the thermal energy gained by the cool water to the thermal energy lost by the warm water. Qnet = 0 Q1 + Q2 = 0 Q1 = –Q2 Thermal energy gained by cool water = Thermal energy lost by warm water: cm1∆T = –(cm2∆T2) Note: Thermal energy transfers from the warm water to the cool water. Because the cool water will absorb thermal energy, Tfinal will be greater than 25 ˚C so: ∆T1 = Tfinal – ∆Tinitial = ∆Tfinal – 25 °C Because the warm water will release thermal energy, Tfinal will be less than 40 ˚C so: ∆T2 = Tfinal – ∆Tinitial = ∆Tfinal – 40 °C As the system of units is homogenous, there is no need for unit conversion. c(100 g)(Tfinal – 25 °C) = c[(75 g))(Tfinal – 40 °C)] 100 gTfinal – 2 500 °C = –75 gTfinal = (3 000 + 2 500)g °C 175 gTfinal + 75 gTfinal = (3 000 + 2 500)g °C Tfinal = 5 500 g °C Tfinal = 5 500g °C 175 g Tfinal = 31,4 g °C In the workshop, you submerge nails of a mass of 100 g at a temperature of 30 ˚C in 200 g of water at a temperature of 20 ˚C Equate the heat gained by the water to the heat lost by the nails and show that the final temperature of the water is about 23,3 ˚C. 5.4.1 Calorimetry But how are the values for specific heat capacity of different substances determined? A device called a calorimeter is used in the laboratory to measure specific heat capacities of different substances. A calorimeter is a well-insulated container used to measure changes in thermal energy or heat transfer. To reduce the loss or gain of energy to or from the surroundings, a calorimeter should be well-insulated. Figure 5.13 shows a calorimeter used in the laboratory. Energy always flows from an object with a higher temperature to an object with a lower temperature. The thermal energy gained by the Figure 5.13 Calorimeter used in the laboratory Chapter 5 Heat and thermodynamics 137 Science_GR11_LB.indb 137 2018/02/02 17:00

cooler substance equals the thermal energy lost by the warmer substance, if we assume no loss of thermal energy to the surrounding environment. Thermal energy lost = Thermal energy results A calorimeter operates on the principle of the law of conservation of energy. This is demonstrated in the practical activity that follows. EXPERIMENT 13 In this experiment, you will determine the specific heat of a metal sample more specifically Lead (Pb). The metal sample is heated to a high temperature and then placed into a calorimeter containing a known quantity of water at a lower temperature. Having measured the mass of the water in the calorimeter, the temperature changes of the water (ΔT) and knowing the specific heat of water (4,184 J·g –1·°C –1) the heat gained by the water (heat lost by the metal) can be calculated as follows: Heat gained by the water = mass of the water (g) × change in temperature (∆T) × specific heat capacity of water (4,184 J·g –1·°C –1) The specific heat of the metal will be calculated as follows: Heat lost by the metal (heat gained by the water) = specific heat of the metal × mass of the metal (g) × change in temperature of the metal (˚C) heat gained by the water Specific heat of the metal = mass of the metal × ∆T of the metal Apparatus For a makeshift calorimeter you will need: • two polystyrene cups with a lid thermometer • thermometer 50 °C • glass rod • retort stand • clamp for the thermometer For experiment you will need: • makeshift calorimeter (Figure 5.14) cardboard • large beaker: heat resistant cover • bunsen burner, stand and wire gauze nested • small beaker styrofoam cups • digital scale • test tube (heat resistant) • retort stand with clamp • thermometer Figure 5.14 Makeshift calorimeter 138   Technical Science | Grade 11 Science_GR11_LB.indb 138 2018/02/02 17:00

Method 1. Fill the large beaker approximately half full of water. Place the beaker of water on the Bunsen burner with wire gauze. Heat the water to boiling point. Adjust the Bunsen burner so that the hottest part of the flame touches the bottom of the beaker through the gauze. 2. Weigh the small beaker (mass 1) and record its mass to two decimal places. Add a piece of metal to the small beaker. Weigh the beaker with the metal in it (mass 2) and subtract to find the mass of the metal (mass 2 – mass 1). 3. Weigh the empty makeshift calorimeter. Record the mass carefully to two decimal places. 4. Fill the calorimeter approximately half full with distilled water at room temperature and record the mass to two decimal places. 5. Measure the temperature of the water in the calorimeter cup and record to two decimal places. 6. Carefully transfer the metal into a large test tube. Clamp the test tube, suspending it in the beaker of water (from step 1). Boil at least for 10 minutes. 7. While the metal is still in the boiling water bath, measure the temperature of the water carefully with a thermometer and record to one decimal place. (It is assumed that the temperature of the metal is the same as that of the boiling water.) 8. After the metal has been heating 10 minutes, remove the test tube from the bath by its utility clamp. Immediately pour the metal into the calorimeter cup so that the metal is covered by the water. Cover the calorimeter with its cover. Turn off the Bunsen burner. 9. Carefully insert the thermometer through the small hole in the calorimeter lid. Stir very slowly by shaking, and record the highest temperature reached by the water. 10. Remove the calorimeter lid, retrieve and dry the metal. Return metal to its original place. Pour the water down the sink. Rinse and dry the calorimeter cup Use the data obtained from the experiment to calculate the specific heat of the metal. Show all units in your calculations. To determine the heat gained by water: use the first equation in the introduction To determine the experimental specific heat of the metal: use the third equation in the introduction Follow-up questions 1. Why is water an excellent material to use in the calorimeter? 2. What substance lost heat in this experiment? 3. What substance gained heat? 4. Metals have a lower specific heat than that of water. Was this true in what you observed? 5. Explain what a ‘low specific heat’ means in terms of heat retention and heat requirements. Chapter 5 Heat and thermodynamics 139 Science_GR11_LB.indb 139 2018/02/02 17:00

5.5 Thermodynamics ‘Thermo’ means heat and matter ‘dynamics’ means change. (e.g. vapour) Thermodynamics refers to the study of processes through which thermal energy is transferred. The first law of thermodynamics is the law of conservation of energy adapted to thermodynamic systems. Thermodynamics is one of the energy energy principle branches of Physics and engineering, which is the study and energy energy application of thermal energy of a system. open closed isolated Figure 5.15 Three types of systems 5.5.1 Thermodynamics system A thermodynamic system is the part of the world constituted by a significantly large number of particles (i.e., atoms and molecules). A thermodynamic system is defined as a portion of matter for example, gas enclosed inside a cylinder fitted with a piston. Figure 5.15 shows three types of systems, namely, isolated, closed and open systems. Figures 5.15 also illustrates the difference between the three systems. In the study of thermodynamics, the term ‘system’ is often used to refer to a gas enclosed in some form of container. The term ‘surrounding’ is used to refer to anything outside the system which has some bearing on the behaviour of the system. 5.5.2 Thermodynamic variables The thermal state of a thermodynamic system is defined by its temperature (T), pressure (P) and volume (V). These quantities are called thermodynamic variables. Heat (Q) flowing in or out of the system can have two effects on the system: 1. It can increase or decrease the internal thermal energy of a system (∆U) 2. It can enable the system to do work (W) on its surroundings This statement sums up the first law of thermodynamics. When a system changes from a given initial state to a given final state, both work (W) and heat (Q) depend on the nature of the process. BUT experimentally it has been proven that the quantity (Q – Wsystem) is the same for all processes, it depends only on the initial and final states. The quantity Q – Wsystem represents a change on an intrinsic property of the system called internal energy. 140   Technical Science | Grade 11 Science_GR11_LB.indb 140 2018/02/02 17:00

5.5.3 Internal energy Internal energy is the sum of various kinds of energy that the atom or molecules of a substance possess. This include translational and rotational kinetic energy, vibrational energy and potential energy, chemical energy, and so on. There are thermodynamic systems in which the thermal energy is considered as the internal energy because only the kinetic and potential energy of the particles are taken into consideration. The internal thermal energy of a thermodynamic system is equal to the sum of the kinetic and potential energies of molecules within the system. 5.6 First law of thermodynamics The change in the internal energy of a closed system is equal to the heat supplied to the system, minus the work done by the system: (∆U = Q – W) This equation can be written as Q = ∆U + W W is the work done by the system The first law of thermodynamics can also be stated as follows: The heat given to a system is used to increase the internal energy of the system and to do work against the external pressure. Alternatively: Heat added = heat used to increase the internal thermal energy of the system + heat used by system to do work on its surroundings. ∆Q = ∆U + ∆W When the system does work the volume increases (expansion) and work done by the system is negative. From the equation we can conclude that the internal energy of the system tends to increase if energy is added as heat and tends to decrease if energy is lost as work. piston Work done on the system is always equal but with the opposite sign to the work done by the system, so we can write the equation of the first law of thermodynamics in terms of the work done on the system as: ∆U = Q + W air This tells us that the internal energy of the system tends to increase if heat is absorbed /added or if positive work is done on the system. The first law of thermodynamics can be observed when you apply pressure on the air inside a sealed syringe a shown alongside. Chapter 5 Heat and thermodynamics 141 Science_GR11_LB.indb 141 2018/02/02 17:00

When you push down on the piston Work done on the system Work done by the system you decrease the space occupied by the air particles, that is, you do work on the air particles inside the syringe as shown in Figure 5.16. This energy flowing into this system increases the internal thermal energy of the air molecules and enables the molecules to vibrate faster exerting pressure on the sides of the syringe. Figure 5.16 When the work is Figure 5.17 When the work is Some of the energy entering the done on the system the volume done by the system the volume system is used to increase the decreases (compression) and increases (expansion) and work internal energy of the air molecules work done is positive. done is negative. and some is used to enable the air molecules to do work on its surrounding which is the container. w This is why the piston is automatically pushed back to its w w original position when you stop applying pressure as shown in Figure 5.17. Note: Most of energy that is added onto the closed system will go into doing work than increasing the internal thermal energy of the air molecules. When the internal energy of the molecules decreases the temperature decreases. That is why the sides of the syringe will not feel warm. The temperature of enclosed air is lower when it does work than when work Figure 5.18 When the work is done done on it. To demonstrate this point: Blow air on the back of by the system the volume increases your hand with your mouth wide open, your breath will feel (expansion) and work done is negative. warm. If you press your lips together and blow to get a fast stream of air, now your breath feels cold. Another example: We are often warned not to throw aerosol cans into a fire or leave them in direct sunlight or exposed to direct heat even when they are empty. This is because the can is a closed system. Heating it will add thermal energy to the system. Some of the heat will go into increasing the internal energy of the molecules causing them to vibrate faster and vaporise. The molecules will use some of the thermal energy to do work by applying pressure on the sides of the canister causing it to expand and eventually explode. An aerosol can is shown in Figure 5.19 Visit the website below to view an illustration of what happens in the gas molecules in an aerosol can when it is heated: http://resources.schoolscience.co.uk/BAMA/14-16/aerosch4pg2.html Figure 5.19 Example of aerosol can 142   Technical Science | Grade 11 Science_GR11_LB.indb 142 2018/02/02 17:00

USE THE FIRST LAW OF THERMODYNAMICS EQUATION TO Example 5.3 CALCULATE INTERNAL ENERGY, WORK DONE AND THE AMOUNT OF HEAT SUPPLIED A gas enclosed in a container is heated which causes the lid of the container to rise. The enclosed gas performs 15 Joules of work to raise the lid, leaving the final energy of the system as 3 Joules. Calculate how much energy was added to the system through heating. 1 ANALYSE Given: W = 15 Joules, ∆U = 3 Joules Unknown: Q=? 2 SOLVE Use the first law of thermodynamics equation to calculate the heat added to the system. Q = ∆U + W Q = (3 + 15) Q = 18 J OR 3 = Q –15 Q = 18 J Of the 18 Joules of heat energy added to the system, 15 Joules of energy was used to perform work on the lid of the container and the rest (3 Joules) of the energy was used to increase the internal thermal energy of the system. (Note: this is if we assume there is no exchange of heat with the surroundings, that all the heat added is used to do work and increase the internal energy of the system.) Activity 5.3 Water enclosed in a cylinder with a piston is heated which causes the piston to move. The steam performs 20 Joules of work to move the piston, leaving the final energy of the system as 5 Joules. Calculate how much energy was added to the system through heating. Chapter 5 Heat and thermodynamics 143 Science_GR11_LB.indb 143 2018/02/02 17:00

5.5.4 Heat engine You may have probably seen a steam engine (train) in the movies or on an excursion as depicted in Figure 5.20. This was the type of train used in the past when there was no electricity or alternative fuels to coal. Engines have been used to power machines for many years. From coal-powered engines, petrol and diesel engines to jet engines in airplanes. The basic operation of an engine is based on the transfer of heat energy between high and low temperatures. So what is an engine? Engines that drive cars or industrial machines are examples of what scientists call heat engines. Heat engines transform heat put into the system into work consistent with the laws of thermodynamics. Figure 5.20 Steam engine (train running on steam powered engine) A heat engine is a device designed to convert disordered energy into ordered energy. The net work done by an engine during one cycle is equal to the Heat flow into the engine. net heat that flow into the engine during the cycle (ΔU = 0). ∙Wnet∙ = ∙Qnet∙ Heat engine. Figure 5.21 illustrates how a heat engine works. Work done by the engine. The basic mechanics of most heat engines follow the steps below. Figure 5.22 shows a diagrammatical representation of basic mechanisms for most heat engines. Heat flow out of the engine. An energy-rich fuel (coal, petrol, diesel, and so on) Figure 5.21 How heat engine works burns and supplies heat energy to the engine (heat is transferred from a high temperature burning source to the gas at low temperature in the engine) The gas uses some of the heat energy to do work for example, push a piston, turn a wheel, and drive a machine. (Note: The gas or air in the engine is referred to as the working substance since it is the substance that absorbs the heat energy from the source and uses it to do work on the surrounding) The rest of the input heat energy is transferred to some source at a lower temperature to bring the engine to its initial temperature (heat sink). 144   Technical Science | Grade 11 Science_GR11_LB.indb 144 2018/02/02 17:00

input heat output heat heat heat source engine heat sink energy energy (hot) (cold) T2 steam Expanding steam pushes piston. T1 water work output (moving wheel) heat input Figure 5.22 Mechanism of a heat engine 5.7 Efficiency of a heat engine An efficient heat engine will be the one that absorbs heat from a hot body (source), converts a part of it into work and rejects the rest to a cold body (heat sink) with minimal energy lost to the environment. The efficiency of a heat engine is defined as the ratio of the heat converted to work (W) to the total heat absorbed (Q1). Heat converted to work W Efficiency = Total heat absorbed = Q The equation for efficiency of a heat engine can be written as follows if the engine does not lose any heat to the environment (which all engines do): Q1 – Q2 Efficiency = Q1 Commonly, efficiency is expressed as a percentage (%) then the equation becomes: Q1 – Q2 Efficiency = Q1 × 100% As a result, there is no machine with an efficiency of 100%. It is impossible to convert all the heat into work. Note: It is impossible to get a continuous supply of work from a body by cooling it to a temperature lower than the lowest of its surroundings. According to the law of conservation of energy a perpetual motion machine cannot exist. No system can deliver a continuous supply of work without an external supply of energy. Chapter 5 Heat and thermodynamics 145 Science_GR11_LB.indb 145 2018/02/02 17:00

5.8 Second law of thermodynamics Thermal energy will flow spontaneously from a region of high temperature to a region of low temperature. In winter, heat flows from inside a warm house towards the cold air outside, which is why we try as much as possible to insulate our houses. The second law of thermodynamics states: Heat never spontaneously flows from a substance with low temperature to a substance with high temperature. External work needs to be done on the system to force heat energy to flow from cold to hot. That is why your refrigerator at home will only work once you have plugged it in. 5.9 Refrigerators The working substance in a heat engine is air, but in a refrigerator the substance that does the work is a coolant, which is usually a nontoxic gas CCl2F2 (Freon-12). The cooling effect of a refrigerator works on the principles of condensation and vaporisation of the coolant. An electrically powered compressor does work on the Freon gas. Compressing the gas molecules results in a decrease in volume and an increase in pressure and temperature (pV = nRT). The hot gas enters the coil at the back of the refrigerator. As the hot gas travels up the coil, heat is transferred from the hot gas inside the coil to the cool air in the room outside the fridge. The loss in thermal energy cools down the Freon gas which causes it to slowly condense into a liquid with low temperature. Prior to entering the inside of the refrigerator, the liquid Freon passes through an expansion valve which increases the volume and decreases expansion the pressure and temperature of the liquid even valve further. heat liquid evaporator The low temperature liquid leaves the expansion valve and enters into the coil insulated inside the refrigerator. radiator fins heat The temperature difference between the liquid Freon in the coil and the air inside the refrigerator causes heat (thermal energy) to be condenser low-pressure transferred from the warmer air inside the fridge gas to the cooler liquid Freon in the coil. In this way heat is absorbed from inside the refrigerator by the liquid Freon in the coil, reducing the temperature inside the refrigerator. compressor pump high-pressure gas The heat absorbed by the liquid Freon causes it to vaporise into the gaseous state. Figure 5.23 Mechanism of the engine in a refrigerator 146   Technical Science | Grade 11 Science_GR11_LB.indb 146 2018/02/02 17:00

The cycle starts all over again once all the liquid Freon changes to gas. Mechanism of the engine in a refrigerator is illustrated in Figure 5.23. Note: The cycle described above does not run continuously. A thermostat regulates the process. The thermostat starts the compressor when the temperature inside the refrigerator rises above the set temperature. When the refrigerator cools below the set temperature, the thermostat turns off the compressor. This control mechanism allows the refrigerator to conserve electricity so that the process only takes place as much as is necessary to keep the refrigerator at the set temperature. A refrigerator is the reverse of a heat engine. The working substance (coolant) absorbs heat from a cold body (freezer), with the help of an external agency (electrically-powered compressor), and rejects it to the hot body (atmosphere) as shown in Figure 5.25. Figure 5.24 A refrigerator output heat input heat hot body cold body) heat (atmosphere) engine (fridge freezer) energy energy compressor Figure 5.25 Mechanism of a refrigerator (heat engine in reverse) You can visit http://home.howstuffworks.com/refrigerator.htm to get an idea how a refrigerator works 5.9.1 Evaporative cooling Long before refrigeration, evaporative cooling was used to keep foods cool. Sweating is our body’s way of cooling itself down when it is overheated. This process works because as sweat evaporates, water molecules are converted from liquid to gas. Evaporation is always fueled by kinetic (heat) energy, and as more and more energy is expended, cooling will occur naturally and inevitably. This principle of heat loss is the secret behind evaporative coolers. Evaporative coolers are able to cool any food or other objects placed inside of them by gradually draining away their heat until only a fraction of it remains. Refrigerators function in the same way, of course, but with evaporative coolers, no external electrical input is required. Figure 5.26 A Zeer Pot A Zeer Pot is an invention which works on the principle of evaporative cooling, which was created by a Nigerian pot-maker by the name of Mohammed Bah Abba who wanted to help Sudanese families to preserve their food. His invention won him the Rolex Award in 2000 and the World Shell Award for Sustainable Development in 2001. Chapter 5 Heat and thermodynamics 147 Science_GR11_LB.indb 147 2018/02/02 17:00

Two different sized (small and large) unglazed Terra Cotta pots (clay pots) are placed one inside the other. A space is left between the pots and filled with wet sand. A wet cloth is used to cover the pots. Food that needs to be kept cool is placed inside the smaller pot. As the water in the wet sand and wet cloth evaporates, it pulls the warm air outward, causing the inside of the pot to be cooler than the outside air. For detailed instructions on how you can make your own Zeer pot watch this youtube video: http://www.offthegridnews.com/grid-threats/how-to-keep-your-food-cold-without-a-refrigerator/ Research project Long before refrigeration was introduced what methods did our ancestors in Africa use to preserve foods such as fruit, meat and milk? Find out and present your findings in class. Chapter summary Temperature is a measure of the average kinetic energy of molecular particles that make up a substance. Thermal energy is the total energy (kinetic and potential) of molecular particles that make up a substance possess. Heat is thermal energy in transit. Heat energy transfers in a direction from a warmer object to the cooler object. Specific heat capacity is the amount of heat required to change the temperature of 1 kilogram of substance by 1 ˚C or 1 K (SI unit: J·kg-1·K-1 or J·kg-1·˚C-1). Heat capacity is the amount of heat required to raise the temperature of the whole substance by 1 K or 1 ˚C. Law of conservation of energy (thermal energy) states that the amount of thermal energy lost equals the amount of thermal energy gained, when no thermal energy is transferred to the environment. A calorimeter is a device used in the laboratory to experimentally determine the values of specific heat capacity for different substances. The first law of thermodynamics states that the change in the internal energy of a closed system is equal to the heat supplied to the system, minus the work done by the system (∆U = Q – W). A heat engine is a device that converts heat into mechanical work. The efficiency of a heat engine is defined as the ratio of the heat converted to work to the total heat absorbed. The second law of thermodynamics states that heat never spontaneously flows from a substance with low temperature to a substance with high temperature. Refrigerators are heat engines in reverse. 148   Technical Science | Grade 11 Science_GR11_LB.indb 148 2018/02/02 17:00

Review exercise Multiple choice questions 1. The specific heat capacity of aluminium is more than twice that of copper. If equal quantities of thermal energy are given to equal masses of aluminium and copper, the metal that rapidly increases in temperature is: a) aluminium b) copper c) Both will increase at the same rate d) none of these 2. A bimetallic strip used in thermostats relies on the fact that different metals have different: a) specific heat capacities b) thermal energies at different temperatures c) rates of thermal expansion d) all of the above 3. A gas in a closed container is heated with 10 Joules of energy, causing the lid to rise 2 metres with 3 N force. What is the total change in energy of the system? a) 6 Joules b) 10 Joules c) 4 Joules d) 4 Joules Long questions 4. How does heat differ from thermal energy? 5. In which direction does thermal energy flow between hot and cold objects? 6. How does the specific heat capacity of water compare with the specific heat capacities of other common materials? 7. A hot water bottle is filled with 0.75 g of water at 80 ˚C. By morning, it has cooled down to 30 ˚C. How much energy has it given out? 8. A 100 g block of aluminum at 100.0 ˚C is placed in 100 g of water at 10.0 ˚C. The final temperature of the mixture is 25.0 ˚C. What is the specific heat of the aluminum as determined by the experiment? 9. In terms of thermal expansion, why is it important that a lock and its key be made of the same or similar materials? 10. In a certain process, 600 J of work is done on the system which gives off 250 J of heat. What is the change in internal energy for the process? 11. Heat is converted into mechanical energy in many different ways from steam engines to gas engines. Is this a violation of the second law of thermodynamics? Please explain. 12. On a hot day your air conditioner goes out of service so you decide to cool your room by opening the door of your refrigerator. Use the second law of thermodynamics to explain why this would not work. Devise a plan that might help make it work. Chapter 5 Heat and thermodynamics 149 Science_GR11_LB.indb 149 2018/02/02 17:00

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6 Chemical change Objectives By the end of this chapter you will gain a broad understanding and acquire skills relating to the following concepts: • explain and describe concepts such as reduction, oxidation, a reducing agent, an oxidizing agent • assign oxidation numbers to various compounds and ions • explain and apply the concept of oxidation numbers in redox reactions • describe the electrolysis process and its applications Science_GR11_LB.indb 151 2018/02/02 17:00

Why chemical change matters Oxidation-reduction reactions form the basis of many applications of chemistry in industry and in our daily lives. Some important applications include: • extraction of metals • electrochemical cells or batteries • photosynthesis • the supply of energy • production of chemicals • quantitative analysis. A chemical property relates to a substance’s ability to transform into different substances. Physical properties can be observed without changing the identity of a substance, but properties of chemical change are not easily observable. Chemical properties are easiest to see when substances react to form new substances. One example of an observable property is the ability of (charcoal) carbon to burn in air. When charcoal burns, it combines with oxygen to become carbon dioxide gas. After the chemical reaction occurs, the amounts of the original substances – that is the carbon and oxygen – are not less than before, but they have simply recombined to form different substances with different properties. A change in which one or more substances are converted into different substances is called chemical change or a chemical reaction. The substances that react in a chemical change are called the reactants. The substances that are formed by chemical change are called the products. In the case of burning charcoal, carbon and oxygen are the reactants. Carbon dioxide gas and ashes are products that are clearly different from reactants. Figure 6.1 depicts a chemical reaction of carbon and oxygen to form carbon dioxide gas. symbolic C(s) + 02 (g) CO2(g) Model + Carbon Oxygen Carbon Dioxide Word notation carbon + oxygen  carbon dioxide Figure 6.1 Chemical change (reaction of carbon and oxygen) Another example of chemical change is an industrial preparation of iron (Fe). Iron is very important in the manufacturing of various materials and in construction. It is usually produced from iron ores, which are minerals with high iron content. 152   Technical Science | Grade 11 Science_GR11_LB.indb 152 2018/02/02 17:00

Hematite, Fe2O3 is one of these and it is very similar to the usual iron rust. To produce iron, hematite is treated with carbon monoxide gas (CO) at high temperature to produce elemental iron (Fe) and carbon dioxide (CO2) as shown in (Equation 1): Fe2O3(s) + 3CO(g) heat 2Fe(l) + 3CO2(g) Oxidation describes a reaction in which a substance reacts with and gains oxygen atoms. Reduction describes a reaction in which a substance loses oxygen atoms. In the example, of Hematite, Fe has been ‘reduced’ because it has lost oxygen atoms and the carbon in CO has been oxidised because it has gained oxygen atoms to produce CO2. However, this definition of oxidation and reduction is too restrictive and insufficient, thus a broader definition would be: Oxidation is a reaction in which electrons are transferred from one element to another in a chemical reaction. 6.2 Using rules to assign oxidation numbers Oxidation-reduction reactions is a special chemical reaction that involves the transfer of electrons. Oxidation involves the loss of electrons in a chemical reaction, whereas reduction involves the gain of electrons in a reaction. These processes can be identified through the use of oxidation numbers (oxidation states). The oxidation number is the number of electrons that must be added or removed from an atom in a combined state to convert the atom into the elemental form. The concept of oxidation numbers: • is used to keep track of the number of electrons lost by the oxidised substance and gained by a reduced substance in a reaction. • helps in classifying a chemical change as involving a transfer of electrons or not. • will help in identifying substances that are either oxidised or reduced in a reaction by tracking the change in the oxidation number. Figure 6.1 shows a summary of the rules on assigning oxidation numbers to monoatomic (consisting of only one atom type, for example, Cl– ) or polyatomic species (consisting of many atom types, for example, SO42–). Table 6.1 Rules for assigning oxidation numbers Rule Example 1. Oxidation number of any pure element is 0. The oxidation number of Na(s) is 0 2. The oxidation number of monoatomic ions The oxidation Cl– is –1 equals to the charge on the ion. 3. The more electronegative element in a binary The oxidation number of O in NO is –2 compound is assigned the number equal to the charge it would have if it were an ion. 4. The oxidation number of fluorine in a The oxidation number of F in NaF is –1 compound is always –1. Chapter 6 Chemical change 153 Science_GR11_LB.indb 153 2018/02/02 17:00

Rule Example 5. Oxygen has an oxidation number of –2, 1. The oxidation number of O in NO2 is –2 unless it is combined with F, (not always, only Because oxygen is more electronegative than in O2F2) in which it is +1 or +2, or it is in a N. peroxide, in which it is –1 2. The oxidation number of O in dioxygen difluoride O2F2 is +1. Because F is more electronegative than O. 3. The oxidation number of O in oxygen difluoride is +2 4. The oxidation number of O in H2O2 is –1 6. Hydrogen oxidation state in most compounds The oxidation number of H in NaH is –1 is +1 unless it is combined with metal, in which case it is –1 Only in metal hydrides. 7. In compounds, Group 1 and Group 2 and The oxidation number Na in Na2CO3 is +1 aluminium have oxidation number +1, +2 and +3 respectively. 8. The sum of oxidation numbers of all atoms in The oxidation number of C in CaCO3 is +4. a neutral compound is zero Check 1 9. The sum of oxidation numbers of all atoms in The oxidation number of P in HPO42– is +5. a polyatomic ion equals charge of the ion. Chromium provides a visual example of oxidation numbers. Different oxidation states of chromium have dramatically different colours. +2 +3 +6 +6 Figure 6.2 Oxidation state of chromium 6.2.1 Using oxidation number to identify redox reaction Consider the chemical changes involving electron transfer (redox reactions) and apply these rules to identify substances undergoing oxidation or reduction in a reaction. Consider the reaction of Zn metal with hydrochloric acid (HCl) below: 0 +1 +2 0 Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g) We observe the following: • the oxidation number of Zn increases from 0 to +2 in ZnCl2. This increase in oxidation number means that Zn has lost electrons – it has been oxidised. 154   Technical Science | Grade 11 Science_GR11_LB.indb 154 2018/02/02 17:00

• the oxidation number of H has decreased from +1 in HCl to 0 in H2(g). This decrease in oxidation number means that H+ has gained electrons – it has been reduced. • Chlorine has the oxidation number of –1 before and after the reaction, this means it has neither oxidised nor reduced; it is a spectator ion. Example 6.1 CALCULATING THE OXIDATION NUMBERS Determine the oxidation number of the atoms written in bold. a) H2SO4 b) NO3– Analyse Given: Oxidation number of H = 1 Oxidation number of O in H2SO4 = –2 Unknown: Oxidation number of S Solve a) Assign the oxidation number of S to any variable (for example, x). Since the overall charge is 0, it means that the sum of the oxidation numbers will add up to zero. 2(oxidation number of H) + x + 4 (oxidation number of O) = 0 2(1) + x + 4(–2) = 0 2+x–8=0 x–6=0 x = +6 ∴ the oxidation number of S in H2SO4 is equal to 6 b) Similarly, the oxidation number of O in NO3– is –2. Since the overall charge of NO3– is –1, we will equate the sum of the oxidation numbers to –1 as follows: x + 3(–2) = –1 x – 6 = –1 ∴ x = +5 ∴ the oxidation number of N in NO3– is equal to +5 Activity 6.1 1. Consider the following redox reaction: CrO2– 2– – 4 (aq) + S (aq) + OH  Cr(OH)3(s) + S(s) + H20(l) a) Determine the oxidation numbers of the underlined atoms in the reaction above b) Identify both the oxidising and reducing agents. Justify your answer. 2. Indicate, with reasoning, which of the following is an oxidation-reduction reaction: a) MnO2(s) + 4H+(aq) + 2Cl–(aq)  Mn2+(aq) + 2H2O + Cl2(g) b) H2PO4– (aq) + OH–(aq)  HPO2– 4 (aq) + H2O 3. Write a net ionic equation to represent the reaction of iron (Fe) with hydrochloric acid (HCl) solution to produce H2 (g) and Fe2+. Show both the oxidation and reduction half-reactions. Chapter 6 Chemical change 155 Science_GR11_LB.indb 155 2018/02/02 17:00

4. Identify, with justification, the oxidising and reducing agents in the following redox reaction: 2NO2(g) + 7H2(g)  2NH3(g) + 4H2O(g) 5. Determine the oxidation number for the element underlined. a) CaH2 b) NaIO3 c) C4H12 d) Al2(SO4)3 e) PbSO4 f) ClO3– g) HPO32– h) Na2O2 6.2.2 Oxidation occurs when valence electrons are lost in a chemical reaction Processes in which the atoms or ions of an element experience an increase in oxidation number (state) are oxidation processes. For example, in a chemical change between sodium (Na) and chlorine (Cl2) in the formation of sodium chloride, the formation of sodium ions illustrate oxidation. The chemical equation for this reaction is written as follows: 2Na(s) + Cl2  2NaCl The oxidation state is represented by placing an oxidation number above the simple of an atom and the ion. 0 +1 Na  Na+ + e– The oxidation state of sodium has changed from 0, its elemental state, to the +1 state of the ion (Rules 1 and 7, Table 6.1). A chemical substance whose oxidation number increases is oxidised. A substance that undergoes oxidation is called reducing agent. In our example, the sodium atom is oxidised to a sodium ion. 6.2.3 Reduction occurs when valence electron is gained in a chemical reaction Processes in which the atoms or ions of an element experience a decrease in its oxidation number (state) are reduction processes. Consider the behaviour of chlorine in its reaction with sodium once again. 0 –1 Cl2 + 2e–  2Cl– The oxidation state of chlorine has decreased from 0, its elemental state, to the –1 state of the ion (Rules 1 and 2, Table 6.1). A chemical substance whose oxidation number increases is reduced. The chlorine atom is reduced to an chlorine ion. 6.2.4 Oxidation and reduction are paired reactions Electrons are lost in oxidation and gained in reduction Therefore, for oxidation to occur during a chemical reaction, reduction must also occur. The number of electrons transferred in oxidation must equal to the number of electrons gained in reduction. 156   Technical Science | Grade 11 Science_GR11_LB.indb 156 2018/02/02 17:00

Electrons are negatively charged, and for a charge to be conserved, the number of electrons lost must equal to the number of electrons gained. Mass is conserved in any chemical reaction. Therefore, the electrons exchanged during oxidation and reduction are conserved. But why do we say substance is reduced when it gains electrons? Remember, when electrons are gained, their electric negative charge will cause the overall oxidation number to decrease, that is, to be reduced. A transfer of electrons causes changes in the oxidation states of one or more elements. Any chemical process in which elements undergo changes in oxidation number creates a reduction-oxidation reaction. Did you know? The name for the reduction-oxidation reaction, is often shortened to redox reaction. An example of redox reaction can be seen in Figure 6.3. In this example, copper is being oxidised and NO3– from nitric acid is brown gas being reduced. The part of the reaction involving oxidation or reduction can be written as a half-reaction. The overall equation for a redox reaction is a sum of the half-reactions. Because the number of electrons involved is the same for oxidation and reduction, they cancel each other out and do not appear in the overall chemical equation. copper coin Copper is oxidised and nitrogen dioxide (brown gas) is produced when this copper coin is placed in concentrated nitric copper ions acid solution Figure 6.3 Oxidation of copper The equation for the reaction of copper and nitric acid illustrates the relationship between half- reactions and the overall redox reaction. 0 +2 Cu  Cu2+ + 2e– (oxidation half-reaction) +5–2 +4–2 +1–2 2NO3– + 2e– + 4H  2NO2 + 2H2O (reduction half-reaction) 0 +5 +2 +4 Cu + 2NO–3 + 4H+  Cu2+ + 2NO2 + 2H2O (overall redox reaction) Notice that the electron lost in oxidation appear on the product side of the oxidation half-reaction. Electrons are gained in reduction and appear as reactants in the reduction half-reaction. The procedure for identifying oxidising and reducing agents is outlined below: • whenever reduction takes place, oxidation also takes place. Oxidation and reduction goes hand in hand. To clearly determine which substance is the oxidising agent, we break down this chemical reaction into half reactions, or reactions where one is the reduction component and the other is the oxidising component. Cu2+(aq) + Mg(s)  Cu(s) + Mg2(aq) Chapter 6 Chemical change 157 Science_GR11_LB.indb 157 2018/02/02 17:00

• break down the redox reaction into half-reactions as follows: –– oxidation half reaction: Mg(s)  Mg2+ +2e– –– reduction half reaction: Cu2+ +2e–  Cu(s) • in the half reactions shown above, magnesium loses two electrons and the copper ion (Cu2+) gains two electrons. Since the oxidising agent is the substance that is reduced and reduction is caused by a gain in electrons. We can deduce that, the oxidising agent is the copper ion and the reducing agent is magnesium. Did you know? To recall the definitions of oxidation and reduction the mnemonic OILRIG can be used as outlined below The following mnemonic may be useful in recalling these reactions: • OIL Oxidation I Loose (Oxidation is Loss of electrons) • RIG Reduction I Gain (Reduction is Gain of electrons) To summarise the key concepts we have covered in this unit, let’s pause and look back: • Oxidation: loss of electrons by substance in a reaction; it is accompanied by an increase in oxidation number. A substance that undergo oxidation is called reducing agent • Reduction: gain of electrons by a substance in a reaction; it is accompanied by a decrease in oxidation number. A substance that undergo reduction is called oxidising agent. • Oxidation and reduction must always occur together and the number of associated with oxidation must be the same as those associated with reduction. • Sometimes it can get more confusing since we use similar words. The following mnemonics can be used to clear confusion GEROA LEORA Gain of Electrons Reduction Oxidsing Agent Loss of Electrons Reduction Oxidsing Agent Figure 6.5 Summary 6.3 Application of redox reaction Applications of oxidation-reduction reactions are vast. One example, is the photochromic lenses shown in Figure 6.6. Photochromic eyeglasses darken when exposed to ultraviolet light and become transparent again in the absence of ultraviolet light. This process is the result of oxidation-reduction reactions. Silver chloride (AgCl) and copper(I) chloride (Cu2O) are Figure 6.6 Photochromic lenses embedded in the lenses. The chloride ions absorb photons, and the silver chloride dissociates and form chlorine atoms and silver atoms. The elemental silver darkens the lenses. Chlorine ions are oxidised and silver atoms are reduced. Then the copper(I) ions reduce the chlorine atoms and form copper(II) ions. In the reverse process, the copper(II) ions oxidise to the silver atoms back to the transparent silver ions. 158   Technical Science | Grade 11 Science_GR11_LB.indb 158 2018/02/02 17:00

Activity 6.2 1. Determine if each of the following changes is oxidation, reduction or neither. a) SO32–  SO42– b) CaO  Ca c) CrO42–  Cr2O72– d) CrO42–  Cr3+ e) MnO4–  Mn2+ f) ClO2–  ClO– 2. Write the oxidation and reduction reactions for each redox reaction. The first one has done for you. Fe2+ + Co  Co2+ + Fe Oxidation: Co  Co2+ + 2e– Reduction: Fe2+ +2e–  Fe a) 3 Ag+ + Ni  Ni3+ +3Ag Oxidation: Reduction: b) Cu2+ + Pb  Pb2+ + Cu Oxidation: Reduction: c) O2 + 2Sn  O2– + 2Sn2+ Oxidation: Reduction: Activity 6.3 1. Consider the following ionic reactions and determine the following: a) substance oxidised b) substance reduced c) oxidising agent d) reducing agent (i) Cr2O2– 7 + Fe 2+  Cr3+ + Fe3+ (ii) IO3– + Re  ReO4– + IO– (iii) SO2– 3– – 4 + Br2  S2O 3 + BrO3 (iv) MnO4– + H2S  MnO (v) O2 + Cr3+  H2O2 + Cr2O2– 7 Chapter 6 Chemical change 159 Science_GR11_LB.indb 159 2018/02/02 17:00

Nonspontaneous reactions redox reactions occurs in 6.4 electrolytic cells We have learnt that sodium reacts violently with chlorine to produce sodium chloride. Na(s) + Cl2(g)  NaCl(s) Sodium also reacts violently with oxygen to produce sodium hydroxide. Chemists refer to such reactions as spontaneous reactions. All the reactions we have studied so far in this chapter are spontaneous reactions. A spontaneous reaction is a reaction that doesn’t require, for example, a catalyst to speed up the reaction, nor any heat to initiate the reaction. That is, it happens by itself. For example, when you expose sodium to air, it reacts spontaneously and produces sodium oxide. This is illustrated in the equation below. Na(s) + O2(g)  Na2O(s) So, in practice, sodium is placed under oil so that it doesn’t react to air. Some oxidation-reduction reactions do not occur spontaneously, but they e– + – e– can be driven by electrical energy. If power electrical energy is required to produce a supply anode redox reaction and bring about chemical cathode change, it is called an electrolytic cell. Cl2 You can split a compound (group of elements bounded together in a fixed inert inert ratio, for example, NaCl) using electricity. electrode electrode To explain what happens in an electrolytic Cl – Na + cell, let’s examine the decomposition of molten sodium chloride into sodium electrolyte metal and chlorine gas as shown in Figure 6.7. Figure 6.7 Electrolytic cell The reaction is written below. NaCl(aq)  2Na(s) + Cl2(g) If molten NaCl is placed into the container and inert electrodes of carbon are attached to the positive and negative terminals of a battery, an electrolytic reaction will occur. Molten NaCl consists of negative (Cl–) and positive ions (Na+). The negative ions are called anions and the positive ions are called cations. The anions will be attracted to the electrode that is connected to the positive terminal of the power supply, and the cations will be attracted to the electrode connected to the negative terminal of the battery. Recall that unlike charges attract each other. The electrode that attract the anions is called an anode and the electrode that attracts the cations is called a cathode. A cathode is the electrode where reduction takes place and anode is the electrode where oxidation takes place. Electrons from the negative terminal of a battery flow to the cathode and are used to reduce sodium ions into sodium atoms. The sodium will plate onto the cathode as it forms. The sodium ions are migrating towards the cathode. 2Na+(l) + 2e–  2Na(s) 160   Technical Science | Grade 11 Science_GR11_LB.indb 160 2018/02/02 17:00

The negative chlorine ions migrate towards the anode and release electrons as they oxidise to form chlorine atoms. The chlorine atoms will combine together to form chlorine gas which will bubble away. 2Cl–(l)  Cl2(s) + 2e– The conditions under which the electrolyte cell operates are very important. The substance that is the strongest reducing agent will undergo oxidation. The substance that is the strongest oxidising agent will be reduced. The overall redox reaction is obtained by combining the two half reactions as follows: Oxidation: 2Na+ (aq) + 2e–  2Na(s) Reduction: 2Cl– (aq)  Cl2(g) + 2e– Overall: 2Na+(l) + 2Cl–  2Na(s) + Cl2(g) If an aqueous solution of sodium chloride were used in the above system, hydrogen would undergo reduction instead of sodium, because it is a stronger oxidising agent that sodium. Sodium chloride will dissociate into negative (Cl–) and positive ions (Na+). NaCl(s) + H2O(l)  Na+(aq) + Cl–(aq) cations anions Oxidation: 2Cl–(l)  Cl2(g) + 2e– Reduction: 2H2O + 2e–  H2(g) + 2OH– Overall: 2Cl– + H2O  Cl2(g) + H2(g) + 2OH– If we add in the sodium ions, then the overall reaction becomes: power supply 2NaCl + 2H2O  Cl2(g) + H2(g) + 2NaOH Sodium ions are said to be spectator ions since they did not take part in the chemical reaction – there was no change in their oxidation state. 6.4.1 Electroplating An electrolytic process in which a metal ion is reduced and – + solid metal is deposited on a surface is called electroplating. An electroplating cell contains a solution of a salt plating e– e– metal, an object to be plated (the cathode) and a piece of plating metal (anode). For example, a copper plating cell cathode anode bracelet contains a solution of soluble copper(II), salt and a copper Cu anode. The cathode is the object to be plated. The copper anode is Cu 2+ connected to the positive terminal of the battery or to some other source of direct current. The object to be plated is Cu connected to the negative terminal. 2– SO 4 Figure 6.10 shows how an object (bracelet) is being plated using copper. Figure 6.9 Electroplating of copper on a bracelet Chapter 6 Chemical change 161 Science_GR11_LB.indb 161 2018/02/02 17:00

Copper ions are reduced at the cathode according to the following half-reaction and deposited as metallic copper when electrons flow through the circuit. Cu2+(aq) + 2e–  Cu(s) At the anode, metallic copper is oxidised by the following half-reaction: Cu(s)  Cu2+(aq) + 2e– In fact, copper is transferred from the anode to the cathode of the cell. 6.4.2 Electrolysis Electroplating is an example of electrolysis. Electrolysis is the decomposition of a substance when an electric current is passed. That is, electric energy is used to force a non-spontaneous reaction to occur. For this cell reaction to occur, the external voltage must be greater than the potential that would be produced by the spontaneous reverse reaction. Electrolysis is important in industry. It is used to purify many metals, including the ores which are chemically combined in the Earth’s crust. During the electrolysis of water, water is broken down into its elements, H2 and O2. The following experiment will demonstrate how electrolysis can be used to break NaCl down into its elements. FORMAL EXPERIMENT Aim: Electrolysis of an aqueous NaCl solution Aims 1. To identify both the cathode and the anode. 2. To identify reactions that occur at the cathode and the anode. Apparatus and reagents 2 test tubes 1 Beaker 2 carbon electrodes 6 or 9 V Battery/cell (you can use pencils sharpened at both ends for this) • dilute NaCl solution • concentrated NaCl solution (Brine) 162   Technical Science | Grade 11 Science_GR11_LB.indb 162 2018/02/02 17:00

Precautions • Wear safety goggles • Perform this experiment in a well ventilated area Method 1. Fill the test tubes with the NaCl solution. 2. Insert the carbon electrodes into the test tubes each connected to a wire. 3. Invert the test tube into the beaker half-filled with the NaCl solution. 4. Add a few drops of the Universal indicator to this solution. 5. Connect the wires to the 6V or 9 V cell. 6. Observe what happens in the test tubes as well as noting the indicator colour change. Questions 1. Justify the use of a safety precaution, other than wearing safety glasses, when carrying out this investigation. 2. Describe how the THREE products of this electrolysis could be identified. 3. Identify the anode and the cathode. 4. Write down the half-reactions taking place in each electrode. 5. Write down the overall redox reactions without spectator ions. 6. Write down the overall redox reactions with spectator ions. 7. Write down the name of spectator ion. 8. Write down TWO uses of the product identified in this experiment. Chapter summary Oxidation-Reduction Terminology Term Change in oxidation number Change in electron population Oxidation In a positive direction Loss of electrons Reduction In a negative direction Gain of electrons Oxidising agent In a positive direction Gains electrons Reducing agent In a negative direction Loses electrons Dilute NaCl solution Concentration NaCl solution Reduction (cathode reaction) Reduction (cathode reaction) + – 2H (aq) + 2e  H2(g) 2H+(aq) + 2e–  H2(g) Oxidation (Anode reaction) Oxidation (Anode reaction) – – 4OH (aq)  2H2O(l ) + O2(g) + 4e 2Cl–(aq)  Cl2(g) + 2e– Products: Products: H2(g) and O2(g) NaOH(aq); Cl2(g) and H2(g) Chapter 6 Chemical change 163 Science_GR11_LB.indb 163 2018/02/02 17:00

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Selected answers Chapter 1 Mechanics Activity 1.1 The Cartesian Plane 1. a) Studying the graph shows that the maximum distance reached is 3 km away from the boy’s home. Therefore, the tuck shop is 3 km away from the boy’s home. b) The total journey took 11 minutes, according to the graph. Therefore, the boy was away from home for 11 minutes. c) The maximum distance, 3 km, is maintained between the time interval 3 – 7 minutes. During this time the boy the distance is not changing, meaning the boy must have stopped there during that time interval. Therefore, the boy was in the shop for 4 minutes (during the time interval 3 – 7 minutes). d) 3 km away from home. e) 1 km away from home. f) The flatness means the distance from home is not changing. The boy has stopped. g) C (2 minutes; 2 km) D (7 minutes; 3 km) Activity 1.2 Sign Convention and Vectors 1. a) Scalar quantity is the physical quantity that has only magnitude b) Vector quantity is the physical quantity having magnitude and direction c) Bearing is the degrees away from the true North, measured clockwise, of a distant point relative to the current (fixed) point 2. a) 10 cm N 45° W E S Chapter 1 Answers 165 Science_GR11_LB.indb 165 2018/02/02 17:00

b) Using the scale 1 cm : 5 m.s–1 such that 5 cm : 25 m.s–1 N W E 22° 25 m · s–1 S c) Using the scale 1 cm : 20 km, we obtain that 100 km : 5 cm N 75° 100 km W E S d) Using the scale 1 cm : 25 mm, we get that 205 mm : 8,2 cm 205 mm N W 300° E S 166   Technical Science | Grade 11 Science_GR11_LB.indb 166 2018/02/02 17:00

Activity 1.3 Dependent and independent variables 1. For two variables to be directly proportional: The ratio between the two variables must be constant The graph representing the two variable must be a straight line passing through the origin 2. For two variables to be inversely proportional: The product of the two variables must be a constant The graph representing the variable must be a hyperbola, that doesn’t intersect with the axes 3. a) Number of cement bags Number of Wheel barrows Ratio 1 4 4 2 8 4 3 12 4 4 16 4 5 20 4 b) Independent variable: cement bags Dependent variable: Cement c) 25 number of wheel barrows full of sand 20 15 10 5 0 1 2 3 4 5 6 number of cement bags d) The number of wheel barrows full of sand is directly proportional to the number of cement bags. e) Let x be number of cement bags, and y be the number of wheel barrows full of sand Chapter 1 Answers 167 Science_GR11_LB.indb 167 2018/02/02 17:00

f) y = 4x 30 25 20 width (cm) 15 10 5 0 1 2 3 4 5 6 7 8 9 Length (cm) 4. a) Group Length (cm) Width (cm) Area (cm2) A 1 24 24 B 2 12 24 C 3 8 24 D 4 6 24 E 5 4.8 24 b) c) Since the product of length and width is the constant, therefore the width is inversely proportional to the length. d) A = W × L Wa La = Wb Lb e) W × L = 24 W × 10 cm = 24 cm2 ∴​2,4 cm ​ 168   Technical Science | Grade 11 Science_GR11_LB.indb 168 2018/02/02 17:00

Activity 1.4 Applications of the Pythagoras Theorem 3–4–5 6 – 8 – 10 9 – 12 – 15 The hypotenuse is the longest The hypotenuse is the longest The hypotenuse is the longest side in a right-angled triangle. side in a right-angled triangle. side in a right-angled triangle. We should prove that the sum of We should prove that the sum of We should prove that the sum of squares of 3 and 4 equals to the squares of 6 and 8 equals to the squares of 9 and 12 equals to the square of 5: square of 10: square of 15: 32 + 42 62 + 82 92 + 122 = 16 + 9 = 36 + 64 = 81 + 144 = 25 = 100 = 225 5 = 25 2 102 = 100 152 = 225 ​∴​ 32 + 42 = 52 ​∴​ 62 + 82 = 102 ​∴​ 92 + 122 = 152 12 – 16 – 20 15 – 20 – 25 The hypotenuse is the longest The hypotenuse is the longest side in a right-angled triangle. side in a right-angled triangle. We should prove that the sum of We should prove that the sum of squares of 12 and 16 equals to the squares of 15 and 20 equals to the square of 20 square of 25 122 + 162 152 + 202 = 144 + 256 = 225 + 400 = 400 = 625 202 = 400 252 = 625 ​∴​ 122 + 162 = 202 ​∴​ 152 + 202 = 252 Activity 1.5 1. We can apply the theorem of Pythagoras to calculate the length of the zip line. Let the length of the zip line be x: x2 = 202 + 502 = x2 16 + 9 x2 = 25 ​∴​ x = 53,85 m 2. The cable used must not be too tight to accommodate the mass to be suspended. Activity 1.6 1. AD2 = AB2 + BC2 + CD2 AD2 = 42 + 52 + 62 AD2 = 77 ​∴​ AD = 8,77 cm Chapter 1 Answers 169 Science_GR11_LB.indb 169 2018/02/02 17:00

Activity 1.7 1. This is a rate problem. It is shown that 1 m of an object is reflected as 8 cm of that object’s image. Therefore: ​1 m: 8 cm = x: 24 cm ​  1 _____m  ​ = ______ ​  x  ​ 8 cm 24 cm x = ​  1 _____m  ​  × 24 cm 8 cm x = 3 cm​ 2. We can apply trigonometric ratios that we have learned in Grade 10 Mathematics. Let the height of the tree be x ​  x  ​ ​tan ​44​​  °​ = _____ 90 m ∴ x = 90 m × tan ​44​​  °​ x = 86,91 m​ Therefore, the height of the tree is 86,91 m. Activity 1.8 Pythagoras Theorem 1. The resultant force is a single force which has the same effect as all the forces combined together. 2. Collinear vectors are vectors that act along the same line of action. 3. Calculating resultant force of perpendicular forces. a) ​​    F​  2R ​  = ​F​  21​ ​  + ​F​  22​  FR ​   F​  2R ​  = ​(23, 7 )​​  2​  + ​9​​  2​ F2 = 9 N ​F​  2R ​  = 642, 69 ∴ ​F​  R​​ = 25, 35 N​ F1 = 23.7 N We know the magnitude of the resultant force, now we must determine the angle to find the direction of the resultant force. ​F​  y​​9 ​sin θ = ​ __ ​F​  ​​  ​  = ​ _____ ​  = 0, ​36​​  ∘​ R 25, 35 θ = 20, ​80​​  ∘​ → ​ F​  R​​​ = 25, 35 N at 20, ​80​​ ∘​  to the negative x-direction​ OR ​F​  ​​ 23, 7 ​cos θ = __ ​  ​F​  x ​​ = ​ _____ ​ = 0, ​93​​  ∘​ R 25, 35 θ = 20, ​79​​  ∘​ → ​ F​  R​​​ = 23, 35 N at 20, ​79​​ ∘​  to the negative x-direction​ 170   Technical Science | Grade 11 Science_GR11_LB.indb 170 2018/02/02 17:00

OR θ = ​tan​​  −1​​(__ ​ ​F​   ​​ )​ ​F​  y​​ ​F​  y​​ ​tan θ = __ ​ ​F​   ​  ​​ OR x x θ = ​tan​​  ​​(​  9 23, 7 ) 9 tan θ = ​ ____  ​  OR −1____ ​ ​ 23, 7 θ = 20, ​79​​  ∘​ → ​ F​  R​​​ = 25, 35 N at 20, ​79​​ ∘​  to the negative x-direction​ b) ​​    F​  2R ​  = ​F​  21​ ​  + ​F​  22​  ​   F​  2R ​  = ​(17 )​​  2​  + ​(23 )​​  2​ ​   F​  2R ​  = 818 FR F2 = 23 N ∴ ​F​  R​​ = 28, 60 N​ F1 = 17 N We know the magnitude of the resultant force, now we must determine the angle to find the direction of the resultant force. ​F​  y​​ 23 ​sin θ = ​ __ ​ = ​ _____ ​ = 0,​8042​​  °​ ​F​  R​​ 28, 60 θ = 53,​53​​  °​ → ​ F​  R​​​ = 28, 60 N at 53,​53​​ °​  to the negative x-direction​ OR ​F​  ​​ 17 ​  x ​ = ​ _____ ​cos θ = __  ​ = 0,​5944​​  °​ ​F​  R​​ 28, 60 θ = 53,​53​​  °​ → ​ F​  R​​​ = 28, 60 N at 53,​53​​ °​  to the positive x-direction​ OR θ = ​tan​​  −1​​(​ __ ​)​ ​F​  y​​ ​F​  y​​ ​tan θ = __ ​   ​  OR ​F​  x​​ ​F​  x​​ ​  23 ​  tan θ = ___ OR θ = ​tan​​  −1​​(___ ​  23 ​)​ 17 17 θ = 53,​53​​  ​ ° → ​ F​  R​​​ = 28, 60 N at 53,​53​​  °​  to the positive x-direction​ → → → 4. a) ​​ F​  x​​​ = ​ F​  1​​​  + ​ ​F​  3​​​ ⟶ ​ ​  F​  x​​​ = 2,3 N + (− 2,1 N ) → → ​ F​  x​​​ = +0,2 N OR ​ F​  x​​​ = 0,2 N in the positive x-direction​ → → → b) ​​ ​F​  y​​​ = ​ F​  2​​​  + ​ ​F​  4​​​ → ​ F​  y​​​ = 4 N + ( − 3, 3 N ) → → ​ ​F​  y​​​ = +0, 7 N OR ​ F​  y​​​ = 0,7 N in the positive x-direction​ Chapter 1 Answers 171 Science_GR11_LB.indb 171 2018/02/02 17:00

c) ​​F​  2R ​  = ​F​ 2x​ ​  + ​F​ 2y​ ​ ​    F​  2R ​  = ​(0, 2 )​​  2​  + ​(0, 7 )​​  2​ ​   F​  2R ​  = 0, 53 ∴ ​F​  R​​ = 0, 73 N​ We know the magnitude of the resultant force, now we must determine the angle to find the direction of the resultant force. ​F​  y​​ 0, 7 ​sin θ = __ ​   ​ = ​ ____ ​ = 0, ​96​​  °​ ​F​  R​​ 0, 73 θ = 73, ​52​​  °​ → ​ F​  R​​​ = 0, 73 N at 73, ​52​​ °​  to the positive x-direction​ OR ​F​  ​​ 0, 2 ​cos θ = __ ​  x ​ = ​ ____ ​ = 0, ​27​​  °​ ​F​  R​​ 0, 73 θ = 73, ​4​​  °​ → ​ F​  R​​​ = 0, 73 N at 73, ​4​​ °​  to the positive x-direction​ OR θ = ​tan​​  −1​​(__ ​   ​)​ ​F​  y​​ ​F​  y​​ ​tan θ = __ ​   ​  OR ​F​  x​​ ​F​  x​​ θ = ​tan​​  ​​(​   ​)​ ___0,7 −1 ___0,7 tan θ = ​   ​  OR 0,2 0,2 θ = 74, ​05​​  ​ ° → ​ F​  R​​​ = 0,73 N at 74,​05​​ °​  to the positive x-direction​ 5. a) First calculate the resultant of forces in the x- and y-direction: → → → ​​ F​  x​​​ = ​ F​  1​​​  + ​ ​F​  3​​​ → ​ F​  x​​​ = (–2,3 N) + (3,3 N ) = +1 N​ → → → ​​ F​  y​​​ = ​ F​  2​​​  + ​ ​F​  4​​​ → ​ F​  y​​​ = (4 N) + (2,1 N ) = +6, 1​ ​​F​  2R ​  = ​ F​ 2x​ ​  + ​F​ 2y​ ​ ​F​  2R ​  = ​(1 )​​  2​  + ​(6,1 )​​  2​ ​F​  2R ​  = 38,21 ∴ ​F​  R​​ = 6,18 N​ We know the magnitude of the resultant force, now we must determine the angle to find the direction of the resultant force. ​F​  y​​ 6, 1 ​sin θ = __ ​   ​ = ​ ____ ​ = 0, ​99​​  °​ ​F​  R​​ 6, 18 θ = 80,​77​​  °​ → ​ F​  R​​​ = 6, 18 N at 80,​77​​ °​  to the positive x-direction​ 172   Technical Science | Grade 11 Science_GR11_LB.indb 172 2018/02/02 17:00

OR ​F​  ​​ 1  ​ = 0,​1618​​  °​ __x ​ = ​ ____ ​cos θ = ​ ​F​  R​​ 6,18 θ = 80, ​67​​  °​ → ​ F​  R​​​ = 6,18 N at 80,​67​​ °​  to the positive x-direction​ OR θ = ​tan​​  −1​​(__ ​   ​)​ ​F​  y​​ ​F​  y​​ ​tan θ = __ ​   ​  OR ​F​  x​​ ​F​  x​​ θ = ​tan​​  −1​​(___ ​   ​)​ 6,1 6,1 tan θ = ___ ​   ​  OR 1 1 θ = 80,​69​​  °​ → ​ F​  R​​​ = 6,18 N at 80,​69​​  °​  to the positive x-direction​ Activity 1.9 Forces in 2D (Head-to-Tail Method) 1. a) Using a scale 1 cm: 1 N, we can draw diagram from which we can determine the resultant of the two forces: F1 = 23 N FR = 13 N F2 = 5 N → Measuring the length of the vector representing ​​ F​  R​​​​we get 13 cm. This is exactly an equivalent of 13 N, according to our scale. b) Using the scale 1 cm : 2 N, we can convert our forces as follows: 16 N : 8 cm 12 N : 6 cm F2 = 16 N F2 = 12 N FR → Measuring the length of the vector representing ​​ F​  R​​​​we obtain 10 cm. according to our scale, this is exactly equivalent to 20 N. Chapter 1 Answers 173 Science_GR11_LB.indb 173 2018/02/02 17:00

c) Using the scale 1 cm : 1 N, we can convert our forces as follows: 8 N : 8 cm 6 N : cm FR = 10 N F2 = 6 N F1 = 8 N → Measuring the length of the vector representing ​​ F​  R​​​​ we obtain that FR = 10 cm. This is exactly an equivalent of 10 N, according to the defined conversion scale d) We can use the scale, 1 cm : 2,5 N, such that the forces can be converted as follows: 15 N : 6 cm 20 N : 8 cm Then the scale diagram will be as shown below: FR = 25 N F2 = 20 N F1 = 15 N → Measuring the length of the vector representing ​​ F​  R​​​​we obtained 10 cm. According to the defined conversion scale, this is an equivalent of 25 N 2. Using the theorem of Pythagoras we can verify the correctness of the answers we obtained graphically in number 1 above a) ​​F​  2R ​  = ​F​  21​ ​  + ​F​  22​  ​F​  2R ​ ​  = ​(12 )​​  2​  + ​(5 )​​  2​ ​   F​  2R ​  = 169 ∴ ​F​  R​​ = 13 N​ b) ​​F​  2R ​  = ​F​  21​ ​  + ​F​  22​  ​F​  2R ​  = ​(16 )​​  2​  + ​(12 )​​  2​ ​   F​  2R ​  = 400 ∴ ​F​  R​​ = 20 N​ c) ​​F​  2R ​  = ​F​  21​ ​  + ​F​  22​  ​F​  2R ​  = ​(8 )​​  2​  + ​(6 )​​  2​ ​   F​  2R ​  = 100 ∴ ​F​  R​​ = 10 N​ 174   Technical Science | Grade 11 Science_GR11_LB.indb 174 2018/02/02 17:00

d) ​​F​  2R ​  = ​F​  21​ ​  + ​F​  22​  ​F​  2R ​  = ​(15)​​  2​  + ​(20)​​  2​ ​F​  2R ​  = 400 ∴ ​F​  R​​ = 20 N​ 3. Let 1 cm : 4 N, such that 3 cm : 12 N and 5 cm : 20 N FR F2 = 20 N α θ = 35° F1 = 12 N → Measuring the length of the vector representing ​​ F​  R​​​​we get 7,7 cm. According to the scale this is → an equivalent of 30,8 N. And angular direction of ​​ F​  R​​​​ is 22˚. → ​∴ ​ F​  R​​​​ = 30,8 N at 22˚ 4. Let 1 cm : 6 N such that 3 cm : 18 N and 5 cm : 30 N. F2 = 30 N FR 30º F2 = 18 N θ → Measuring the length representing ​​ F​  R​​​​we get 7,9 cm. Using the conversion scale this is an equivalent of 47,4 N and ​θ​ = 35,6˚. → ​∴ ​ ​F​  R​​​​ = 47,4 N at 35,6˚ to the horizontal/to positive x-direction Activity 1.10 Parallelogram Law of Forces 1. The Parallelogram Law of Forces states that if two forces acting at a point can be represented by the adjacent sides of a parallelogram both in magnitude and direction, then the diagonal from the point gives the resultant of the two forces. 2. Resultant of 2 forces acting at an angle using the parallelogram of forces a) We can use the scale 1 cm : 3 N to find the resultant, such that, F1 is represented by 4 cm : 12 N and F2 by 5 cm : 15 N F2 = 15 N FR F2 = 15 N FR 120º θ = 70.9º F1 = 12 N F1 = 12 N Chapter 1 Answers 175 Science_GR11_LB.indb 175 2018/02/02 17:00

Measuring the length representing FR we obtain 4,6 cm. Using the defined scale, this is an equivalent of 13,8 N. The angle made by the resultant is 70,9°. ​∴​ FR = 13,8 N at an angle of 70,9° → Note: If the ​​ F​  2​​​​is the one drawn horizontal, then the direction of the resultant force will be 120° – 70,9˚ = 49,1° b) Using the scale 1 cm : 40 N, we obtain 5 cm : 200 N and 3 cm : 120 N The length of the vector representing FR = 6,3 cm According to the scale, this is an equivalence of 252 N. The resultant makes an angle of 28°. ​∴​ FR = 252 N at an angle of 28° F2 = 120 N F2 = 120 N FR FR 80º 28º F1 = 200 N F1 = 200 N 3. To simplify the problem, let’s take g = 10 m.s–2. The tension on the string attached to m2 is the same as the weight of m2. Same applies to m1. By applying some simple geometry, we can see that the angle between drawn vertical line that separates the two ropes and the rope connected to m2 is 30˚. The same applies to m1. The angle between the two ropes, where they meet is, therefore 75˚. T1 = m1g T1 = (0,1)(10) = 1 N T2 = m2g T2 = (0,12)(10) = 1,2 N T1= 1.2 N T2= 1 N 45º 30º 176   Technical Science | Grade 11 Science_GR11_LB.indb 176 2018/02/02 17:00

→ → Using the scale 1 cm : 0,2 N, we get that ​​ T​  1​​​​is represented by 5 cm and ​​ T​  2​​​​is represented by 6 cm. F1 = 1.8 N T2= 1.2 N T2= 1 N 93.5º → Measuring the length of the vector representing ​​ F​  R​​​​we get 8,8 cm, which is an equivalent of 1,8 N according to the defined scale. The angle made by the resultant is 93,5°. ​∴ ​FR = 1,76° at an angle of 93,5°. 4. The problem can be represented in a form of a sketch to help us see the situation we are talking about F2 FR 15º F1 45º θ = 53.6º Chapter 1 Answers 177 Science_GR11_LB.indb 177 2018/02/02 17:00

Measuring the length and the angle representing we obtain 13,9 cm at 53,6˚. Using the scale defined, this is an equivalent of 41,7 N. ∴ ​FR = 41, 7 N at 53,6˚ to the horizontal/to the positive x-direction. ​ 5. This is the problem situation in a sketched drawing F2 = 1 N 100º 125º F2 = 15 N Using the scale 1 cm : 2,5 N, we get that 4 cm : 10 N and 6 cm : 15 N. F1 = 10 N θ = 188.6º FR = 16.5 N F2 = 15 N → Measuring the length and the angle representing ​​ F​  R​​​​we get: 6,6 cm at an angle of 188,6°. According to the scale this is an equivalent of 16,5 N. → ​∴ ​​​ F​  R​​​​ = 16,5 N at 188,6° to the positive x-direction. 178   Technical Science | Grade 11 Science_GR11_LB.indb 178 2018/02/02 17:00

6. The problem situation in sketched diagram F1 = 21 N 150º 75º F2 = 14 N Using the scale 1 cm : 3,5 N, we get that 6 cm : 21 N and 4 cm : 14 N. Such that the resultant can be obtained from the scale drawing: F1 = 21 N FR = 28 N 179º F2 = 14 N Measuring the length and the angle representing the resultant, we obtain 8 cm at an angle of 179°. According to the scale this is an equivalent of: → ​​ F​  R​​​​ = 29,75 N at an angle of 179° to the positive x-direction. 7. This is the direct application of the law of parallelograms. Hence we know that, since the system is in equilibrium, then the resultant of the forces has the same magnitude as the third force; that is the unknown weight. Using the scale 1 cm : 6 N we get that 4 cm : 24 N and 5 cm : 30 N. This then in a scale drawing gives: T1= 24 N T1= 30 N 30º 40º ω Chapter 1 Answers 179 Science_GR11_LB.indb 179 2018/02/02 17:00

FR = 31.5 N T1= 30 N T2= 24 N 86º ω = –31.5 N → Measuring the length of the vector representing ​​ F​  R​​​​we get 10,5 cm. According to the defined → → scale, this is equivalent to ​​ F​  R​​​​= 31,5 N. The equilibrant, which has the same magnitude as ​​ F​  R​​​​, but in the opposite direction, has the same magnitude as the unknown weight. ​∴ → ​ w​ = − 31,5 N​ OR ∴→ ​​ w​ = 31, 5 N downwards​ 8. We can use the scale 1 cm : 10 kN such that 5 cm : 50 kN. T1 T2 45º 30º ω = 50 kN Step 1: Draw the vertically downward vector → ​​ w​​. ω = 50 kN 180   Technical Science | Grade 11 Science_GR11_LB.indb 180 2018/02/02 17:00

Step 2: From the head measure an angle of 30˚ anticlockwise, and from an angle of 45˚ clockwise. Put the dots to indicate where your angles are formed. {Note that, even when done in reversed order this will not have any impact on the final answer.} 45º ω 30º Step 3: Draw a line segment from joining the head of the vector → ​​ w​​and point formed by the angled measured from the head. Do the same with the tail of the vector → ​​ w​​. Your longs line segments must be long enough that they cross each other. Where your lines intersect put a point. 45º ω 30º Step 4: Draw an arrow which points to the intersection point from the head of the vector → ​​ w​​, → and another arrow pointing to the tail of ​​ w​​from the intersection point. Erase the part where the lines cross each other from the intersection point. 45º T2= 34 kN ω T1= 36 kN 30º → → Measuring the length of the vectors representing ​​ T​  1​​​​ and ​​ T​  2​​​​we get 3,6 cm and 4,5 cm respectively. This according to our scale drawing means: T1 = 36 kN and T2 = 45 kN 9. Follow the same procedure outlined in the 8. Using the scale 1 cm : 300 N, we get that 5 cm : 1 500 N. The only difference is with the orientation of the angles, but the problem is still the same as the previous. Chapter 1 Answers 181 Science_GR11_LB.indb 181 2018/02/02 17:00

T2 T1 20º T2 = 975 N 30º 20º ω = 1500 N 30º ω T2 = 675 N → → Measuring the length of the vectors representing ​​ T​  1​​​​ and ​​ T​  2​​​​we get 2,3 cm and 3,3 cm respectively. This according to our scale drawing means: T1 = 660 N and T2 = 990 N 10. The problem situation presented here is exactly the same as the previous one. Using the scale 1 cm : 30 N we get that 4 cm : 120 N. Hence the scale diagram: 25º T 1 ω = 120 N 30º T1 → → Measuring the length of the vectors representing ​​ T​  1​​​​ and ​​ T​  2​​​​we get 2,1 cm and 2,5 cm respectively. This according to our scale drawing means: T1 = 63 N and T2 = 75 N 11. This problem is more or less the same as the previous. The only difference is that magnitude if the weight is unspecified. Use a scale 1 cm : 500 N, such that 8 cm : 4 000 N. Step 1: Draw a vector facing downward with no specified magnitude. This vector represent the unknown weight of the load. 182   Technical Science | Grade 11 Science_GR11_LB.indb 182 2018/02/02 17:00

Step 2: At the head of the vector, measure the angle of 20˚ clockwise from the vertical. Draw the tension T1 20° ω T1 Step 3: with the aid of geometry, we can show that the angle formed between the vectors, when they joined from head-to-tail without changing their angular orientation, is 140˚. Measure this → → angle clockwise, relative to the head of ​​ T​  1​​​​ and draw ​​ T​  2​​​​ in that direction. → {Note that if done correctly, the head ​​ T​  ​​​​ should lie exactly on → 2 ​​ w​​} T2 = 4000 N ω 140° 20° T1 = 400 N → Step 4: Erase the part of → ​​ w​​that extends beyond the head of ​​ T​  2​​​​, and measure the length of → ​​ w​​. → In this case, the length of the vector representing ​​ w​​is about 15 cm. Using the defined scale: w = 7 500 Activity 1.11 Resolving forces 1. We know the formulas to calculate the components of the force, that is: → → ​​ F​  x​​​ = ​ F​cos θ 45° → → Fx ​​ ​ ​F​  y​​​ = ​ F​sin θ​ → → a) ​​ ​F​  x​​​ = ​ F​cos θ → ​ ​F​  x​​ = 205 cos ​45​​  °​ → ​ ​F​  x​​​ = 144, 96 N to the left​ F = 205 N → → ​​ ​F​  y​​​ = ​ F​sin θ → ​ ​F​  y​​​ = 205 sin ​45​​  °​ → Fy ​ ​F​  y​​ = 144,96 N downwards​ Chapter 1 Answers 183 Science_GR11_LB.indb 183 2018/02/02 17:00

→ → b) ​​ ​F​  x​​​ = ​ F​cos θ Fx → ​ ​F​  x​​​ = 8 cos ​30​​  °​ 30° → ​ ​F​  x​​​ = 6,93 N to the right​ F=8N Fy → → ​​ ​F​  y​​​ = ​ F​sin θ → ​ ​F​  y​​​ = 8 sin ​30​​  °​ → ​ ​F​  y​​​ = 4 N downwards​ → → c) ​​ ​F​  x​​​ = ​ F​cos θ Fx → ​ ​F​  x​​​ = 400 cos ​15​​  °​ 15° → Fy ​ ​F​  x​​​ = 386,37 N to the left​ F → → ​​ ​F​  y​​​ = ​ F​sin θ → ​ ​F​  y​​​ = 400 sin ​15​​  °​ → ​ ​F​  y​​​ = 103,53 N downwards​ → → d) ​​ ​F​  x​​​ = ​ F​cos θ → Fy ​ ​F​  x​​​ = 1 500 cos ​30​​  °​ → ​ ​F​  x​​​ = 1 299,04 N to the left​ → → ​​ ​F​  y​​​ = ​ F​sin θ F → ​ ​F​  y​​​ = 1 500 sin ​15​​  °​ → Fx 30° ​ ​F​  y​​​ = 750 N upwards​ → → e) ​​ ​F​  x​​​ = ​ F​cos θ → ​ ​F​  x​​​ = 90 cos ​90​​  °​ → ​ ​F​  x​​​ = 0 N​ F → → ​​ ​F​  y​​​ = ​ F​sin θ → ​ ​F​  y​​​ = 90 sin ​90​​  °​ y → ​ ​F​  y​​​ = 90 N upwards​ 184   Technical Science | Grade 11 Science_GR11_LB.indb 184 2018/02/02 17:00

2. Scale drawing to resolve forces into their perpendicular components a) Using the scale 1 cm : 25 N, so that 205 N : 8,2 cm Fx 45° F Fy → → Measuring the length representing ​​ F​  x​​​​ and ​​​ F​  y​​​​we obtained that they are both equals to 5,8 cm Using the specified conversion scale we found that these are an equivalent of 145 N. → → ​∴ ​​ F​  x​​ = 145 N, to the left ​​ F​  y​​​​ = 145 N downwards b) We can use the scale 1 cm : 1 N 30° Fx Fy F → → Measuring length representing ​​ F​  x​​​​ and ​​ F​  y​​​​we obtained 6,9 cm, and 4 cm, respectively. Use the defined conversion scale, we can see that: → → ​∴ ​ F​  x​​​​ = 6,9 N, to the right, and ​​ F​  y​​​​ = 4 N downwards Chapter 1 Answers 185 Science_GR11_LB.indb 185 2018/02/02 17:00

c) Use the scale 1 cm : 50 N such that 8 cm : 400 N Fx 15° Fy F → → Measuring the lengths representing ​​ F​  x​​​​ and ​​ F​  y​​​​ we obtain that: → → → → ​​ ​F​  x​​​​ = 7,7 cm and ​​ F​  y​​​​ = 2,1 cm. These are equivalent to ​​ F​  x​​​​ = 385 N, to the left, and ​​ F​  y​​​​ = 105 N downwards d) We can use the scale 1 cm : 250 N, such that 1 500 N : 6 cm. Fy F Fx 30° → → → → The lengths of ​​ F​  x​​​​ and ​​ F​  y​​​​were measured and obtained to be ​​ F​  x​​​​ = 5,2 cm and ​​ F​  y​​​​ = 3 cm. According to the scale we can see that: → → ​​ ​F​  x​​​​ = 1 300 N, to the left, and ​​ F​  y​​​​ = 750 N upwards e) Use the scale 1 cm : 10 N such that 9 cm : 90 N F = Fy → → From the diagram above, ​​ F​  x​​​​= 0 N, while ​​ F​  y​​​ = ​F​  y​​​ = 90 N upwards 186   Technical Science | Grade 11 Science_GR11_LB.indb 186 2018/02/02 17:00

3. All the three forces can be resolved into their horizontal and vertical components. The diagram above shows the horizontal components of each of the three forces and their directions. Let’s take direction to the right as positive, and the direction upward as positive: F 1y F1 30° F3x F1x F2x 60° Fxy 10° F3 F2 Fxy → ⟶ ⟶ ⟶ a) ​​ ​F​  x​​​ = ​ F​  1x​​​  + ​ ​F​  2x​​​  + ​ ​F​  3x​​​ → ​ F​  x​​​ = 15 cos ​150​​  °​  + 20 cos ​240​​  °​  + 10 cos ​350​​  °​ → ​​ F​  x​​ = − 13,14 N​ → ⟶ ⟶ ⟶ b) ​​ ​F​  y​​ = ​ F​  1y​​​  + ​ ​F​  2y​​​  + ​ ​F​  3y​​​ → ​ F​  x​​​ = 15 sin ​150​​  °​  + 20 sin ​240​​  °​  + 10 sin ​350​​  °​ → ​​ F​  x​​ = − 11, 56 N​ → → c) The simplified diagram showing only ​​ F​  x​​​​ and ​​ F​  y​​​​: θ Fx FR Fy ​​F​  2R ​  = ​F​ 2x​ ​  + ​F​ 2y​ ​ ​F​  2R ​  = ​(−13,14)​​  2​  + ​(−11,56)​​  2​ ​F​  2R ​  = 306,2932 ∴ ​F​  R​​ = 17,50 N​ ​R​  y​​ ​tan θ = __ ​   ​ ​R​  ​​ x 11,56 tan θ = ​ _____ ​ 13,14 ∴ θ = 41,​33​​  °​​ ​∴​ FR = 17,50 N at 41,33˚ below the horizontal to the left Chapter 1 Answers 187 Science_GR11_LB.indb 187 2018/02/02 17:00

→ 4. Let the vertical component of the force be ​​ F​  y​​​​. → ​​ ​F​  y​​​ = F sin ​60​​  °​ → ​ ​F​  y​​​ = 1 000 sin ​60​​  °​ → ​ F​  y​​​ = 866 N vertically up​ 5. http://www.physicsforarchitects.com/Sailing_against_the_wind.php Activity 1.12 Friction 1. Advantages of friction: »» It is difficult to move on slippery floor because of low friction. For example, walking on ice is very difficult because of low friction. »» Friction enables object, such as cars, to stop over a short distance. The more friction there is the shorter the stopping distance. »» Friction between a match and a box causes fire »» Friction enables to pen to be gripped on to the paper when writing. it would be impossible to write on a paper if there was no friction between the pen (or pencil) and the paper. »» Friction keeps objects in their positions. If it was not for friction object would easily slide away from their (fixed) positions. 2. The coefficient of friction is the ratio between the frictional and the normal force. ​ ​  ​Ff​   ​​​​ μ = __ N 3. The force required must at least be equal to the maximum static frictional force, otherwise the box won’t be displaced. ​  ​ = ​μ​  s​​ ​F​  N​​ ​​f​  max s ​  ​ = ​μ​  s​​  mg ​f​  max s ​  ​ = 0,72 × 85 × 9,8 ​f​  max s ​  ​ = 599,76 N​ ​f​  max s 4. The force required is equals to the maximum static frictional force. ​  ​= ​μ​  s​​ ​F​  N​​ ​​f​  max s ​  ​ = ​μ​  s​​  mg ​f​  max s ​f​  s​ max​ = 0,8 × 80 × 9,8 ​  ​ = 627,20 N​ ​f​  max s 188   Technical Science | Grade 11 Science_GR11_LB.indb 188 2018/02/02 17:00

Review exercise Multiple Choice Questions: 1. A 2. C 3. D 4. C 5. D Long Questions: 6. a) (i) Total mass of the trolley and the hanging mass. (ii) Pulling force. (iii) Acceleration. b) 0,6 0,5 acceleration (m·s²) 0,4 0,3 0,2 0,1 0 0,05 0,1 0,15 0,2 0,25 0,3 0,35 force (N) c) The acceleration is directly proportional to the pulling force. → → → 7. a) ​​ F​  x​​​ = ​ F​  2​​​  + ​ ​F​  4​​​ → ​ F​  x​​​ = − 10 N − 15 N → ​ F​  x​​​ = − 25 N → ∴ ​ ​F​  x​​​ = 25 N to the left​ → → → b) ​​ ​F​  y​​​ = ​ F​  1​​​  + ​ ​F​  3​​​ → ​ F​  y​​​ = 25 N − 10 N → ​ ​F​  y​​​ = 15 N → ∴ ​ ​F​  y​​​ = 15 N to the right​ c) Scale 1 cm : 5 N 25 N : 5 cm FR 15 N : 3 cm Fy The length representing FR = 5,8 cm This is an equivalent of 29 N Fx d) ​   F​  ​  ​  = ​F​  ​ ​  + ​F​  ​ ​ 2 R 2 x 2 y ​   F​  2R ​  = ​(−25 )​​  2​  + ​(15 )​​  2​ ​F​  2R ​  = 850 ∴ ​F​  R​​ = 29,15 N​ Chapter 1 Answers 189 Science_GR11_LB.indb 189 2018/02/02 17:00

8. Let 1 cm : 3 N 18 N : 6 cm 15 N : 5 cm F2 FR 33° F1 Measuring the length representing FR we obtain 9,54 cm. Using the conversion scale, this is an → equivalent of 28.5 N. The angle made by the resultant force is 33°. Therefore, ​​ F​  R​​​​= 28,5 N at 33° measured anticlockwise from the positive horizontal axes. → → 9. a) ​​ ​F​  x​​​ = ​ F​cos ​40​​  °​ → ​ F​  x​​​ = 100 cos ​40​​  °​ → ​ ​F​  x​​​ = 76, 60 N to the left​ → → b) ​​ ​F​  y​​​ = ​ F​sin ​40​​  °​ → ​ F​  y​​​ = 100 sin ​40​​  °​ → ​ ​F​  y​​​ = 64, 28 N upwards​ c) Looking at the free-body diagram FR FN Fk Fx Fg 190   Technical Science | Grade 11 Science_GR11_LB.indb 190 2018/02/02 17:00

10. Since the resultant forces is zero, taking the direction up as positive, we get: 200° Fx F Fy → → → → Measuring the lengths of ​​ F​  x​​​​ and ​​ F​  y​​​​ we obtain that ​​ F​  x​​​​ = 2,1 cm and ​​ F​  y​​​​ = 5,6 cm. Using the conversion factor, we get that: → ​​ F​  x​​​​ = 52,5 N to the left → ​​​ F​  y​​ = 140 N downwards Chapter 1 Answers 191 Science_GR11_LB.indb 191 2018/02/02 17:00

192   Technical Science | Grade 11 Science_GR11_LB.indb 192 2018/02/02 17:00

Chapter 2 Magnetism Experiment 1 Determining the north pole of the Earth using a bar magnet Results Magnet always come to rest in the same direction Conclusion and discussion 1. The bar magnet always points towards the north pole of the earth. 2. Yes, magnets always come to rest in the same direction. 3. No, the other materials don’t always come to rest in the same direction. Practical investigation: Making a simple electromagnet Observations – the pins or paper clips stick to the nail. Conclusion – The results show that in general the more batteries you add to the electromagnets, the more weight it can hold. The more wire you wrap around the nail, the stronger your electromagnet will be. Activity 2.1 1. A magnet always has two poles, namely, the north and the south pole 2. The like poles of a magnet repel each other when brought closer to one another, whereas the unlike poles will attract one another when brought closer to each other. 3. The force of a magnet is strongest where the magnetic field lines are closest to one another. 4. A permanent magnet is one that retains its magnetic properties for a long period of time. A temporary magnet only maintains its magnetism while in the magnetic field produced by a permanent magnet or an electric current. 5. By coiling a thin coated copper wire around an and connected the ends of the wire to a battery (or a cell). To verify if the electromagnet works, it can be used to pick paper clips. 6. An electromagnet is a temporary magnet because it only retains its magnetic properties when it is within the magnetic field create by the electric current. 7. Differences between a bar magnet and an electromagnet: Permanent magnet Electromagnet A permanent magnet is an object made from a An electromagnet is made from a coil of wire material that is magnetized and creates its own which acts as a magnet when an electric current persistent magnetic field passes through it. Permanent magnet strength depends upon the The strength of an electromagnet can be adjusted material used in its creation by the amount of electric current allowed to flow into it Chapter 2 Answers 193 Science_GR11_LB.indb 193 2018/02/02 17:00

Similarities between a bar magnet and an electromagnet: The two are similar in that while the electromagnet is magnetised it acts as a magnet in that it can attract magnetic materials, just like a bar magnet. And like a bar magnet, an electromagnet also has the north and the south poles. S N Experiment 2 Magnetic field lines Results Redraw your magnetic field map here Magnetic field lines Compass needle N N S North Pole South Pole Compass Image adapted from Phet simulations Discussion and conclusion The filed lines should be somewhat symmetric and balanced and never cross. Magnetic field lines move from north to south pole. Questions? 1. Yes, the lines are more concentrated in some areas than in others. 2. The relative strength is high. 3. If two field lines crossed, then that is saying that the magnetic field points in two different directions, at one place. There’s only one direction to the magnetic field at any place at any time, which is north to south pole, so this doesn’t happen. 194   Technical Science | Grade 11 Science_GR11_LB.indb 194 2018/02/02 17:00

Review exercise Multiple choice questions 1. D 2. D 3. E 4. B 5. B 6. E Long questions 7. A magnet is a piece of metal that pulls other magnetic metal objects towards it. 8. A magnetic field is the region of space within which the magnetic materials will experience the magnetic force. 9. Permanent magnets have regions, called domains, which have a magnetic field aligned in a certain direction. These regions give rise to a magnetic field by aligning themselves towards the same direction when magnetised by another magnet, or an electric current. 10. Ferromagnetic materials include iron, nickel and cobalt. 11. When a magnet is cut into two pieces two new magnets will be formed. The cut out end of one magnet will be a new south pole, while the other cut out end will be the new north pole. The old poles will be retained as they in the old, uncut, magnet. 12. The like poles of a magnet will repel each when brought closer to one another. 13. The opposite poles of a magnet will attract one another when brought closer to one another. 14. Magnetic field lines have the following properties: »» The direction of the magnetic field is tangent to the field line at any point in space. A small compass will point in the direction of the field line »» The strength of the field is proportional to the closeness of the lines. It is exactly proportional to the number of lines per unit area perpendicular to the lines (called the areal density). »» Magnetic field lines can never cross, meaning that the field is unique at any point in space. »» Magnetic field lines are continuous, forming closed loops without beginning or end. They go from the north pole to the south pole. 15. Magnetic field of a bar magnet: S N Chapter 2 Answers 195 Science_GR11_LB.indb 195 2018/02/02 17:00

16. The needle of a compass when placed in a magnetic field will point away from the north pole, and towards the south pole. 17. The Earth’s magnetic field is similar to that of a bar magnet tilted 11 degrees from the spin axis of the Earth. The north pole of a compass needle is a magnetic north pole. It is attracted to the geographic North Pole, which is a magnetic south pole (opposite magnetic poles attract). 18. A compass is always attracted towards the south pole of a magnet. Oppositely, it is attracted to the earth’s north pole. So the geographic north pole is actually a magnetic south pole. 19. Aurora, solar winds 20. Earth’s magnetic field North Pole Earth axis South Pole Image adapted from: Easy Science for Kids. http://easyscienceforkids.com/ all-about-earths-magnetism/ 09 September 2016 21. Compare the geographic North Pole with the magnetic north pole. 22. Explain the difference between the geographical North pole and the magnetic North pole of the Earth. 23. Discuss qualitatively how the Earth’s magnetic field provides protection from solar winds. 196   Technical Science | Grade 11 Science_GR11_LB.indb 196 2018/02/02 17:00

Chapter 3 Waves and sound Practical investigation Demonstrate a single vibration in different media 1. The pattern produce in the water by the coin is circular. 2. The energy comes from the gravitational potential energy of the stones. 3. An object that moves downwards has a lot of energy when it hits the water. The energy from its downward motion changes when it hits the water and turns in part into energy that pushes outward from the center where the object has landed. The energy from the object moves out equally in all directions. As the energy moves out from the object, it makes little waves. These waves are called ripples. Since the energy moves out equally in all directions, the ripples are always in a circle Practical investigation Generate a longitudinal wave in a slinky or spring 1. A pulse is a single disturbance in a medium. 2. The pulse moves forward. 3. The particle in the medium moves back and forth. 4. The particles move parallel to the particles of the medium. 5. A series of repeating or periodic disturbances that move through a medium from one location to another is called waves. Activity 3.1 1. A medical imaging device produces ultrasound by oscillating with a period of 0,400 ms What is the frequency of this oscillation? 2. The frequency of middle D on a typical musical instrument is 294 Hz. What is the time for one complete oscillation? 3. Identify an event in your life (such as receiving a monthly salary) that occurs regularly. Identify both the period and frequency of this event. Activity 3.2 The relationship between speed, frequency, period and wavelength 1. a) wave speed – is a distance travelled by a wave in one second b) frequency – is the number of waves passing a point per second c) period – is the time taken to complete one wave. d) wave velocity – is a wave speed in a particular direction 2. a) ​c = fλ 340 = f × 0,77 f = 441 56 Hz​ ​  1 ​ T = __ b) ​ f 1  ​ T = ​ ______ 221,56 T = 2,26 × ​10​​  −3​  s​ Chapter 3 Answers 197 Science_GR11_LB.indb 197 2018/02/02 17:00

Practical activity (Project) How to make a string telephone (project) Questions 1. If the string hangs loose, you can hardly hear your friend. 2. If you pinch the string, you stop the movement of particle and your friend will not hear you. 3. If you connect three or four string telephones together you can hardly hear your friend. Conclusion The phone works best when the string is tight, in line, and free of contact with any other object. Formal Experiment Determining the difference between pitch and loudness using an oscilloscope Conclusion Frequency is directly proportional to the pitch. Amplitude is directly proportional to the loudness. Activity 3.3 Transversal waves a) Transverse pulse is a pulse in which the particles of the medium vibrate at right angles to the direction of propagation of the pulse. b) Points that are in phase move exactly at the same time in the same direction. c) B and C; G and H; C and D; F and G d) B and C; G and H; C and D; F and G e) Rest position/ position of rest/ equilibrium line f) amplitude = 1 cm ​  3  ​ = 10 Hz ​  1 ​  , but f = ___ g) ​ T = __ f 0,3 1  ​ = 0,1 s​ ∴ T = ​ ___ 10 OR time 0,3 ​ T = ​ ______________     ​ = ​ ___ ​ = 0,1 s​ number of waves 3 number of waves ______________ h) ​ f = ​    time  ​ 3 f = ​ ___ ​ 0,3 f = 10 Hz​ distance AE ______________ 120 λ = ​    i) ​     ​ = ​ ____  ​ = 40 cm number of waves 3 v = f × λ = 10 × 40 = 400 ​ m.s​​  −1​​ Practical demonstration Make a pulse on s string and on a rope Questions 1. The coloured string moves back and forth as the pulse passes. 2. The particles of the medium move back and forth as the longitudinal wave propagates through the slinky spring 198   Technical Science | Grade 11 Science_GR11_LB.indb 198 2018/02/02 17:00

Activity 3.4 Calculations involving the wave equation 1. Using the wave equation a) v=f×λ 345 = f × 25 ∴ f = 345 = 13,8 Hz 25 b) v = f × λ 345 = 790 × λ ∴ λ = 345 = 0,44 m 790 c) v=f×λ v = 3 × 10 = 30 m·s-1 d) v = f × λ 100 = 3 × λ λ = 100 = 33.33 m 3 e) v=f×λ v = 345 × 1 = 345 Hz 2. v=f×λ 300 000 000 = 97 100 000 × λ ∴ λ = 300 000 000 = 3,09 m 97 100 000 3. Assuming the speed of the air in whistle to be, then we can calculate the wavelength. v=f×λ 343 = 320 × λ λ = 343 = 1.49 m 320 We have been told that the length of the whistle must be exactly 1 of the wavelength. 4 ∴ length = 1 × 1.49 m= 0.37 m 4 4. C4 note: v=f×λ 343 = 261.6 × λ ∴ λ = 343 = 1.31 m 261.6 D4 note: λ = 343 = 1.17 m 293.6 E4 note: λ = 343 = 1.06 m 323.6 F4 note: λ = 343 = 0.98 m 349.2 Chapter 3 Answers 199 Science_GR11_LB.indb 199 2018/02/02 17:00

G4 note: λ = 343 = 0.875 m 392 A4 note: λ = 343 = 0.78 m 440 B4 note: λ = 343 = 0.69 m 493.9 C5 note: λ = 343 = 0.66 m 523.2 Practical activity Interference of pulses a) In the situation of figure a, the displacements of the two pulses are in out of phase directions. When the two pulses meet, there is a large / small / no displacement of the medium. It seems that when the two pulses meet we can find the total displacement by subtracting (fill in: the right mathematical operation) the displacements of the individual pulses. b) In the situation of figure b. the displacements of the two pulses are in the in phase direction. When the two pulses meet, there is a large / small / no displacement. It seems that when the two pulses meet, we can find the total displacement by adding (fill in: the correct mathematical operation) the displacements of the individual pulses. Activity 3.5 The ripple tank experiment 1. To be recorded as observed during performing the experiment. 2. To be recorded as observed during performing the experiment. 3. Because of the waves overlapping one another. The type of interference responsible for this is the destructive interference. 4. When the crest of one pulse coincides with a trough of another pulse, it results in the destruction of the pulse as they move over each other. If the pulses are of equal amplitude, the result will be _____ at that point. We call this destructive interference. When the trough of one pulse coincides with the trough of another, it results in a bigger trough. This is visible on the paper as a region of lighter bands. This is known as constructive interference Activity 3.6 Superposition of waves a) b) 200   Technical Science | Grade 11 Science_GR11_LB.indb 200 2018/02/02 17:00

c) In the top example, constructive interference produces a louder sound. d) In the bottom example, destructive interference produces a softer sound Activity 3.7 1. Define echo and mention its application. 2. A SONAR receives a strong echo of ultrasonic signal it sent downwards into the sea 3,6 seconds after the transmission. Calculate the depth of the sea-bed if the speed of the ultrasonic waves in sea-water is 1 250 m.s–1. speed (v) = distance travelled (∆x) time travelled (∆y) Note that we have been given the total time (to and fro), so to determine the depth of the sea we only use half the time (i.e. 3,62 =1,8 s). 2 depth of the sea ∴ 1250 = 1,8 ∴ depth of the sea = 2250 m 3. a) pitch b) frequency c) loudness 4. You hook up a stereo system. When you test the system, you notice that in one corner of the room, the sounds seem dull. In another area, the sounds seem excessively loud. Describe how the sound moving about the room could have resulted in this. 5. You observe two musical instruments that you cannot identify. One plays high-pitched sounds and the other plays low-pitched sounds. How could you determine which is which without hearing either of them play? 6. Imagine that you observe two fireworks displays. You hear the explosion of one as soon as you see it. However, you see the other display several milliseconds before you hear the explosion. Explain why this is so. Review exercise Multiple choice questions 1. A 2. C 3. A 4. C 5. A 6. B 7. D 8. D 9. A 10. B Chapter 3 Answers 201 Science_GR11_LB.indb 201 2018/02/02 17:00

Long questions 11. There are no atoms to vibrate to carry the sound waves. It needs a medium to travel through. 12. The speed of a sound wave depends on what type of medium it is traveling through, and if that medium is a gas or liquid, a significant change in temperature can also affect its speed. It does not depend on the frequency or wavelength of the sound wave. 13. Bats: higher frequency means shorter wavelength. 14. With headphones and louder personal mp3 players our ears are generally more stressed than they used to be. 15. No change in speed – speed is determined by the material through which the wave travels, not its frequency. The wavelength will decrease by a factor of 2. 16. v = f × λ v = 0,100 × 15 v = 0,15 m∙s–1 17. v = f × λ v = 94 500 000 × 3,17 v = 3,00 × 108 m∙s–1 18. v = f × λ 343 = 20 × λ 343 λ = 20 = 17.15 m 343 λ = 20 000 = 0.01715 m = 17.15 m 202   Technical Science | Grade 11 Science_GR11_LB.indb 202 2018/02/02 17:00

Chapter 4 Waves and sound Chapter 4 Electrostatics Solutions-part 1 Activity 4.1: Application of Coulomb’s Law 1. Coulomb’s law states that the magnitude of the electrostatic force that a charged particle exert on another charged particle is directly proportional to the product of the two charges, and inversely proportional to the square of the distance between them. 2. a) b) c) F F F 1 Q1 Q2 r2 r2 k ​Q​  ​​ ​Q​  ​​ 3. a) ​F = ​ ______ 1  ​ 2 2 ​r​​  ​ k​( 2 ​Q​  A​​ ​Q​  B​​)​ ________ ​F​  new​​  = ​   ​ (​​ 2r)​​​  2​ 2k ​Q​  ​​ ​Q​  ​​ ​F​  new​​  = ​ _______ A  ​ B 2 4 ​r​​  ​ ( ​ k ​ Q​  A​​ ​Q​  B​​)​ 1 ​ ​ _______ ​F​  new​​  = ​ __  ​ 2 ​r​​  2​ 1 ​  F​ ∴ ​F​  new​​  = ​ __ 2 Hence the new force will be decreased by the factor of 2 k2 ​Q​  ​​ ​Q​  ​​ b) ​​F​  new​​  = ​ _______ A  ​ B ​r​​  ​ 2 ​F​  new​​  = 4​(______ ​r​​  ​ ) k ​ Q​  ​​ ​Q​  B​​ ​  A2 ​ ​ ∴ ​F​  new​​  = 4F​ Therefore, doubling the charges on the two spheres increases the force by the factor of 4. k ​Q​  ​​ ​Q​  ​​ c) ​F​  new​​  = ​ ______ A  ​ B 2 ​r​​  ​ k ​Q ______​  A​​ ​Q​  B​​ ​F​  new​​  = ​   ​ ​ ( ) 2 ​  1 ​  r ​​​  ​ ​​ __ 3 ​  ​​ ​Q​  ​​ k ​Q ​F​  new​​  = ​ ______ 1 A __ 2  ​ B ​   ​ ​r​​  ​ 9 ​F​  new​​  = 9​(______ ​r​​  ​ ) k ​Q​  ​​ ​Q​  B​​ ​  A2 ​ ​ ∴ ​F​  new​​  = 9F​ Decreasing the distance by the factor of 3, increases the force by the factor of 9. Chapter 4 Answers 203 Science_GR11_LB.indb 203 2018/02/02 17:00

4. a) The charges are like, both positively charged, and by the very nature they repel each other. So, QB will be pushed away from QA by force F. Q B + F b) Like charges will repel each other, as depicted in the force diagram in (a) above. c) The force the charges exert on each other is equal in magnitude. The force depends on the product of the charges, as well as the distance between the charges k ​Q​  ​​ ​Q​  ​​ d) ​F = ​ ______ A  ​ B 2 ​r​​  ​ ​(9 × ​10​​  9​)​​(2, 6 × ​10​​  −6​)​​(3, 2 × ​10​​  −6​)​ _________________________     F = ​      ​ ​​(0, 05)​​​  2​ F = 29, 952 N​ k ​Q​  ​​ ​Q​  ​​ 5. a) ​F = ​ ______ m  ​ n 2 ​r​​  ​ (​ 9 × ​10​​  9)​ ​​(2, 5 × ​10​​  −6​)​​(3, 5 × ​10​​  −6​)​ _________________________     F = ​      ​ ​​(0, 045)​​​  2​ F = 38, 89 N​ b) Doubling the distance will result in the new force being: k ​Q​  ​​ ​Q​  ​​ ​F = ​ ______ m  ​ n ( )2 ​​ 2r ​​​  ​ k ​Q ______ ​  m​​ ​Q​  n​​ F = ​   ​ 4 ​r​​  2​ 1 ​  F __ ∴ F = ​  4 1 ​​ 38, 89 N ​ ​F​  new​​  = ​ __ 4 ( ) ​F​  new​​  = 9, 72 N​ 6. The spheres are carrying equal charges. Therefore, we have Q1 = Q2 = Q: k ​Q​  ​​ ​Q​  ​​ ​ F = ​ _______ 1  ​ 2 2 ​r​​  ​ k ​Q​​  2​ _____ F = ​   ​ 2 ​r​​  ​ 9 × ​10​​  9​ ​Q​​  2​ 3, 6 × ​10​​  −4​  = ​ ___________  ​ ​​(0, 04)​​​  ​ 2 ​Q​​  2​  = 6, 4 × ​10​​  −17​ ∴ Q = 8 × ​10​​  −9​  C or Q = 8 nC​ 7. a) There are three charges in the system. Each will experience a force due to the two charges within its vicinity (i) Two Forces (ii) Two Forces (iii) Two Forces 204   Technical Science | Grade 11 Science_GR11_LB.indb 204 2018/02/02 17:00

b) Sphere A will experience the attractive force due to sphere B, and the repulsive force due to the sphere C. The following is the force diagram, showing the forces acting on sphere A: Q A FCA + FBA k ​Q​  ​​ ​Q​  ​​ ​​F​  BA​​  = ​ ______ A  ​ B 2 ​r​​  ​ ( 9)( −6)( −6) ​ 9 × ​10​​  ​ ​​ 3 × ​10​​  ​ ​​ 5 × ​10​​  ​ ​ _____________________ ​F​  BA​​  = ​         ​ 2 ​(0, 12 )​​  ​ ​F​  BA​​  = 9, 375 N​ k ​Q​  ​​ ​Q​  ​​ ​​F​  BA​​  = ​ ______ A  ​ C 2 ​r​​  ​ ( 9)( −6)( −6) ​ 9 × ​10​​  ​ ​​ 3 × ​10​​  ​ ​​ 2 × ​10​​  ​ ​ ​F​  BA​​  = ​ _____________________         ​ ​(0, 07 + 0, 12 )​​  ​ 2 ​F​  CA​​  = 1, 50 N​ Take the direction to the right as positive: → ​​ F​​  net​​  = 9, 375 N − 1, 50 N → ​​ F​​  net​​  = +7, 875 → ∴ ​​ F​​  net​​  = 7, 875 N to the right​ c) Q B FAB + FCB k ​Q​  ​​ ​Q​  ​​ ​​F​  BA​​  = ​ ______ A  ​ B 2 ​r​​  ​ ( 9)( −6)( −6) ​ 9 × ​10​​  ​ ​​ 3 × ​10​​  ​ ​​ 5 × ​10​​  ​ ​ ​F​  BA​​  = ​ _____________________         ​ 2 ​(0, 12 )​​  ​ ​F​  BA​​  = 9, 375 N​ k ​Q​  ​​ ​Q​  ​​ ​​F​  CB​​  = ​ ______ B  ​ C 2 ​r​​  ​ ( 9)( ​ 9 × ​10​​  ​ ​​ 5 × ​10​​  _____________________ −6)( −6) ​ ​​ 2 × ​10​​  ​ ​ ​F​  CB​​  = ​         ​ ​(0, 07 )​​  ​ 2 ​F​  CB​​  = 18, 39 N​ Let the direction to the right be positive → → → ​​​ F​​  net​​  = ​​ F​​  BA​​  + ​​ F​​  CA​​ → ​​ F​​  net​​  = 17, 367 N+​(− 9, 375 N)​ → ​​ F​​  net​​  = +8, 992 → ∴ ​​ F​​  net​​  = 8, 992 N to the right​ Chapter 4 Answers 205 Science_GR11_LB.indb 205 2018/02/02 17:00

8. a) Identical spheres will equally share their when brought into contact. So the new charge will be: ​Q​  ​​ ​​Q​  new​​  = ​ ____ ​ total 2 +4, 8 × ​10​​  −9​  C + ​(− 3, 2 × ​10​​  −9​  C)​ ​Q​  new​​  = ​ ________________________         ​ 2 ​Q​  new​​  = +8, 0 × ​10​​  −10​​ C k ​Q​  ​​ ​Q​  ​​ b) ​F = ​ ______ A  ​ B 2 ​r​​  ​ (9 × ​10​​  9​)​​(8, 0 × ​10​​  −10​)​​(8, 0 × ​10​​  −10​)​ ​__________________________     F = ​      ​ ​(0, 05 )​​  2​ F = 2, 304 × ​10​​  −6​  N​ Activity 4.2: Calculating the electric field strength 1. a) Electric field is a region of space in which an electric charge experiences a force. b) The electric field strength at a point is defined as the magnitude of the force experienced by a Coulomb of charge, divided by the charge. 2. a) ​E = ​  F __ ​ Q 4 × ​10​​  ​ _______ −6 E = ​   ​ 5 × ​10​​  ​ −9 E = 800 N ∙ ​C​​  −1​​ b) The charge of an electron is Q e = 1,6 × 10–19 C. The force experienced by an electron due to the field will be: F = QE F = 1,6 × 10–19 × 800 F = 1,28 × 10–16 N 3. The charge of an alpha particle is: ​​Q​ α​​  = 2 × 1, 6 × ​10​​  −19​  = 3, 2 × ​10​​  −19​  C​ The force that an alpha particle will experience due to the given field, assumed to uniform, and the alpha particle assumed to be behaving like a point charge: F = QE F = (300)(3,2 × 10–19) F = 9,6 × 10–17 N Activity 3: Application of Electric Field Strength 1. a) b) c) _ –Q –Q +Q –Q Electric field of a positive point charge 206   Technical Science | Grade 11 Science_GR11_LB.indb 206 2018/02/02 17:00

2. a) + + + + + + + + + + + – – – – – – – – – – – b) ​ V ​ __ E = ​  d 15 ____ E = ​   ​ 0, 02 E = 750 N ∙ ​C​​  −1​​ c) Since the electric field between two parallel plates is inversely proportional to the distance between the plates, doubling the distance will half the electric field of the two plates. 1 ​ × 750 N ∙ ​C​​  −1​ __ ​​E​  new​​ = ​  2 ​ E​  new​​ = 375 N ∙ ​C​​  −1​​ 3. E d 4. a) + + + + + + + + + + + – – – – – – – – – – – b) ​ V ​ __ E = ​  d E = ​  1 000 _________  ​ ​(3, 2 × ​10​​  ​)​ −2 E = 3, 125 × ​10​​  4​  N ∙ ​C​​  −1​​ c) (i) The charge is positively charged. It is being pushed away from the positive terminal of the parallel plates, while simultaneously attracted to terminal A. (ii) F = QE F = (3,2 × 10–12) (3,125 × 104) F = 1,00 ×10–7 N Activity 4.4 Electric circuits 1. A circuit will only work if it is complete. This means that it must make a complete loop, otherwise the electrical energy be transferred. A circuit is a path that allows charges to pass through it. The conventional direction of the electric current is from the positive terminal to the negative terminal. Chapter 4 Answers 207 Science_GR11_LB.indb 207 2018/02/02 17:00

2. When the switch is closed a path for electrons to flow is created. The battery, which stores chemical energy, acts as the ‘electron-pump’, converts the chemical energy to electrical energy. The evidence of this is the light given off by the bulb. The electron pressure will start to accumulate at the negative terminal. As a result; they will start to flow towards the positive terminal through the wires. By convention, the electrical current flows from the positive terminal through to the negative terminal EXPERIMENT 10 Variables Independent: current Dependent: Potential difference (voltage) Results Trial No. Ammeter reading (A) Voltmeter reading (V) 1 0,02 6 2 0,04 12 3 0,06 18 4 0,08 24 5 0,10 30 Graph of potential difference versus current The graph is a straight line, which is a graph of a linear function with the following linear equation: y = mx + c Making y = ΔV x = ΔI​ Extrapolating we get c = 0 ΔV ​​ ___ m = gradient = ​  ΔI The gradient of the graph represent the resistance of the resistor Calculating gradient (slope) of the graph: ​gradient = ​  ΔV ___ ​ ΔI 30 − 6 _________ 24  ​ gradient = ​   ​  = ​ ____ 0, 10 − 0, 02 0, 08 gradient = 300​ Therefore, the resistance of the resistor is R = 300 ​Ω​ 208   Technical Science | Grade 11 Science_GR11_LB.indb 208 2018/02/02 17:00

Conclusion The graph is a straight line that extends through the origin. Therefore, the potential difference across a resistor is directly proportional to strength of the electric current through it, provided that the temperature remains constant (resistant is constant). ​​ ΔV Hence R = ___ΔI  ​​ = constant Activity 4.5 Ohms Law 1. a) 0.007 0.006 0.005 Current (A) 0.004 0.003 0.002 0.001 0 0 1 2 3 4 5 6 7 Potential difference (V) ΔV ​  = ​  1 b) The gradient of the graph is ​m = gradient = ​ ___ __ ​​. Therefore, the resistance will be the ΔI R reciprocal of the gradient of the graph _____ ​I​  ​​  − ​I​  ​​ c) ​m = ​  ​V​  2 1 ​​  − ​V​  ​  ​ 2 1 1, 5 − 0, 375 _________ m = ​   ​ 12 − 3 _____ 1, 125 m = ​   ​ 9 1 __ ∴ R = ​  m ​ 9 _____ R = ​   ​ 1, 125 R = 8Ω​ d) The graph is a straight line that extends through the origin. Therefore, the current is directly proportional to the voltage. V ​ 2. ​R = ​ __ I 3 ___ R = ​   ​ 0, 4 R = 7, 5Ω​ V ​ 3. a) ​R = ​ __ I V  ​ ___ 10 = ​  0, 4 V = 2, 1 V​ Chapter 4 Answers 209 Science_GR11_LB.indb 209 2018/02/02 17:00

b) ​ Q = IΔt Q = (0, 21 ) (10 ) Q = 2, 1 C​ w __ ​ V = ​  Q w  ​ ___ 2, 1 = ​  2, 1 w = (2, 1 ) (2, 1 ) w = 4, 41 J​ 4. V ​ a) ​I = ​ __ R ___ ​4, 5 I = ​  15 I = 0, 3 A​ b) V = IR V = (3)(12) V = 36 V V ​ c) ​R = ​ __ I 9 ___ R = ​   ​ 0, 5 R = 18Ω​ d) V = IR V = (2,5)(3) V = 7,5 V V ​ __ e) I = ​  R 220 ____ I = ​  ​  ​ 80 I = 2, 75 A​ Activity 4.6 Series and parallel combination 1. a) R1 and R2 are connected in parallel. _____ ​R​  ​​ ​R​  ​​ ​​R​  12​​  = ​  ​R​  ​​  1 2 + ​R​  ​  ​ 1 2 (3 ) (6 ) ​R​  12​​  = ​ _____ ​ 3+6 18 ​R​  12​​  = ​ ___  ​ 9 ​R​  12​​  = 2Ω​ ​∴​ Rtotal = R12 + R3 Rtotal = 2​Ω​+ 2​Ω​ Rtotal = 4​Ω​ b) The ammeter A1 measures the total current of the circuit. ​​I​  total​​  = ​  ​RV ____ ​ ​  ​​ total 210   Technical Science | Grade 11 Science_GR11_LB.indb 210 2018/02/02 17:00

12 ​ ​I​  total​​  = ​ ___ 4 ​I​  total​​  = 3 A​ c) The voltmeter V2 measures the potential difference across resistor R3. Through that resistor flows the total current of the circuit. ​∴​ V2 = ItotalR3 V2 = (3)(2) V2 = 6 V d) The reading on the voltmeter V1 will be: V1 = V – V2 V1 = 12 V – 6 V V1 = 6 V This reading will be the same across both the resistors connected in parallel, that is R1 and R2. Therefore, the current flowing through the resistor R2 will be: ​V​  ​​ ​​I​  2​​  = ​ ___ 1 ​R​  ​​  ​ 2 ​I​  2​​  = ​  6 __ ​ 6 ​I​  2​​  = 1 A​ e) Decrease. If R1 burns out, the total resistance of the circuit will decrease. According to Ohm’s Law, the current flowing in the circuit is inversely proportional to the resistance of the circuit. Therefore, the total current, measured by the ammeter A1 will decrease. 2. a) The potential difference across the resistor R1 is: V1 = I1R1 V1 = (0,5)(5) V1 = 2,5 V b) The series combination of the resistors R1 and R2 is connected in parallel to the resistor R3. The current flowing through the resistor R1 is the same as the current flowing through the resistor R2, and is calculated as follows: I1 + I3 = Itotal I1 + 1,5 A = 2 A I1 = 0,5 A c) I2 = I1 = 0,5 A d) The resistance of the resistor R2 is: ​V​  ​​ ​​R​  2​​  = ​ ___ 2 ​I​  ​​  ​ 2 3, 5 ​R​  2​​  = ​ ___ ​ 0, 5 ​R​  2​​  = 7Ω​ Chapter 4 Answers 211 Science_GR11_LB.indb 211 2018/02/02 17:00

e) The potential difference across the resistor R3 is the same as the combined potential difference across the resistors R1 and R2: V3 = V1 + V2 V3 = 2,5 V + 3,5 V V3 = 6 V f) The resistance of the resistor R3 is: ​V​  ​​ ​​R​  3​​  = ​ ___ 3 ​I​  ​​  ​ 3 6 ​R​  3​​  = ​ ___ ​ 1, 5 ​R​  3​​  = 4Ω​ g) The resistors R4 and R5 are connected in parallel. We now the current flowing through the resistor R5, we can calculate the current flowing through the resistor R4: Itotal = I4 + I5 2 A = I4 + 1 A ​∴​ I4 = 1 A h) The resistance of the resistor R4 will be: ​V​  ​​ ​​R​  4​​  = ​ ___ 4 ​I​  ​​  ​ 4 4 ​ ​R​  4​​  = ​ __ 1 ​R​  4​​  = 4Ω​ i) The potential difference across the resistor R5 will be the same as that across the resistor R4, the resistors are connected in parallel. ​∴​ V5 = 4 V j) The resistance of the resistor R5 will be same as that of the resistor R4, because the two resistors are equally sharing their current and are connected in parallel: ​∴​ R5 = R4 = 4​Ω​ k) The potential difference across the resistor R6 is: V6 = I6R6 V6 = (2)(2) V6 = 4 V l) The current flowing through the resistor R6 is the total current of the circuit. ​∴​ I6 = 2 A m) The emf of the battery, which is the total potential difference of the circuit will be: VT = V5 + V6 + V3 VT = 4 V + 4 V + 6 V VT = 14 V 212   Technical Science | Grade 11 Science_GR11_LB.indb 212 2018/02/02 17:00

n) The effective resistance of the circuit will be: ​V​  ​​ ​​R​  T​​  = ​ ___ T ​I​  ​​  ​ T 14 ​ ​R​  T​​  = ​ ___ 2 ​R​  T​​  = 7Ω​ Activity 4.7 Ohmic and non-Ohmic conductors 1. a) Ohm’s law states that the current flowing through the conductor is directly proportional to the potential difference across the ends of the conductor, given that the temperature of the conductor stays constant. b) An ohmic conductor is the conductor that obeys Ohm’s law in all aspects. c) Potential difference is the work done (or energy transferred) per unit charge d) Electric current is the rate at which charges flow in the circuit e) Resistance is the opposition to the flow of charges in the electric circuit 2. ohmic conductors: Copper, Nichrome, platinum, graphite Non-ohmic conductors: acid battery, semiconductors, ionized gases, vacuum tubes Review activity Multiple Choice Questions 1. D 2. B 3. D Long Questions k ​Q​  ​​ ​Q​  ​​ ______ 4. ​ F = ​  1  ​ 2 2 ​r​​  ​ (​ 9 × ​10​​  9)​ (​​ 3 × ​10​​  _____________________ −9)( ​ ​​ 5 × ​10​​  −9)​ ​     F = ​      ​ ​​(0, 4)​​​  2​ F = 8, 44 × ​10​​  −7​  N​ 5. a) Coulomb’s law states that the electrostatic force between two charged objects is directly proportional to the product of the charges, and inversely proportional to the square of the distance separating the two charges. Chapter 4 Answers 213 Science_GR11_LB.indb 213 2018/02/02 17:00

b) +Q –Q k ​Q​  ​​ ​Q​  ​​ c) ​F = ​ ______ 1  ​ 2 2 ​r​​  ​ (9 × ​10​​  9​)​​(6 × ​10​​  −9​)​​(1, 10 × ​10​​  −8​)​ ​________________________     F = ​      ​ ​​(0, 02)​​​  2​ F = 1, 35 × ​10​​  −3​  N​ d) During this physical process, the two spheres will equally share their total charge, such the new charge on each sphere will be: ______ ​Q​  ​​  + ​Q​  ​​ ​​Q​  new​​  = ​  1  ​2 2 6 × ​10​​  −9​ + ​(− 1, 0 × ​10​​  −8​)​ ​Q​  new​​  = ​ _________________     ​ 2 ​Q​  new​​  = − 2 × ​10​​  −9​  C​ e) The new force after separating the two charges will then be: k ​Q​  2 ​ ​F = ​ _____  ​ new 2 ​r​​  ​ ( ​ 9 × ​10​​  ​ ​ ​​ _______________ 9) ( −9) 2 2 × ​10​​  ​ ​​​  ​    F = ​      ​ ​​(0, 06)​​​  ​ 2 F = 1 × ​10​​  −5​  N​ 6. W ​ a) ​V = ​ __ Q V = ​  5 000 ________  ​ 2, 5 × ​10​​  ​ −6 V = 2 × ​10​​  9​  V​ V ​ __ ​E = ​  d E = ​ 2 × ​10​​  ​ 9 _______  ​ −3 5 × ​10​​  ​ E = 4, 00 × ​10​​  11​  N ∙ ​C​​  −1​​ b) F = QE F = (1,6 × 10–19)(4,00 × 1011) F = 6,4 × 10–8 N 7. ​E = ​  V __ ​ d 300 ____ E = ​   ​ 0, 02 E = 1, 5 × ​10​​  4​  N ∙ ​C​​  −1​  from positive plate to negative plate.​ 214   Technical Science | Grade 11 Science_GR11_LB.indb 214 2018/02/02 17:00

8. A 10Ω R1 R2 25Ω 24V 35Ω R3 R4 25Ω Note in the diagram above, we can combine R1 and R2 and treat them as one resistor whose effective resistance is given by: ​R​  ​​ ​R​  ​​ ​​R​  ​(12)​​  = ​ _____ 1 2 ​R​  ​​  + ​R​  ​​  ​​ 1 2 We can also do the same with R3 and R4: _____ ​R​  ​​ ​R​  ​​ ​​R​  ​(34)​​  = ​  ​R​  ​​  3 4 + ​R​  ​​  ​​ 3 4 Now the circuit looks like a simple series circuit with two resistors R12 and R34 A R12 24V R34 _____ ​R​  ​​ ​R​  ​​ a) ​​R​  ​(12)​​  = ​  ​R​  ​​  1 2 + ​R​  ​​  ​ 1 2 ​R​  ​(12)​​  = ​  25 × ______10  ​ 25 + 10 ∴ ​R​  ​(12)​​  = 7, 14Ω​ _____ ​R​  ​​ ​R​  ​​ ​​R​  (​ 34)​​ ​ = ​  ​R​  ​​  3 4 + ​R​  ​​  ​ 3 4 ​R​  ​(34)​​  = ​  35 × ______20  ​ 35 + 20 ∴ ​R​  ​(34)​​  = 12, 72Ω​ ​​R​  T​​  = ​R​  ​(12)​​ ​ + ​R​  ​(34)​​​ ​R​  T​​  = 7, 14 + 12, 72 ∴ ​R​  T​​  = 19, 86Ω​ Chapter 4 Answers 215 Science_GR11_LB.indb 215 2018/02/02 17:00

b) To calculate the total current, we need to know what the total resistance of the circuit is, and also what is the total voltage supplied to the external circuit. In this case we know both the quantities. So: ​V​  ​​ ​​I​  T​​  = ​ ___ T ​R​  ​​  ​ T 24  ​ ​I​  T​​  = ​ _____ 19, 86 ∴ ​I​  T​​  = 1, 21 A​ c) (i) For R12, R1 and R2 the voltage is the same. If the current and the voltage that passes through either one of the 3 quantities are known, we can with ease calculate the voltage of the three (which has the same value). The current passing through R12, and R34 is the same as the total current, (resistors in series have the same current). V12 = IR12 V12 = (1,21)(7,14) ∴ V12 = 8,63 V Now that we know the voltage of R1, we can easily calculate three flowing through R1. ​V​  ​​ ​​I​  1​​  = ​ ___ 1 ​R​  ​​  ​ 1 8, 63 ​I​  1​​  = ​ ____ ​ 10 ∴ ​I​  1​​  = 0, 86 A​ (ii) V34 and V12 are in series. Therefore: V34 + V12 = VT and I34 = I12 = IT = 1,21 A V34 + 8,63 = 24 ​∴​ V34 = 15,37 V ​V​  ​​ ​​I​  3​​  = ​ ___ 3 ​R​  ​​  ​ 3 15, 37 ​I​  3​​  = ​ _____ ​ 25 ​I​  3​​  = 0, 44 A​ Resistors R3 and R4 are connected in parallel to each other. Their voltages have the same value as V34. ​∴​ V3 = 15,37 V Alternative solution V3 = I34 R34 V3 = (1,21)(12,72) ​∴​ V34 = 15,39 V Resistors R3 and R4 are connected in parallel to each other. Their voltages have the same value as V34. ∴​ V3 = 15,39 V ​ ​V​  ​​ ​​I​  3​​  = ​ ___ 3 ​R​  ​  ​ 3 15, 39 ​I​  3​​  = ​ _____ ​ 35 ​I​  3​​  = 0, 44 A​ 216   Technical Science | Grade 11 Science_GR11_LB.indb 216 2018/02/02 17:00

(iii) We have already established that V1 = V2 = V12 ∴​ V2 = 8,63 V ​ And we also know that for resistors in parallel: I1 + I2 = I12 ∴​ I2 + 0,86 = 1,21 ​ ∴​ I2 = 0,35 A ​ Alternative solution We have already established that: V1 = V2 = V12 ∴​V2 = 8,63 V ​ ​V​  ​​ ​​I​  2​​  = ​ ___ 2 ​R​  ​​  ​ 2 8, 63 ​I​  2​​  = ​ ____ ​ 25 ∴ ​I​  2​​  = 0, 345 A​ (iv) We have already established that: V3 = V4 = V34 ∴​ V4 = 15,37 V ​ And we also know that for resistors in parallel: I3 + I4 = I34 ∴​ I4 + 0,44 = 1,21 ​ ∴​ I4 = 0,77 A ​ Alternative solution We have already established that: V3 = V4 = V34 ∴​V4 = 15,37 V ​ ​V​  ​​ ​​I​  4​​  = ​ ___ 4 ​R​  ​​  ​ 4 15, 39 ​I​  4​​  = ​ _____ ​ 20 ∴ ​I​  4​​  = 0, 77 A​ IT 9. This question can best be understood by tracing the path of IT the electric current. Consider the diagram alongside: A1 20V V1 R2 IT IT IT I2 R2 A2 I3 V2 V3 R3 I 3 A3 I3 Chapter 4 Answers 217 Science_GR11_LB.indb 217 2018/02/02 17:00

We can see by following the arrows that the total current is divided at the junction, such that I3 flows towards R3, and I2 flows towards R2. At the second junction we see the currents diverting and combining again, such IT flows through R1. From this statement, it is evident that the resistors R2 and R3 are parallel to each other, at their junction the current divides. Therefore, we can combine R2 and R3, and represent them with their effective resistance R23. Obviously: _____ ​R​  ​​ ​R​  ​​ ​​R​  23​​  = ​  ​R​  ​​  2 3 + ​R​  ​​  ​​ 1 3 IT IT A1 A 1 20V V1 RI IT I 23 IT IT R23 A V 23 Now that there is no junction, it should be clear that I23 = IT. Since R2 and R3 are parallel to each other, it can be established that V2 = V3 = V23. _____ ​R​  ​​ ​R​  ​​ a) ​​R​  23​​  = ​  ​R​  ​​  2 3 + ​R​  ​​  ​ 2 3 ​R​  23​​  = ​  10 ×8 _____  ​ 10 + 8 ​R​  23​​  = 4, 44Ω ∴ ​R​  T​​  = ​R​  23​​  + ​R​  1​​ ​R​  T​​  = 4, 44 + 5 ​R​  T​​  = 9 . 44Ω​ b) The ammeter A1 measures the total current flowing in the external circuit. ​V​  ​​ ​​I​  T​​  = ​ ___ T ​R​  ​​  ​ T 20 ​I​  T​​  = ​ ____  ​ 9, 44 ​I​  T​​  = 2, 12 A​ c) Across the end of the potential difference (voltage) is given by: V1 = IR1 V1 = (2,12)(5) V1 = 10,6​Ω​ d) We have already established that V2 = V3 = V23 From the second diagram above, it is clear that I23 = I1 = IT I23 and I1 are connected in series ​∴​ V23 = IR23 V23 = (2,12)(4,44) V23 = 9,41 V 218   Technical Science | Grade 11 Science_GR11_LB.indb 218 2018/02/02 17:00

​∴​ V2 = 9.41 V Alternative solution V23 + V1 = VT V23 + 10,6 = 20 ​∴​ V23 = 9.4 V e) The ammeter A3 measures the current that flows through the resistor R3. ​V​  ​​ ​​I​  3​​  = ​ ___ 3 ​R​  ​​  ​ 3 9, 41 ​I​  3​​  = ​ ____ ​ 8 ​I​  3​​  = 1, 18 A​ f) V3 = V23 = 9,41 V g) A2 measures the current flowing through R2. ​V​  ​​ ​​I​  2​​  = ​ ___ 2 ​R​  ​​  ​ 2 9, 41 ​I​  2​​  = ​ ____ ​ 10 ​I​  2​​  = 0, 941 A​ Alternative solution Since R2 and R3 are in series, therefore: I2 + I3 = IT I2 + 1,18 = 2,12 I2 = 0,94 A 10. Brightness of a bulb is associated to the current strength flowing the bulb. The higher the current, the brighter the bulb a) The bulbs are identical and have the same current strength flowing through them. Their brightness will be the same. But, because of increased resistance, they will glow dimmer. A=B=C b) The current is divided at the junction, such that, a certain portion goes toward C and the another portion towards A and B. But, since the resistance of the branch C, is higher than that of A and B, the current at C is higher that of A and B branch. Therefore, A and B will be dimmer, but have the same dimness, and B will be the brightest. A = B, C c) All three bulbs are identical and connected parallel to each other. This reduces their effective resistance, thereby increasing the total current strength. The bulbs will equally share the current. Therefore, they will glow equally brighter, and very brighter, relative to the bulbs is 3.1. A=B=C d) The connection of the wire between the two bulbs means the bulbs are short-circuited. This is because electric current will take the route with the lower resistance, in this case the wire has a comparable lower resistance. Therefore, the bulbs will not glow at all Chapter 4 Answers 219 Science_GR11_LB.indb 219 2018/02/02 17:00

10. Ohm’s law states that the current flowing through the conductor is directly proportional to the potential difference across the ends of the conductor, given that the temperature of the conductor stays constant. 11. V (V) I (A) 12. The main difference between an ohmic and a non-ohmic conductor is whether they follow Ohm’s law. An Ohmic conductor would have a linear relationship between the current and the voltage. With non-Ohmic conductors, the relationship is not linear. More information; follow the link below: http://www.differencebetween.net/technology/difference-between-ohmic-and-non-ohmic- conductors/#ixzz489W82boh 13. ohmic conductors: Copper, Nichrome, platinum, graphite Non-ohmic conductors: acid battery, semiconductors, ionized gases, vacuum tubes 14. Length of the conductor Thickness of the conductor Temperature Nature of the material 15. a) Number of Cells V (V) I (A) ​R = ​  V __ ​​(Ω)​​ I 1, 44 1 1.44 0.14 ​R = ​ ____ ​ 0, 14 R = 10, 3Ω​ 2 2.88 0.29 9,9​Ω​ 3 4.33 0.43 10,1​Ω​ 10, 3 + 9, 9 + 10, 1 b) ​​R​  avg​​  = ​ _____________     ​ 3 ​R​  avg​​  = 10, 1Ω​ 220   Technical Science | Grade 11 Science_GR11_LB.indb 220 2018/02/02 17:00

c) 0,5 0.45 0,4 0,35 stroom (V) 0,3 0,25 0,2 0,15 0,1 0,05 0 0 1 2 3 4 5 spanning (V) ​V​  ​​  − ​V​  ​​ d) ​m = ​ _____2 ​I​  ​​  − ​I​  ​​  ​ 1 2 1 0, 43 − 0, 14 _________ m = ​   ​ 4, 33 − 1, 44 m = 0, 100 A ∙ ​V​​  −1​ 1 __ ∴ R = ​  m ​ = 10Ω​ e) The answers are approximately the same. Due to some experimental errors, they are not exactly the same. 16. Where the current branches, that’s where we find resistors connected in parallel. In our case that’s R2 and R3. Combining the two resistors gives us the following diagram: IT IT R1 = 2.2Ω IT IT R 4 = 1Ω R23 IT IT R5 = 3.3Ω IT IT 9V V I R R1 V1 = I1R1 ​∴​ I1 = 1,02 A 2,2​Ω​ V1 = 1,02 × 2,2 V1 = 2,244 V Chapter 4 Answers 221 Science_GR11_LB.indb 221 2018/02/02 17:00

V I R ​V​  ​​ R2 V23 = V2 = V3 ​​I​  2​​  = ​ __ 2 ​R​  ​​  ​ 4,7​Ω​ 2 2, 39 V23 + V1 + V4 + V5 = VT ​I​  2​​  = ​ ____ 4, 7  ​ V23 + 2,244 + 1,02 + 3,366 = 12 ​I​  2​​  = 0, 51 A​ ​∴​ V23 = 2,39 V ​∴​ V2 = 2,39 V ​V​  ​​ R3 ​∴​ V3 = 2,39 V ​​I​  3​​  = ​ __ 3 ​R​  ​​  ​ 4,7​Ω​ 3 2, 39 ​I​  3​​  = ​ ____ 4, 7  ​ ​I​  3​​  = 0, 51 A​ R4 V4 = I4R4 I4 = 1,02 A 1​Ω​ V4 = 1,02 × 1 V4 = 1,02 V _____ ​R​  ​​ ​R​  ​​ R5 V5 = I5R5 I5 = 1,02 A ​​R​  23​​  = ​  ​R​  ​​  2 3 + ​R​  ​​  ​ 2 3 4, 7 × 4, 7 V5 = 1,02 × 3,3 ​R​  23​​  = ​ ________ 4, 7 + 4, 7  ​ V5 = 3,366 V ​R​  23​​  = 2, 35Ω ​​R​  T​​  = ​R​  1​​  + ​R​  23​​  + ​R​  4​​  + ​R​  5​​ ​R​  T​​  = 2, 2 + 2, 35 + 1 + 3, 3 ​R​  T​​  = 8, 85Ω​ 17. IT IT I2 I2 R1 = 2.2Ω R 2 = 2.2Ω IT I3 I3 R4 = 1Ω R3 = 4.7Ω IT I2 I3 IT IT IT 9V From the sketch, it is evident that the current is divided at the branch of R2 and R3. Therefore, these are the two resistors which are parallel. We then combine the represent them with their effective resistance, as shown in the sketch below: 222   Technical Science | Grade 11 Science_GR11_LB.indb 222 2018/02/02 17:00

IT IT R1 = 2.2Ω IT R4 = 1Ω R23 IT IT IT IT 9V The circuit is now reduced to a circuit with three resistors, R1, R23 and R4, connected in series. To complete the table, it is clear that the easiest quantity to calculate is the total resistance of the circuit, followed by the total current flowing in the circuit. But first we need to calculate the effective resistance of R2 and R3. V I R R1 V1 = I1R1 ​∴​ I1 = 2,01 A 2,2Ω​ V1 = 2,01 × 2,2 V1 = 4,42 V R2 V23 + V1 + V4 = VT ​V​  ​​ 6,8​Ω​ ​​I​  2​​  = ​ __ 2 ​R​  ​  ​ 2 V23 + 4,42 + 2,01 = 12 5, 57 ​I​  2​​  = ​ ____ ​ ​∴​ V23 = 5,57 V 6, 8 ​∴​ V2 = 5,57 V ​I​  2​​  = 0, 82 A​ R3 V3 = 5,57 V ​V​  ​​ 4,7​Ω​ ​​I​  3​​  = ​ __ 3 ​R​  ​​  ​ 3 5, 57 ​I​  3​​  = ​ ____ 4, 7  ​ ​I​  3​​  = 1, 19 A​ Alternatively I2 + I3 = IT 0,82 + I3 = 2,01 ​∴​ I3 = 1,19 A R4 V4 = IT × R4 I4 = 2,01 A 1​Ω​ V4 = 2,01 × 1 V4 = 2,01 V Chapter 4 Answers 223 Science_GR11_LB.indb 223 2018/02/02 17:00

V I R Total 12 V ​V​  ​​ _____ ​R​  ​​ ​R​  ​​ ​​I​  T​​  = ​ __ T ​R​  ​  ​ ​​R​  23​​  = ​  ​R​  ​​  2 3 + ​R​  ​  ​ T 2 3 12  ​ ​I​  T​​  = ​ ____ 4, 7 × 6, 8 5, 98 ​R​  23​​  = ​ ________ ​ 4, 7 + 6, 8 ​I​  T​​  = 2, 01 A​ ​R​  23​​  = 2, 78Ω ​R​  T​​  = ​R​  1​​  + ​R​  23​​  + ​R​  4​​ ​R​  T​​  = 2, 2 + 2, 78 + 1 ​R​  T​​  = 5, 98Ω​ 18. a) emf (electromotive force) is the potential difference across the battery when there is no current flowing in the circuit. b) Internal resistance is the resistance within a battery, or other voltage source, that causes a drop in the source voltage when there is a current. c) The electrical resistance of an electrical conductor is a measure of the difficulty to pass an electric current through that conductor. d) The energy transferred to an electrical device per unit charge e) Electric current is the flow of electric charges through a conductor. Or Electric current is a flow of charge carriers through a conducting media. 19. a) In a series circuit, voltage is divided between the resistors as the current flows. b) In a series circuit, current flowing through the resistors is the same c) In a series circuit, resistance increases as more resistors are added 20. a) In a parallel circuit, voltage across all the resistors is the same. b) In a parallel circuit, current is divided between the resistors c) In a parallel circuit, resistance decreases as more resistors are added. 224   Technical Science | Grade 11 Science_GR11_LB.indb 224 2018/02/02 17:00

Chapter 5 Heat and thermodynamics Activity 5.1 Calculations for the amount of heat Q = cm∆T Q = 450 J∙kg –1∙K–1 × 10 kg × (100 – 0)K Q = 450 000 J Experiment 1 1. Water is an excellent material to use in a calorimeter because it has a very high specific heat, making it very hard for water to raise its temperature even by 1 degree Celsius, but as hard as it was to raise the temperature of the water, it is equally as hard to lower the temperature of the water-making it able to effectively retain heat and allowing the other material inside of the calorimeter to absorb that heat. 2. Metal (Pb) 3. Water 4. Yes 5. Substances with a low specific heat capacity heat up quickly - but then they lose their heat quickly. Substances with a high specific heat capacity require a large amount of heat to be added to change their temperature – but then they hold their heat much longer. This should be the case in this experiment. Activity 5.2 Q = ∆U + W Q = 5 + 20 Q = 25 J Review exercise Multiple choice questions 1. B 2. C 3. D Long questions 4. Heat is thermal energy in transit and thermal energy is the average kinetic energy of molecules in a substance. is the total energy (kinetic and potential) that molecular particles that which make up a substance possess. 5. Thermal energy flows from hot to cold objects 6. The specific heat capacity of water is higher than most substances 7. ​​Q​  released​​ = cmΔT = ( )​(0,75 g) ​(80 − 30 )​​  ​  C = 156,9 J​ 4, 184 J ​ ______ ​  ∘  ​ ° g ​.​​  ​  C 8. ​​Q​  gained by water​​ = ​c​  water​​  × ​m​  water​​  × Δ​T​  water ​​ = ( )​(100 g ) ​(25.0 − 10.0 )​​  ​  C = 6 276 J​ 4,184 J ​ ______ ​  ∘  ​ ∘ g ​.​​  ​  C ​Q​  gained by water​​ 6276 J 0,84 J ​​C​  Al​​ = _________ ​   ​ = ​ _________________     ​ = 0,8368 ≈ ~0,84 J⋅​g​​  −1​  ​⋅​​ ∘​C = _____    ​  ∘  ​​ ​m​  Al​​  × Δ ​T​  Al​​ (100 g ) (100,0 − 25,0 ) g ​.​​  ​  C Chapter 5 Answers 225 Science_GR11_LB.indb 225 2018/02/02 17:00

9. It is important that a lock and its key be made of the same or similar materials so that both materials can contract and expand at the same rate with changing temperature conditions. Otherwise, the key would not open the lock if one expands or contracts more than the other. 10. ​Δ​U = Q – W = 250 J – 600 J = –350 = 350 J 11. This is not a violation of the second law of thermodynamics. The second law of thermodynamics states that heat can never spontaneously flow from a substance with low temperature to a substance with high temperature. In steam engines, thermal energy from the heat source transfers to the water molecules at low temperature. The thermal energy enables the molecules now in gaseous phase to do work on the piston which produces motion. 12. According to the second law of thermodynamics thermal energy can never spontaneously flow from the refrigerator with low temperature to your room at a higher temperature. An external device would have to be used to force the cool air to move from the refrigerator into your room. 226   Technical Science | Grade 11 Science_GR11_LB.indb 226 2018/02/02 17:00

Chapter 6 Chemical change Activity 6.1 1. a) Let the unknown oxidation number be x The oxidation number of Cr in ​​CrO​ 2− 4 ​  ​​ x + 4(–2) = –2 x – 8 = –2 x = +6 Therefore, the oxidation number of Cr in ​​CrO​ 42−​  ​​ is +6 The oxidation number of Cr in Cr(OH)3 is 3, knowing that the charge of the OH – is –1: x + 3(–1) = 0 x = +3 Therefore, the oxidation number of Cr in Cr(OH)3. The oxidation number of sulphur in S2– is –2 Sulphur in S(s) has the oxidation number of zero The oxidation state of hydrogen in OH– is: –2 + x = –1 x = +1 Therefore, the oxidation number of H in OH– is +1 Hydrogen in H2O: 2x + (–2) = 0 x = +1 Therefore, hydrogen on H2O has the oxidation number +1 b) From the calculations above, we can conclude that Cr has been reduced from +6 in ​​ ​  ​​ to +3 in Cr(OH)3. Therefore ​​CrO​  2− CrO​  2− 4 4 ​  ​​is an oxidising agent, because it has undergone reduction. The oxidation number of S has increased from –2 in S2– to 0 in S(s). This indicates that S2– has undergone oxidation, and therefore, is an oxidising agent. 2. a) The oxidation number of Mn in MnO2 is: x + 2(–2) = 0 x = +4 While in Mn2+, the oxidation number of Mn is +2. This indicates that MnO2 has undergone reduction. And so, this is a redox reaction. OR The oxidation number of Cl in Cl– is –1, while in Cl2, the oxidation number of Cl is zero. Again the change in oxidation of a species in a chemical reaction indicates that such a chemical reaction is a redox reaction. Chapter 6 Answers 227 Science_GR11_LB.indb 227 2018/02/02 17:00

3. In this chemical reaction, all the species maintain their oxidation states. And for that reason this is not a redox reaction. Starting the half reactions, we have: Fe → Fe2+ + 2e– (Oxidation) 2H+ + 2e –→ H2 (Reduction) _____________________________________ Fe + 2H+ → Fe2+ + H2 (Net Reaction) 4. The oxidation number of N in NO2: x + 2(–2) = 0 x = +4 While in NH3 the oxidation number of N is: x + 3(–1) = 0 x = +3 The calculation reveals that NO2 has undergone reduction because the N atom has been reduced from +4 to +3. Hence, NO2 is the oxidising agent. On the other hand, H in H2 has an oxidation number of zero. While in H2O, the oxidation number of H is: 2x + (–2) = 0 x = +1 Indicating that the hydrogen atom has been oxidised. Therefore H2 is a reducing agent. 5. a) H in CaH2 has an oxidation number calculated as follows: 2 + 2x = 0 x = –1 Therefore, in CaH2 hydrogen has an oxidation number of –1 b) I in NaIO3 has an oxidation number calculated as follows: 1 + x + 3(–2) = 0 x = +5 Therefore, the oxidation number of I in NaIO3 is +5 c) C in C4H12 4x + 12(1) = 0 x = –3 Carbon in C4H12 has the oxidation number of –3 d) Sulphur in Al2(SO4)3: 2(+3) + 3x + 12(–2) = 0 3x = 18 x = +6 So, sulphur in C4H12 has an oxidation state of +6 228   Technical Science | Grade 11 Science_GR11_LB.indb 228 2018/02/02 17:00

e) Lead in PbSO4: x + 6 + 4(–2) = 0 x = +2 In, PbSO4 has the oxidation state of +2 f) Cl in ​​Cl​  −3​ ​​: x + 3(–2) = –1 x = +5 g) P in ​​HPO​  2− 3 ​  ​​: 1 + x + 3(–2) = –2 x – 5 = –2 x = +3 h) O in Na2O2 2(1) + 2x = 0 x = –1 Activity 6.2 1. a) S in ​​SO​  2− 3 ​  ​​ x + 3(–2) = –2 x = +4 In ​​SO​  4​ 2−​​ x + 4(–2) = –2 x = +6 Sulfur has been oxidised, and so this is an oxidation half reaction b) Ca in CaO: x + (–2) = 0 x = +2 But the oxidation number of Ca atom, in Ca, is zero. c) Cr in ​​Cr​  2− 4 ​  ​​ x + 4(–2) = –2 x = +6 While in ​​Cr​  2​​ ​O​  2− 7 ​  ​​ 2x + 7(–2) = –2 2x = +12 x = +6 No change oxidation number for Cr atom. Hence, this is not a redox reaction d) In ​​Cr​  4​ 2−​​, the oxidation number of Cr is: x + 4(–2) = –2 x = +6 Chapter 6 Answers 229 Science_GR11_LB.indb 229 2018/02/02 17:00

While in Cr3+ the oxidation number of Cr is +3, indicating that Cr has been reduced. So this is reduction half reaction. e) ​  ​​ Mn in ​​MnO​  2− 4 x + 4(–2) = –2 x = +6 Mn is reduced. Hence this is reduction half reaction f) Cl in ​​ClO​  −2​ ​​ x + 2(–2) = –1 x = +3 while in ClO– x + (–2) = –1 x = +1 Therefore, this is a reduction half reaction 2. a) ​ 3 ​Ag​​  +​  + Ni → ​Ni​​  3+​  + 3Ag​ Reduction: 3Ag + + 2e– → 3Ag Oxidation: Ni → Ni3+ + 3e– b) ​​Cu​​  2+​  + Pb → ​Pb​​  2+​  + Cu​ Reduction: Cu2+ + 2e– → Cu Oxidation: Pb → Pb2+ + 2e– c) ​​O​​  + 2Sn → ​O​2−​  + 2 ​Sn​​  2+​​ 2 Reduction: O2 + 2e – → O2– Oxidation: Sn → Sn2+ + 2e– Activity 6.3 ​  ​  + ​Fe​​  2+​ → ​Cr​​ 3+​  + ​Fe​​  3+​​ (i) ​​Cr​  2​​ ​O​  2− 7 Oxidation number of Cr atom in ​​Cr​ 2​​ ​O​  2− 7 ​  ​​ 2x + 7(–2) = –2 2x = +12 x = +6 The calculation reveals that the Cr atom has been reduced from +6 to +3 in Cr3+. On the other hand, the Fe atom has been oxidised from +2 to +3. The reducing agent is Fe2+, while the oxidising agent is ​​Cr​  2​​ ​O​  2− 7 ​  ​​ (ii) ​​ClO​  −3​ ​  + Re → Re​O​  −4​ ​  + ​ClO​​  −​​ The oxidation number of Re atom in ​Re ​O​ −4​ ​​ x + 4(–2) = –1 x = +7 This shows that the Re atom has been oxidised from the oxidation state of zero in to the oxidation number of +7 in ​Re ​O​  −4​ ​​. Therefore, Re is a reducing agent. On the other hand, the oxidation number of the Cl atom in ​​ClO​ −3​ ​​: 230   Technical Science | Grade 11 Science_GR11_LB.indb 230 2018/02/02 17:00

x + 3(–2) = –1 x = +5 While in ClO–: x + (–2) = –1 x = +1 The Cl atom has been reduced from the oxidation state of +5 in ​​ClO​  −3​ ​​to the oxidation state of +1 ClO–. Therefore, ​​ClO​  −3​ ​​is an oxidising agent. ​  ​  + ​Br​  2​​ → ​S​  2​​  ​O​  2− (iii) ​​SO​  2− 4 3 ​  ​  + ​BrO​  −3​ ​​ ​  ​​: The oxidation number of S atom in ​​SO​ 2− 4 x + 4(–2) = –2 x = +6 While in ​​S​  2​​ ​O​  2− 3 ​  ​​ 2x + 3(–2) = –2 2x = +4 x = +2 Therefore, S atom has been reduced from the oxidation number of +6 in ​  ​​ to +2 in ​​S​  2​​ ​O​  2− ​​SO​  2− 4 3 ​  ​​. Therefore, ​​SO​  2− 4 ​  ​​is an oxidising agent. The oxidation number of Br atom in Br2 is zero. While in ​​BrO​ −3​ ​​ x + 3(–2) = –1 x = +5 This shows that the Br atom has been oxidised. Therefore, Br2 is a reducing agent. (iv) ​​MnO​  −4​ ​  + ​H​  2​​  S → S + MnO​ The oxidation number of Mn in ​​MnO​ −4​ ​​ is: x + 4(–2) = –1 x = +7 While in MnO x + (–2) = 0 x = +2 Therefore, Mn atom has been reduced from +7 to +2. Therefore, ​​MnO​  −4​ ​​is an oxidising agent. On the other hand, the oxidation number of S in H2S 2(+1) + x = 0 x = –2 Therefore, S atom has been oxidised from –2 to zero. H2S is a reducing agent. (v) ​​O​  2​​  + ​Cr​​  3+​ → ​H​  2​​  ​O​  2​​  + ​Cr​  2​​  ​O​  2− 7 ​  ​​ O atom in H2O2 has an oxidation number: 2(1) + 2x = 0 x = –1 While in O2 it is zero. Therefore, O atom has been reduced from zero in O2 to –1 in H2O2. Therefore, O2 is an oxidising agent. Chapter 6 Answers 231 Science_GR11_LB.indb 231 2018/02/02 17:00

The oxidation number of Cr in ​​Cr​ 2​​ ​O​  2− 7 ​  ​​ 2x + 7(–2) = –2 2x = +12 x = +6 The oxidation number of Cr atom has increased from +3 in Cr3+ to +6 in ​​Cr​  2​​ ​O​  2− 7 ​  ​​. Implying that Cr3+ has been oxidised Therefore, Cr3+ is a reducing agent. Formal experiment Questions 1. Justify the use of a safety precaution, other than wearing safety glasses, when carrying out this investigation. 2. Describe how the THREE products of this electrolysis could be identified. 3. Identify the anode and the cathode. 4. Write down the half-reactions taking place in each electrode. 5. Write down the overall redox reactions without spectator ions. 6. Write down the overall redox reactions with spectator ions. 7. Write down the name of spectator ion. 8. Write down TWO uses of the product identified in this experiment. 232   Technical Science | Grade 11 Science_GR11_LB.indb 232 2018/02/02 17:00

Review exercise 1. Oxidation is loss of electrons by substance in a reaction. It is accompanied by an increase in oxidation number. A substance undergoing oxidation is called a reducing agent. 2. Reduction is gain of electrons by a substance in a reaction; it is accompanied by a decrease in oxidation number. A substance undergoing reduction is called an oxidizing agent. 3. Redox or not: a) 2Na + Cl2 ​→​ 2NaCl The oxidation number Na atom in NaCl x + (–1) = 0 x = +1 There oxidation number of Na atom has increased from zero to +1. Therefore, this is a redox reaction. b) C + O2 → CO2 The oxidation number of C in CO2 x + 2(–2) = 0 x = +4 Since there is a change in oxidation number, this is a redox reaction. c) NaCl + AgNO3 → AgCl + NaNO3 In this reaction all the species retained their oxidation states. This is therefore, not a redox reaction. d) 2H2O → 2H2 + O2 Oxidation number of H atom in H2O 2x + (–2) = 0 x = +1 While H atom in H2 has the oxidation number of zero. This change in oxidation number indicates that this is a redox reaction e) Zn + CuSO4 → ZnSO4 + Cu The oxidation number of Cu atom in CuSO4 is, (knowing that the charge of ​S ​O​ 2− 4 ​  ​​ is –2) x + (–2) = 0 x = +2 The oxidation number of atom has changed from to 0. This is therefore, a redox reaction f) H2 + Cl2 → 2HCl Cl atom in Cl2 has the oxidation number of zero. While in HCl 1+x=0 x = –1 Once again, a change in oxidation number implies that this is a redox reaction. g) H2SO4 + 2KOH → K2SO4 + 2H2O In this reaction there has only been a displacement of ions, but no change in oxidation number of any of the species. This is therefor, not a redox reaction. Chapter 6 Answers 233 Science_GR11_LB.indb 233 2018/02/02 17:00

h) 2KClO3 → 2KCl + 3O2 The oxidation number of Cl atom in 2KClO3 1 + x + 3(–2) = 0 x –5 = 0 x = +5 While in KCl 1+x=0 x = –1 Change on oxidation number of any of the species in a chemical reaction means that the reaction is a redox reaction. i) NH3 + HCl → ​N ​H​  +4​  ​​ + Cl– None of the species has changed its oxidation state. This is therefore, not a redox reaction. Marking guidelines for electrolysis of sodium chloride 1. Chlorine gas is produced in this reaction which is toxic. The reaction must be carried out in the fume cupboard. Answer could include: Explosive mixture of H2 and Cl2 and Electrical hazard 2. Cl2 gas Cl2 gas bubbling at anode Universal indicator added was bleached colourless from green near anode by Cl2 Colourless I– ions added were oxidised to brown I2 molecules near anode by Cl2(g). Prescence of I2 confirmed by adding starch (blue/black colour) [any (2)] H2 gas Bubbling at cathode and Gas collected pop test OH– Universal indicator changed from green to purple near cathode indicating OH– ion present 3. Anode is connected to the positive terminal of the battery. Cathode is connected to the negative terminal of the battery 4. Anode: 2Cl – (aq) → Cl2(g) + 2e– Cathode: 2H2O + 2e – → H2(g) + 2OH – 5. Overall reaction: 2Cl – + 2H2O → Cl2(g) + H2(g) + 2OH – 6. 2NaCl(aq) + 2H2O(l ) → Cl2(g) + H2(g) + 2NaOH(aq) 7. Sodium ions 234   Technical Science | Grade 11 Science_GR11_LB.indb 234 2018/02/02 17:00

8. Chlorine gas Chlorine is used for producing safe drinking water. Chlorinated compounds are used mostly for sanitation, pulp bleaching, disinfectants, and textile processing. Chlorine is also used for the manufacture of chlorates and it is important in organic chemistry, forming compounds such as chloroform, carbon tetrachloride, polyvinyl chloride, and synthetic rubber. Other uses of chlorine compounds include dyestuffs, petroleum products, medicines, antiseptics, insecticides, foodstuffs, solvents, paints and plastics. Any 2 Hydrogen gas Hydrogen is widely used for the hydrogenation of vegetable and animal oils and fats. Hydrogen also finds uses in the metallurgy field because of its ability to reduce metal oxides and prevent oxidation of metals in heat treating certain metals and alloys. Hydrogen is extensively used in the synthesis of ammonia and in petroleum refining operations. Sodium hydroxide Sodium hydroxide is used in drain and oven cleaners, chemical manufacturing, oil refining, hydraulic fracturing, water treatment and metal processing. It is also commonly used in the manufacture of fabric, plastic wrap, paper and soap. Chapter 6 Answers 235 Science_GR11_LB.indb 235 2018/02/02 17:00

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