PHY-NOV-P1-QP and MEMO_hlayiso.com_.pdf

basic education ) y Department: (Gy Basic Education 5 REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE | @ OSS SS SSSSSSSSSSSSS5555 005555555555 ec5555555555 m5) : PHYSICAL SCIENCES: PHYSICS (P1) " | n NOVEMBER 2018 : MARKS: 150 TIME: 3 hours This question paper consists of 15 Pages and 2 data sheets. | Copyright reserved Se | Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 2 DBE/November 2018 CAPS — Grade 11 INSTRUCTIONS AND INFORMATION 1. 10. a. 12. Write your name and class (e.g. 11A) in the appropriate spaces on the ANSWER BOOK. This question paper consists of 12 questions. Answer ALL the questions in the ANSWER BOOK. Start EACH question on a NEW page in the ANSWER BOOK. Number the answers correctly according to the numbering system used in this question paper. Leave ONE line between two subquestions, e.g. between QUESTION 2.1 and QUESTION 2.2. You may use a non-programmable calculator. You may use appropriate mathematical instruments. You are advised to use the attached DATA SHEETS. Show ALL formulae and substitutions in ALL calculations. Round off your FINAL numerical answers to a minimum of TWO decimal places. Give brief motivations, discussions, etc. where required. Write neatly and legibly. Copyright reserved NAAM FA I AAT Pte ase tre ever

Downloaded from hlayiso.com Physical Sciences/P1 3 DBE/November 2018 CAPS — Grade 11 QUESTION 1: MULTIPLE-CHOICE QUESTIONS Various options are provided as possible answers to the following questions. Choose the answer and write only the letter (A-D) next to the question numbers (1.1 to 1.10) in the ANSWER BOOK, e.g. 1.11 D. 11 Two forces, F; and F2, act on a point. If F; and Fy» act in the same direction the maximum resultant has a magnitude of 13 N. If forces F, and F2 act in opposite directions the magnitude of the minimum resultant is 3 N. The magnitude of the two forces, in newton, is ... A 8and5. B16 and 10. C 3and 10. D 10 and 7. (2) 1.2 A _free-moving block slides down an inclined plane at a CONSTANT VELOCITY. This means that the ... A _ frictional force acting on the block is zero. B net force acting on the block is in the direction down the slope of the plane. C net force acting on the block is zero. D component of weight parallel to the plane is greater than the frictional force, (2) 13 A trolley is pushed along a horizontal surface with a force of 150 N at an angle of 45° to the horizontal. The trolley experiences a constant frictional force of 60 N. 150 N 60 N The NET FORCE acting on the trolley: (i) Causes the trolley to accelerate horizontally (ii) Is equal to the applied force (iii) Is horizontally forward Which of the statements above are CORRECT? A (i) and (ii) B (ii) and (iii) C (i) and (iii) D (i), (ii) and (iii) Copmintresaved YAMA ANAM, P2256 oe

Physical Sciences/P1 Downloaded from hlayiso.com DBE/November 2018 CAPS — Grade 11 1.4 A man ina lift is moving upwards ata CONSTANT SPEED. The weight of the man is W. According to Newton's Third Law, the reaction force of the weight W is the force of ... A _ the floor on the man. B Earth on the man. C the man on the floor. D the man on Earth. (2) 415 The optical density of a medium ... A will be high if the refraction of light is less. Bis ameasure of the refracting power of the medium. C __ is less when light bends towards the normal when entering the medium. D will be high if light moves faster through the medium. (2) 1.6 In which ONE of the graphs below will the gradient represent the refractive index of a material when light passes from the air through the material? A 0,4 5 sin 0,4 Li or sin 0; D . 8; # sin 0,4 oy 6; sin 9, (2) Copyright reserved NAY AANA ATHY Pease tarm oven

Downloaded from hlayiso.com Physical Sciences/P1 5 DBE/November 2018 id 1.8 1.9 CAPS — Grade 11 Every point on a wave front acts as a point source of spherical, secondary waves that move forward at the same speed as the wave. This statement represents ... A Snell's law. B_ Huygens' principle. C refraction. D _ the law of reflection. Three charges of magnitudes +2q, +2q and -2q are shown in the sketch below. “24 @ di -2q 4) -0,,, Which arrow CORRECTLY indicates the direction of the NET FORCE acting on the -2q charge? a a Which ONE of the sketches below represents the CORRECT magnetic field pattern around a straight current-carrying conductor? Coowrahresered NAUMANN AIA! e951 ove (2) (2)

Physical Sciences/P 1 CAPS — Grade 11 Downloaded from hlayiso.com DBE/November 2018 1.10 Which ONE of the graphs below CORRECTLY represents the relationship between potential difference and current in a non-ohmic resistor? A VA (Vv) Cc VA (Vv) Copyright reserved ——— > 1 (A) 1A) (2) [20} HNN AOA AG AAA UMN] Pte @se arm oven

Downloaded from hlayiso.com Physical Sciences/P1 DBE/November 2018 CAPS — Grade 11 QUESTION 2 (Start on a new page.) Two forces, of magnitudes 50 N and 80 N, act at a point on a Cartesian plane in the directions shown in the sketch below. vw 50)N 30 TN] s0N 2.1 Give the correct term for the following description: A single vector having the same effect as two or more vectors together (1) 2.2 Calculate the: 2.2.1 Magnitude of the vertical component of the 50 N (2) 2.2°2 Magnitude of the resultant (net) force (5) 2209 Direction of the resultant (net) force (2) [10] Copy eseres NANA sem oe

Downloaded from hlayiso.com Physical Sciences/P1 DBE/November 2018 CAPS — Grade 11 QUESTION 3 (Start on a new page.) A box, with a mass of 45 kg, is pulled with a force of 90 N at an angle of 50° to the horizontal. The box moves at a CONSTANT VELOCITY. 90 N 45 kg — f--==-------- 3.1 Define the term kinetic frictional force. (2) 3.2 State Newton's First Law of Motion in words. (2) 33 Calculate the magnitude of the horizontal component of the applied force. (2) 3.4 Calculate the magnitude of the normal force. (4) 3.5 Calculate the coefficient of kinetic friction. (4) 3.6 Will the coefficient of kinetic friction change if the angle of the applied force is decreased? Write only YES or NO and give a reason. (2) [16] Conyrahtreseved ANNAN AMINA) 2521 ve

Downloaded from hlayiso.com Physical Sciences/P1 DBE/November 2018 CAPS — Grade 11 QUESTION 4 (Start on a new page.) Learners investigate the relationship between the mass of an object and the acceleration it experiences when a constant net force is applied on the object. They use their results to draw the graph below. Graph of the inverse of acceleration versus mass Hoo EEEREEEEEEE PEPER EE LE eet et + ¢ + Fisescencerseeratrasscseseeree ja S25 FEE ipeerese as 7 het HE = oooh t Hae ro Se t =! wo 2 : cot | Lee i Soeee -4 FH + rt rf 1 T T T t t A HH TE rt FEEL PEPE Trt t | a Ht 14. LI | cH + + i im im i= 1 foapcaa Ett Cee 0 FERRER EEE : 0) 0,25 0,75 1 1,25 1,5 m (kg) 41 State Newton's Second Law of Motion in words. (2) 4.2 Calculate the gradient of the graph. (3) 4.3 Hence, determine the net force applied on the object during the experiment. (2) 44 Write down a conclusion for this experiment. (2) [9] Conwiontesored ANTAL AAA HAT, 282 oe

Downloaded from hlayiso.com Physical Sciences/P1 DBE/November 2018 CAPS — Grade 11 QUESTION 5 (Start on a new page.) A crate, with a mass of 25 kg, slides down a plane which is inclined at 15° to the horizontal. During the first part of the motion, from A to B, there is no friction between the crate and the plane, but part BC has a rough surface. 54. Draw a free-body diagram of ALL the forces acting on the crate while it moves from B to C. (3) 5.2 Calculate the magnitude of the acceleration of the crate while it moves from A to B. (4) 53 Write down the direction of the acceleration of the crate while it slows down from B to C. Write only UP THE SLOPE or DOWN THE SLOPE. (1) 5.4 The magnitude of the net acceleration from B to C is 1,2 m-s™. Calculate the magnitude of the frictional force acting on the crate. (4) [12] QUESTION 6 (Start on a new page.) The gravitational force on a probe, called Curiosity, on the surface of Mars is 3 338 N. The radius of Mars is 3 390 km and the mass of the planet is 6,39 x 10% kg. 6.1 State Newton's Law of Universal Gravitation in words. (2) 62 Calculate the mass of the probe. (4) 6.3 Calculate the weight of the probe on Earth. (2) Copyrontresores MINNA MANIA L252 ve

Physical Sciences/P1 a DBE/November 2018 Downloaded from hlayiso.com CAPS — Grade 11 QUESTION 7 (Start on a new page.) A glass prism is placed at the bottom of a container filled with water. A light ray passes from the air through the water into the glass prism. The light ray changes direction every time it passes into a new medium. hel ad 7.3 14 75 7.6 air water glass Name the phenomenon described by the underlined words above. If the refractive index of water and air is 1,33 and 1 respectively, calculate the angle 8 between the light ray and the SURFACE OF THE WATER if the angle of refraction in the water is 40°. The angle of refraction in the glass is 35°. Calculate the refractive index of glass. Draw the sketch below and complete the diagram of the path of the light ray from the air to the water to the glass. Show ALL the values of the angles of incidence, angles of refraction and normal in EACH medium. \ air water glass Calculate the speed of light through the glass prism if the refractive index of glass is 1,5. Is it possible that total internal reflection of the light ray can occur in the above situation? Write only YES or NO. Commienresees NAAT, 256 on (1) (4) (3) (5) (3) (1) [17]

Downloaded from hlayiso.com Physical Sciences/P1 DBE/November 2018 CAPS — Grade 11 QUESTION 8 (Start on a new page.) An experiment is performed to investigate the effect of wavelength on the degree of diffraction. Monochromatic light shines through a slit with a width of 0,002 mm and the pattern produced is shown on a screen. Screen Slit Monochromatic 4 light ray 8.1 Define the term diffraction. (2) 8.2 Write an investigative question for this experiment. (2) The degree of diffraction is recorded for different colours of monochromatic light and the results are shown on the graph below. Degree of diffraction > Wavelength (nm) 8.3 Write the mathematical relationship between wavelength and the degree of diffraction. (2) 8.4 Which colour of light, RED or GREEN, has the largest degree of diffraction? (1) The experiment is repeated with only green light with a wavelength of 560 nm, but the slit width is changed and the degree of diffraction is recorded. 8.5 Copy the set of axes below into your ANSWER BOOK and draw a graph showing the relationship between slit width and degree of diffraction. Degree of diffraction > Slit width (mm) (2) [9] Conyrontrserve NTL MUNI! 29st ver

Downloaded from hlayiso.com Physical Sciences/P1 13 DBE/November 2018 CAPS - Grade 11 QUESTION 9 (Start on a new page.) A small isolated sphere A, with a mass of 0,2 g, carrying a charge of +7 x 10° C, is suspended from a horizontal surface by a string of negligible mass. A second sphere B, carrying a charge of -8 x 10° C, on an isolated stand, attracts sphere A so that the string forms an angle of 20° to the vertical. The horizontal distance between the centres of the two spheres is 3 cm. Refer to the diagram below. --3cm___ 9 ~8x10°C +7x10°C ©) 9.1 State Coulomb's law in words. (2) 6.2 Draw a VECTOR DIAGRAM of the forces acting on sphere A. Indicate at least ONE angle. (4) 9.3 Calculate the magnitude of the electrostatic force that sphere B exerts on sphere A. (4) 9.4 Calculate the magnitude of the tension force in the string. (3) [13] QUESTION 10 (Start on a new page.) Two points, P and T, are situated 3 mm apart in the electric field of positive charge Q, as shown below. Q P T @ occu MHP EAe een eee ° *—~ 3mm — 10.1 Draw the electric field pattern around charge Q. (2) The magnitude of the electric field at point P is 4 x 10° N-C” and at point T the magnitude is 2,5 x 10°N-C"1. 10.2 Calculate: 10.2.1 The ratio of the electric field at point P to the electric field at point T. Write the answer as Ep : Ey. (1) 10.2.2 The distance between charge Q and point P (4) 10.2.3 The magnitude of charge Q (2) [9] Cooyrinresereg HNN ANA) 286m oer

Downloaded from hlayiso.com Physical Sciences/P1 14 DBE/November 2018 CAPS — Grade 11 QUESTION 11 (Start on a new page.) A SQUARE induction coil with a side length 3 cm and 400 windings, is placed perpendicularly in a uniform magnetic field and then rotated through an angle of 45° in 0,08 s. b= ---fe---- >| }----f----- >| N Lu ----5| 8 } ---— ---->| An emf of 7 V is induced in the coil. 11.1 State Faraday’s law of electromagnetic induction in words. (2) 11.2 Calculate the change in the magnetic flux. (3) 11.3 Calculate the magnitude of the magnetic field. (4) The coil is now rotated through an angle of 45° in 0,05 s. 11.4 How will the induced emf be affected? Write only INCREASE, DECREASE or STAY THE SAME. (1) 11.5 Explain the answer to QUESTION 11.4. (1) The north pole of a bar magnet is pushed into a solenoid, as shown in the sketch below. AAAAAN N |X VV VVV | eer, 11.6 Which pole will be induced at point X? Write only NORTH or SOUTH. (1) 14.7 In which direction will the induced current flow? Write only FROM A TO B or FROM B TOA. (1) [13] conwoteseres MINA MUTA NPP =5 © ver

Downloaded from hlayiso.com Physical Sciences/P 1 15 DBE/November 2018 CAPS ~ Grade 11 QUESTION 12 (Start on a new page.) Consider the circuit diagram below. The internal resistance of the battery and any resistance in the wires can be ignored. 4R a & 2R 12.1 Calculate the value of resistor R if the total resistance of the circuit is 4,8 Q. (3) 12.2 Calculate the reading on the voltmeter if the current through the 4R resistor is 1,8 A. (5) 12.3 Calculate the energy converted in resistor 4R in 2 minutes. (3) The 4R resistor is replaced with an ammeter. 12.4 How will the reading on the voltmeter be influenced? Write only INCREASE, DECREASE or STAY THE SAME. (1) 12.5 Explain the answer to QUESTION 12.4. (2) [14] TOTAL: 150 cooniditoomauat DUE A A A

Physical Sciences/P1 Downloaded from hlayiso.com CAPS — Grade 11 DBE/November 2018 DATA FOR PHYSICAL SCIENCES GRADE 11 PAPER 1 (PHYSICS) GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 11 VRAESTEL 1 (FISIKA) TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES NAME/NAAM SYMBOL/S/MBOOL VALUE/WAARDE a van pa Aarde Re 6,38 x 10° m Mass of Earth M 5,98 x 10% kg Massa van die Aarde TABLE 2: FORMULAE/TABEL 2: MOTION/BEWEGING FORMULES Vv, =v, +adt Ax = v,At+ at? 2 2 Vy =V, +2aAx i (“4 At 2 FORCE/KRAG Fret = ma w=mg Fr om i" = “neti Copyright reserved KWAZULU-NATAL Please turn over

Downloaded from hlayiso.com Physical Sciences/P1 DBE/November 2018 CAPS — Grade 11 WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG v=fa T= n, sin 0; =n, sin 0, n= ELECTROSTATICS/ELEKTROSTATIKA kQ,Q Fae r (k= 9,0 x 10°N-m?-C?) | E = =A E r2 (k = 9,0 x 10°N-m?-C) | n= ELECTROMAGNETISM/ELEKTROMAGNETISME e=- Nee @®=BAcos0 At ELECTRIC CIRCUITS/ELEKTRIESE STROOMBANE ja ane At | es Dg tens R= Gt fe Hay tes Romy fh Ww = P=— W=Vq At W=VIAt P=VI W=FPRAt 4 P=IR VAt We 2 R piv Genistein OANA AA

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Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe Vi 3 DBE/November 2018 CAPS/KABV — Grade/Graad 11 - Marking Guidelines/Nasiennigiyne QUESTION 2/VRAAG 2 24 221 2.2.2 22.3 Copyright reserved/Kopiereg voorbehou Resultant (net) vector/Resultante (netto) vektor ¥ (ty Fy=Fsin @ = 50sin 30° — ORIOF 80cos 60° =25NY (2) POSITIVE MARKING FROM QUESTION 2.2.1 POSITIEWE NASIEN VANAF VRAAG 2.2.1 Fx = 50cos 30° ¥ =43,3N > PA =80- v ae! ea Substitution marks awarded within the question 2 even if calculations for Fx and Rx are wrong Fret? = Re + = Substitusiepunte toegeken in die vraag selfs indien = 36,72 +252 berekeninge vir F, en Rx verkeerd bereken word. =4441NV - _ (5) POSITIVE MARKING FROM QUESTION 2.2.1 AND 2.2.2 POSITIEWE NASIEN VANAF VRAAG 2.2.1 EN 2.2.2 OPTION 1/OPSIE 1 SRA a 25 S o , 0=5574°v ed .. scone af SATE GAG Raed TRETORA TOOT” a Rr= 36,7.N 8 -i1- 12 APPROVED MARKING GUIDELIN? OPTION 2/OPSIE 2 0 = 55,74° v OPTION 3/OPSIE 3 36,7 sind = maa v _ Accept direction as /Aanvaar rigting as ‘ = 90° 6 = 55,749 ¥ Loe | 34,269 vw OPTION 4/0PSIE 4 - 25 eee. yh cos 0 = a 6=55,74° v (2) [10] Please turn over/Blaai om assebliet \ gv Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/Noverber 2018 CAPS/KABY ~ Grade/Graad 11 — Marking Guidelines/Nasienrigtyne QUESTION 3/VRAAG 3 3.1 3.2 3.3 34 3.6 3.6 Copyright reserved/Kopiereg voorbehou The force that opposes the motion of a moving object relative to a surface.¥ “ Die krag wat die beweging van 'n bewegende voorwerp relatief tot 'n opperviak teenwerk. [2 orfof 0} (2) A body will remain in its state of rest or motion at constant velocity “unless a non-zero resultant/net force acts on it. ~ bly/volhard tensy 'n nie-nul resulterende/netto krag daarop inwerk. [Penalise -1 if key words/phrase is omitted/ Penaliseer -1 indien sleutelwoorde/frase is uitgelaat] (2) Fe=90cos 50°“ OR/OF 90sin 40° =57.85N ¥ (2) N=Fo-Fy ¥ .) NOTE/NOTA: N = 45(9,8) v - 90sin 50° v : Weight and the vertical component can be N = 372,06 N¥ calculated separately, award one mark each even (4) if the formula for N is incorrect Gewig en vertikale kemponent kan apart bereken word, een punt elk selfs indien die formule vir N verkeerd is. POSITIVE MARKING FROM QUESTION 3.3 and 3.4 POSITIEWE NASIEN VANAF VRAAG 3.3 en 3.4 fc = peN Y 57,85 ¥ = UK(372,06) ¥ bk = 0,16 (4) No ¥ The coefficient is dependent on the (nature of) the surfaces / type of material in contact. ¥ Nee. Die koéfiisiént is athanklik van die (tipe) opperviakke / soort material in kontak. (2) [18] Please tum over/Blaai om assebiie Gf

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Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/November 2018 CAPS/KABV ~ Grade/Graad 11 — Marking Guidelines/Nasienriglyne QUESTION 6/VRAAG 6 61 6.2 63 Copyright reserved/Kopiereg voorbehou Each particle in the universe attracts every other particle with a gravitational force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. v Elke deeltjiie in die heelal trek elke ander deeltjie aan met 'n krag wat direk eweredig fs aan die produk van hulle massas en omgekeerd ewerediy is aan die kwadraat van die afstand tussen hulle middeipunte [Penalise -1 if key words/phrase is omitted/ Penaliseer -1 indien sleutelwoordefirase is uitgelaal| ) - omm, aD " 8 PS as (667 x10 Dest xie }im) 9 (3.390 x 10°) FE v m= 900 kg ¥ OR/OF ; (6.67 x 10"')(6,39 x 10%) ge (3390 x 10) g=3,71m-s? v Fo= mg 3338 =m(3,71) ¥ m= 900 kg“ (899,73 kg) (4) POSITIVE MARKING FROM QUESTION 6.2 POSITIEWE NASIEN VANAF VRAAG 6.2 w=mg = 900(9,8) ¥ =8820NY (2) [8} 2H eas Private enc Mas gpg 2018 -11- 49 " 2PROVED al MARKING a Please tum over/Blaai om asseblict Physical Sciences P1/Fisiese Wetenskappe V1 8B DBE/November 2018 CAPSIKABV ~ Grade/Graad 11 — Marking Guidelines/Nasienriglyne QUESTION 7/VRAAG 7 uP 7 Refraction/Refraksie ¥ (1) 7.2 | OPTION 1/OPSIE1 OPTION 2/OPSIE 2 nisin 8 = nesin 8 7 ne 208 y tsin 6 = 1,33sin 40° ¥ sin@; Oi= 58,75° sing, ’ 1,33= a" gnag ~ sin 8: = 1,33sin 40° 6\= 58,75° Therefore the angle between ray and surface/Daarom is die hoek tussen invallende straal er opperviak © = 90°- 58,75° v* =31,25°% i (4) 7.3 ni Sin@| = nrsind: 1,33sin 40° v = nisin 35° v m= 149 4 (3) 74 & iefle he a] i Be ut Sik ja 2 3k wb we Sle aT Bl aye 2 Ely eR BS 8 z|8' ie es Allocation of marks/Toekenning van punte: (5) Copyright reserved/Kopiereg voorbehou _Ligstraal breek nog meer na die normaal in glas _Hoeke in glas (35°) . | Ifnormal lines are not indicated, penalise with one mark “een punt | ight ray bends towards normal in water v Ligstraal breek na die normaal in water ight ray bends further towards normal in glass vo Angle of incidence 58,75° shown (OR 31,25°) aa _lnvalshoek 58, 75° aangedui (OF 31,25°) / “Angles in water (40°) a floeke in water (409) i i “Angle in glass (35°) v Indien normaal lyne nie aangedui is nie, penaliseer met. if arrows are omitted, penalise -1 (maximum 4/s) Indien pylpunte weggelaat word, penaliseer -1 (maks “/s) Please lum over/Blaai om assebtiot \\; i

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Downloaded from hlayiso.com Physical Sciences P1/Fisiese Welenskappe V1 4 DBE/November 2018 CAPS/KABV ~ Grade/Graad 11 — Marking Guidelines/Nasienriglyne 93 OPTION 1/OPSIE 4 If Fa and Fe were used/indien Fa en Fs gebruik word Fe 1a.Q, “ NOTE/NOTA: ‘i Due to information given in the question, | = 2x 10° x 107 x 10%) ¥ accept all possible options : 0,03? v | As gevolg van die inligting indie vraag = 5,60 x 104 Nv | gegee, aanvaar alle moontlike opsies | OPTION 2/OPSIE 2 If Fy and Fe were used/Indien Fy en Fe gebruik word Fy =mg = (0,2 x 105)(9,8) v “HehxAteN Fe = (1,96 x 10)tan 70° ¥ alg =713x104N v Biz N DIC — O72 OPTION 3/OPSIE 3 wa SE If Fy and Fe were used/Indien Fy en Fe gebruik word 7 BIS Fy = mg gs Bi = (0,2 x 10°)(9,8) ~ & ay =196x10°N 3a! \ ai?- sin20° sin70®/ a, Y Fe _ (1,96x 10") a sin20° sin 70° Fe=7,13 x 104N¥ (4) 9.4 POSITIVE MARKING FROM QUESTION 9.3 POSITIEWE NASIEN VANAF VRAAG 9.3 OPTION 4/OPSIE 1 OPTION 2/0PSIE 2 i Using Fy and Fe/Gebruik Fg en Fe | Using Fa and angle/Gebruik Fz en hoek | Fg =mg =" = (0,2 x 10°)(9,8) ~ sin 70° = 1,96 x10? N pa 198% 10? ¥ T?= (1,96 x 1097 +(5,6x 104? vo! Sin7os = a (T =2,04x103Nv T#2,09 x10°N v OPTION 3/OPSIE 3 i — ‘ Using Fe and angle/Gebruik Feen NOTEINOTA: = / hoek Due to information given in the / Fe question, accept all possible i ie 08 70° options i As gevolg van die inligting in die 5ex10¢ % vraag gegee, aanvaar alle i Ts 0" ¥ moontlike opsies i T=1,64 x109N v ou a e a ae . (3) Copyright reserved/Kopiereg voorbehou 113] Please tum overBleaiom asseblot f Physical Sciences P1/Fisieso Wetenskappe V1 12 DBE/November 2018 CAPS/KABY — Grade/Graad 11 — Marking Guidelines/Vasienriglyne QUESTION 10/VRAAG 10 10.1 Criteria for marking/Nasienkriteria Shape of the field (minimum of 4 field lines) _Vorm van veld (minimum van 4 veldlyne) Direction of the field | Rigting van veld | _ Lines don’t touch charge/lines cross etc. (maximum ¥%) | | Lyne raak nie lacing/yne kris ens. (maksimum 4) (2) 10.2.1 16:17 (1) o A ccs 2018 -1- 12 PROVED MARKING GUIDELIN., (Uit_ic EXAMINATION Copyright reserved/Kopicreg voorbehou Please turn over/Blaai om asseblief

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Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wotenskappe V1 CAPSIKABV ~ Grade/Graad 11 ~ Marking Guidelines/Nasienriglyne QUESTION 12/VRAAG 12 124 OPTION 1/OPSIE 1 DBE/November 2018 "OPTION 2/OPSIE 2 = RR, oR, FR, R v v @) 12.2 POSITIVE MARKING FROM QUESTION 12.1 POSITIEWE NASIEN VANAF VRAAG 12.1 OPTION 1/OPSIE 1 Ver = [Rar = 1,8(4)(2) v =14,4V NATION 2D MARKING GUIDEUN: TS Examen zoe -11- 12 OPTION 3/OPSIE 3 Ri: Re 4:6 Isle 6:4 Copyright reserved/Kopiereg voorbehou OPTION 2/0PSIE 2 Var = [Rar = 1,8(4)(2) ¥ =144V , y be=3-1,8 =12A Ver = IR ¥ ] = 1,2(4)°¥ =48V v OPTION 4/0PSIE 4 Vae = IRar = 1,8(4)(2) ¥ = 144s R:2R:3R 1:2:3 (Ve: Var: Var fh BUS 2 ¢ =f v Ver v6 x14,4 =4eavv (5) Please turn over/Bisai om asseblief ve Physical Sciences P1/Fisiese Wetenskappe V1 16 OBE/November 2018 CAPS/KAB// ~ GradefGraad 11 — Marking Guidelines/Nasienriglyne 125) POSITIVE MARKING FROM QUESTION 12.1 AND 12.2 _POSITIEWE NASIEN VANAF VRAAG 12.1 EN 12.2 OPTION 1/OPSIE 1 "OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 W=PRAt ¥ W=Viat~ Seat = 1,82(8)(120) ¥ = (14,4)(1,8)(120) v Waa = 10368) ¥ =31104J7 | : w= 4) 20) y i W=3110,.40¥ 12.4 Decrease/Neem af v 12.5 © The ammeter has such a low resistance It short circuits the parallel part and all current flows through the ammeter. “ OR The ammeter short circuits the resistors “ No current flows through resistor 2R ¥ Die ammeter het so ‘n fae weerstand Dit kortsluit die paralleigedeelte en al die stroom vioei deur die ammeter. OF Die ammeter kortsluit die resistors Daar vioei geen stroom deur resistor 2R nie TOTALITOTAAL: 2018 -11- 12 “PROVED MARKING GUIDELIN SUBIC EXAMINATION Copyright reservediKopiorog voorbeliow (3) (1) (2) [14] 150