Technical Sciences Grade 12 P1 Eastern Cape Mock 2020 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 SEPTEMBER 2020 TECHNICAL SCIENCES P1/ TEGNIESE WETENSKAPPE V1 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 8 pages./ Hierdie nasienriglyn bestaan uit 8 bladsye.

Downloaded from hlayiso.com 2 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/SEPTEMBER 2020) QUESTION/VRAAG 1 1.1 C  (2) 1.2 D  (2) 1.3 C  (2) 1.4 A  (2) 1.5 C  (2) 1.6 A  (2) 1.7 D  (2) 1.8 D  (2) 1.9 B  (2) 1.10 A  (2) [20] QUESTION/VRAAG 2 2.1 (2) 2.2 An object will continue its state of rest or uniform motion in a straight line unless it is acted upon by a net force.  ʼn Voorwerp sal is sy toestand van rus bly of aanhou beweeg met uniforme beweging in ʼn reguitlyn tensy ʼn netto krag daarop inwerk.  (2) 2.3 Block will continue its state of uniform motion (constant velocity).  Die blok sal aanhou beweeg met uniforme beweging (konstante snelheid).  (2) [6] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2020) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 3 QUESTION/VRAAG 3 3.1 When a net force acts on an object, the object will accelerate in the direction of the force. This acceleration is directly proportional to the net force and inversely proportional to the mass of the object.  Indien ʼn netto krag op ʼn voorwerp inwerk, sal die voorwerp in die rigting van die krag versnel. Die versnelling is direk eweredig aan die netto krag en omgekeerd eweredig aan die massa van die voorwerp.  (2) 3.2 No net external force act on the system.  Daar is geen eksterne krag wat op die stelsel inwerk nie.  (2) 3.3 (5) 3.4.1 Fnet = ma  Any one / Enige een T – f = ma 50 – (0,2) (20)(9,8) = 20a  a = 0,54 m.s-2 left/west / links/ wes (Positive marking from QUESTION 3.4.1)(Positiewe merk vanaf VRAAG 3.4.1) (4) 3.4.2 OPTION/OPSIE 1 Fnet = ma Fcosθ – T – µkN = ma Any one / Enige een  FH – T – f = ma Fcos 30°  – 50 – (0,2)( 30x9,8 – Fsin30) = (30)(0,54)  F(cos30° + (0,2) sin30°) = 125 F = 129,4 N  OPTION/OPSIE 2 Fnet = ma  Any one / Enige een FH – f1 – f2 = ma Fcos30° – (0,2)[(30x9,8) – Fsin30]  – (0,2) (20)(9,8) = (50)(0,54) F = 129,4 N  (5) 3.5 3.5.1 Decreases / Neem af  (1) 3.5.2 Decreases / Neem af  (1) 3.6 Fnet decreases, therefore acceleration decreases.  Fnet neem af, dus neem die versnelling ook af.  (2) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 4 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/SEPTEMBER 2020) QUESTION/VRAAG 4 4.1 Momentum is the product of mass of the object and its velocity.  Momentum is die produk van die massa van die voorwerp en sy snelheid.  (2) 4.2 Momentum (1) 4.3 ∑pi = ∑pf East = +ve)(Oos = +) Any ONE / Enige EEN  m1v1i + m2v2i = m1v1f + m2v2f (5)(4) + (3)(0)  = (5)( - 0,25) + (3)(v2f) v2f = 7,08 m.s-1 east / oos  (4) 4.4 Law of conservation of momentum / Wet van behoud van momentum  (Negative marking)(Negatiewe merk) Total linear momentum in a closed system remains constant.  Totale lineêre momentum in ʼn geslote sisteem bly konstant.  (3) [10] QUESTION/VRAAG 5 5.1 100 000 N  due east / reguit ooswaarts  (2) 5.2 -ve Newton’s Third law of Motion / Newton se Derde Bewegingswet  When object A exerts a force on object B, object B simultaneously exerts an oppositely directed force of equal magnitude on object A.  Indien voorwerp A ʼn krag op voorwerp B uitoefen, oefen voorwerp B ʼn krag uit wat in die teenoorgestelde rigting maar van gelyke grootte op voorwerp A uit.  (3) 5.3 It is the product of the resultant/net force acting on an object  and the time the resultant/net force acts on the object. Dit is die produk van die resultante/netto krag wat op ʼn voorwerp inwerk  en die tyd wat die resultante / netto krag op die voorwerp inwerk.  (2) 5.4 OPTION/OPSIE 1 West is taken as positive / Wes word as positief geneem. Fnet ∆t = ∆p  Fnet(truck) ∆t = ( mvf – mvi) (- 100000)(0,4) = (5000)[(vf) – 20]  vf = 12 m.s-1 west / wes  OPTION/OPSIE 2 East is taken as positive / Oos word as positief geneem. Fnet ∆t = ∆p  Fnet(truck) ∆t = ( mvf – mvi) (100000)(0,4) = (5 000)[(vf) – (-20)]  vf = - 12 m.s-1 vf = 12 m.s-1 west / wes  (4) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2020) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 5 QUESTION/VRAAG 6 6.1 6.1.1 Work done is defined as the product of the force applied on an object and the displacement in the direction of the force.  Die arbeid verrig word gedefinieer as die produk van die toegepaste krag op die voorwerp en die verplasing in die rigting van die krag.  (2) 6.1.2 Tension  Force of gravity  Spanning  Gravitasiekrag  (2) 6.1.3 Fnet = 0  Block of bricks is moving with constant velocity Houer met die bakstene beweeg met konstante snelheid  (2) 6.1.4 P = Fv  F = mg = (50)(9,8) = 490 N  P = (490)(5)  P = 2 450 W  (4) 6.2 6.2.1 Total mechanical energy of a closed system remains constant. Totale meganiese energie van ʼn geslote sisteem bly konstant. (2) 6.2.2 ME at a height h above the ground: ME op hoogte h bokant die grond: MEh = ½mv2 + mgh  MEh = ½(0,5)(10)2 + (0,5)(9,8)h  …….. (1) ME⅔h = ½(0,5)(20,34)2 + (0,5)(9,8)(⅔h )  …… (2) (1) = (2) ½(0,5)(10)2 + (0,5)(9,8)h = ½(0,5)(20,34)2 + (0,5)(9,8)(⅔h ) (⅓h)(9,8)(0,5) = 78,43 h = 48,02 m  ME = ½(0,5)(10)2 + (0,5)(9,8)(48,02)  ME = 260,3 J  (6) 6.2.3 (ME)at h = (ME)ground  260,3 = mgh + ½mv2 260,3 = 0 + ½(0,5)(v2)  v = 32,27 m.s-1  (3) 6.3 6.3.1 0J Normal force is perpendicular to the direction of the displacement.  Normaalkrag is loodreg aan die rigting van verplasing.  (3) 6.3.2 W = F∆xcosθ  W = (10)(4)(cos0°)  W = 40 J  (3) [27] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 6 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/SEPTEMBER 2020) QUESTION/VRAAG 7 7.1 A body which regains its original shape and size completely when the deforming force is removed.  Dit is ʼn liggaam wat sy oorspronklike grootte en vorm volkome herwin as die vervormingskrag verwyder word. (2) 7.2 The internal restoring force per unit area is called stress.  Die interne herstelkrag per eenheidsoppervlakte van die liggaam word druk genoem.  (2) 7.3 7.3.1 Stress is directly proportional to strain.  Druk is direk eweredig aan rekking.  (2) 7.3.2 Hooke’s law  Hooke se wet  (1) 7.3.3 Tungsten  Graph of tungsten is steeper than steel.  Grafiek van tungsten (wolfram) is steiler as staal.  (2) 7.3.4 5,4 ×109 −0 m=  3×10−2 −0 m = 1,8 × 10 11 Modulus of elasticity / Elastisiteitsmodulus = 1,8 × 1011 Pa  (3) 7.4 7.4.1 The property of the fluid to oppose relative motion between the two adjacent layers.  Die eienskap van ʼn vloeistof om die relatiewe beweging tussen twee aangrensende vlakke teen te werk.  (2) 7.4.2 Water  The lower the viscosity, the faster the flow.  Hoe laer die viskositeit – hoe vinniger die vloei. (2) 7.4.3 As the temperature of the fluid increases, its viscosity decreases.  Indien die temperatuur van die vloeistof toeneem, verminder sy viskositeit.  (2) 7.5 7.5.1 Pascal's law states that in a continuous liquid at equilibrium, the pressure applied at a point is transmitted equally to the other parts of the liquid.  Pascal se wet stel dat in ʼn kontinue vloeistof in ewewig die druk wat by enige punt toegepas word eweredig na die ander dele van die vloeistof versprei word.  (2) 7.5.2 F1 F2 =  A1 A2 1765 F2 =  284×10−6 507×10−6 F2 = 3 150,9 N  (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/SEPTEMBER 2020) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 7 7.5.3 Increases / Neem toe  (1) 7.5.4 Bulldozer’s working systems, dentists’ chairs, hydraulic lifts used to lift heavy loads, car jacks  (ANY TWO) Stootskraper se werkende sisteme, tandarts-stoele, hidrouliese hysers om swaar vragte te lig, motors se domkragte.  (ENIGE TWEE) (2) [27] QUESTION/VRAAG 8 8.1 8.1.1 The process of adding impurities to intrinsic / inpure semiconductors. Die proses waardeur onsuiwerhede by intrinsieke / onsuiwer halfgeleiers gevoeg word.  (2) 8.1.2 5 (1) 8.1.3 n-type / n-tipe  (2) 8.1.4 In n-type material current is due to electrons whereas in p-type the current is due to holes.  In n-tipe materiaal is die stroom as gevolg van elektrone terwyl in p- tipe is die stroom as gevolg van holtes.  (2) 𝜀 𝐴 8.2 8.2.1 C= 0  𝑑 8,85 ×10−12 ×10×10−4 C =  1×10−2 C = 8,85 × 10 F  -13 Q = CV  Q= (8,85 × 10-13)(12)  Q = 1,06 × 10-11 C  (6) 8.2.2 Zero / Nul  (1) 8.2.3 Area of the plate / Oppervlak van die plate Distance between the plate / Afstand tussen die plate. ( Any TWO) Dielectric used / Diëlektrikum gebruik (Enige TWEE) (2) 8.3 8.3.1 Energy transferred per second. / Energie oorgedra per sekonde.  (2) 8.3.2 1 1 1 𝑅1 𝑅2 = +  Rp =  𝑅 𝑅1 𝑅2 𝑅1 + 𝑅2 1 1 1 26 × 52 = +  OR/OF =  𝑅 26 52 26 + 52 1 3 = 𝑅 52 = 17,33 Ω  R = 17,33 Ω  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 8 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/SEPTEMBER 2020) 8.3.3 OPTION/OPSIE 1 OPTION/OPSIE 2 𝑉2 𝑉 230 P=  I= = = 8,85 A  𝑅 𝑅 26 (230)2 P = VI P=  26 P = 2 034,62 W  = (230)(8,85)  = 2 034,62 W  (3) 8.3.4 OPTION/OPSIE 1 OPTION/OPSIE 2 W = Pt  W = VIt  = (2 034,62)(2)(60)  = (230)(8,85)(120)  = 244 154,4 J  = 244 154,4 J  (3) [27] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief