Physical Sciences Grade 12 P1 Gauteng Mock 2022 Memo (Afr)

Downloaded from hlayiso.com PREPARATORY EXAMINATION/ VOORBEREIDENDE EKSAMEN 2022 MARKING GUIDELINES/ NASIENRIGLYNE 10841 PHYSICAL SCIENCES: PHYSICS/ FISIESE WETENSKAPPE: FISIKA PAPER/VRAESTEL 1 21 pages/bladsye

Downloaded from hlayiso.com 10841/22 QUESTION/VRAAG 1 1.1 C  (2) 1.2 B  (2) 1.3 D  (2) 1.4 B  (2) 1.5 B  (2) 1.6 A  (2) 1.7 D  (2) 1.8 D  (2) 1.9 A  (2) 1.10 Accept any answer. All answers correct. If learner gives no answer, allocate two marks as well.  (2) (10 x 2) [20] 2

Downloaded from hlayiso.com 10841/22 QUESTION/VRAAG 2 2.1 T Tcos18o OR/OF Tsin18 mg/w/Fg mg/w/Fg If learner does free body diagram Indien leerder vryliggaamdiagram op vertically up and down, also accept the en af teken, aanvaar. Vraag het nie answer. Question did not stipulate in posisie van bal aangedui nie. which position the ball was. T w Marking Guidelines: Nasienriglyne: ✓ for each force with label ✓ vir elke krag met ʼn byskrif Arrow with line (2) 3

Downloaded from hlayiso.com 10841/22 2.2 2.2.1 OPTION/OPSIE 1 Marking Guidelines:/ Fg = mg  Nasienriglyne: = 0,05 x 9,8  ✓ Formula = 0,49 N ✓ Fg = 0,49 N (substitution) ✓ Trig calculation/trigonometrie 𝐹 bewerking tan  = 𝐹𝑥 ✓ Answer: Fy = 0,159 N 𝑔 tan 18 = Fx  0,49 ✓ Antwoord: Fy = 0,159 N Fx = 0,159 N  OPTION/OPSIE 2 Tcos 18 =mg ✓ T cos 18 = 0,05 x 9,8 ✓ T = 0,5152N Fx = T sin 18 = 0,5152 x sin18✓ = 0,1592N ✓ (4) 2.2.2 Fnet = ma  If carried over 0,16 instead of 0,159 = 0,05 x a  0,159 a = 3,1842 m.s-2  Final answer 3,2 m.s-2 Learner gets full marks if calculation is correct. Marking Guideline: Nasienriglyne: Positive marking from 2.2.1 Positiewe nasien vanaf 2.2.1 ✓ Fnet = ma ✓ Fnet = ma ✓ Substitution ✓ Invervanging ✓ Final answer ✓ Finale antwoord (3) 2.3 2.3.1 𝐹𝑛𝑒𝑡𝐴 = 𝑚𝑎  = +𝑇 − 𝑓𝑘𝐴 5 x 2,17 = +26 - fkA  fkA = 15,15 N  𝐹𝑛𝑒𝑡𝐵 = 𝑚𝑎 = +𝐹𝑥 − 𝑇 − 𝑓𝑘𝐵 5 x 2,17 = +(65 x cos 40)– 26 - 𝑓𝑘𝐵  𝑓𝑘𝐵 = 12,94 N  ✓ Formula ✓ Whole substitution ✓ Answer ✓ Whole substitution ✓ Answer (5) 4

Downloaded from hlayiso.com 10841/22 2.3.2 𝑓𝑘 = 𝜇𝑘 N 𝜇𝑘 is a constant for a specific surface.  The surfaces can differ. Force (Block B) at an angle will decrease the normal force.  Normal force is directly proportional to the kinetic frictional force, thus if the normal force decreases then the frictional force will decrease.  𝑓𝑘 = 𝜇𝑘 N 𝜇𝑘 is ʼn konstante vir ʼn spesifieke oppervlakte. ✓ Die oppervlaktes kan verskil. Krag (Blok B) teen ʼn hoek sal die normaalkrag verminder. ✓ Normaalkrag is direk eweredig aan die kinetiese wrywingskrag, dus indien die normaalkrag verminder dan sal die wrywingskrag verminder. ✓ (3) [17] QUESTION/VRAAG 3 3.1 An object which has been given an initial velocity and then it moves under the influence of the gravitational force only  ʼn Voorwerp wat ʼn aanvanklike snelheid gegee is en dan beweeg dit onder slegs die invloed van die gravitasiekrag OR/OF Moving body where the only force acting on the object is the force of gravity Bewegende liggaam waar die enigste krag wat op die voorwerp inwerk, gravitasiekrag is Marking criteria/Nasienriglyne If any of the underlined words/phrases in the correct context is omitted deduct 1 mark./ Indien enige van die onderstreepte woorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. (2) 3.2 3,43 m.s-1. ✓ penalise if there is no unit / penaliseer indien geen eenheid (1) 5

Downloaded from hlayiso.com 10841/22 3.3 NOTE: ACCEPTED LABELS/ MARK/ AANVAARDE BYSKRIFTE PUNT w Fg/Fw/weight/0,49N/gravitational ✓✓  force Fg Notes:/Notas: • Mark awarded for label and arrow./Ken punt toe vir byskrif en pyl. • Do not penalise for length of arrows, drawing is not to scale./Moenie penaliseer vir lengte van pyle, nie op skaal geteken nie. • Any other additional force(s) Max. ½ /Enige ander addisionele kragte Max. ½ • If force(s) do not make contact with body. Max. ½ /Indien krag nie kontak maak met kol nie Maks ½ • If no arrows indicated and all forces correctly drawn deduct 1 mark./Indien geen pyle nie maar alle kragte korrek getrek is, trek een punt af (2) 3.4 3.4.1 OPTION/OPSIE 1 ∆𝑦 Gradient/Gradiënt = acceleration/versnelling = ∆𝑥 = 3,92 – 0✓ 0,4 ✓ = 9,8 m.s-2 ✓ OPTION/OPSIE 2 ∆𝑦 Gradient/Gradiënt = acceleration/versnelling = ∆𝑥 = 3,43 – 0 ✓ 0,77 – 0,42 ✓ = 9,8 m.s-2 ✓ OPTION/OPSIE 3 𝑣𝑓 = 𝑣𝑖 + 𝑎𝛥𝑡 3,92✓ = 0 + a(0,40) ✓ a = 9,8 m.s-2 ✓ any equation of motion can be used (3) Option 4 If learner uses the values for the collision during the 0,02 seconds when the ball is in contact with the ground. 𝑣𝑓 = 𝑣𝑖 + 𝑎𝛥𝑡 -3,43 ✓ = +3,92 + a(0,02)✓ a = -367,5 m.s-2 Answer will be 367,5 m.s-2 ✓ Give learner the credit if following this path and doing the correct science and calculations. (3) 6

Downloaded from hlayiso.com 10841/22 3.4.2 OPTION/OPSIE 1 Height/Hoogte = area under the graph/ area onder die grafiek = ½ bh  = ½ x 0,4 x 3,92✓ = 0,784 m  Do not penalise if formula is not given OPTION/OPSIE 2 𝑦 = 𝑣𝑖 𝑡 + ½ g𝑡 2  = 0✓ + ½ (9,8)(0,4)2 = 0,784 m ✓ OPTION/OPSIE 3 Downwards as positive/Afwaarts as positief 𝑣𝑓2 = 𝑣𝑖2 + 2𝑎𝑦  3,922  = 0 + 2 x (9,8)y  y = 0,784 m (downwards/afwaarts)  (4) OPTION/OPSIE 4 Downwards as negative/Afwaarts as negatief 𝑣𝑓 2 = 𝑣𝑖 2 + 2aΔy✓ (-3,92)2 = (0)2 + 2(-9,8) Δy✓ ∴ Δy = -0,784 m ∴ Δy = 0,784 m downwards/afwaarts The height from which the ball was dropped = 0,784 m.✓ Die hoogte waarvan dan die bal laat val is = 0,784 m. OPTION/OPSIE 5 𝑣 +𝑣 𝑦 = ( 𝑓 2 𝑖 ) 𝑡 3,92 + 0 = ( ) 0,4 2 = 0,784 m 7

Downloaded from hlayiso.com 10841/22 3.5 (Downwards taken as positive)/(afwaarts geneem as positief) positive marking on the height / positiewe nasien vir die hoogte OR/OF (Downward taken as negative)/(Afwaarts geneem as negatief) positive marking on the height / positiewe nasien vir die hoogte +0,784 0,77 1,12 Marking Criteria/Nasienriglyne Mark/Punt Graph starting from the maximum height (0,784 m) ✓ Grafiek begin vanaf maksimum hoogte (0,784 m) Time taken to reach the ground indicated ✓ Tyd geneem om grond te bereik aangedui Correct shape of the graph/Korrekte vorm van ✓ grafiek Ground taken as zero/Grond geneem as ✓ nulverwysing (4) 8

Downloaded from hlayiso.com 10841/22 3.6 Net force/resultant force.  Netto krag/resultante krag (2) 3.7 𝐹𝑛𝑒𝑡 = 𝐹𝑔 + 𝐹𝑓𝑙𝑜𝑜𝑟  −18,375 = 0,49 + Ffloor  𝐹𝑛𝑒𝑡 = 18,865 N  Rounding-off will have to be 18,87 N./Afronding sal 18,87 N moet wees. Minus must not be in the final answer. (3) 3.8 DECREASES  𝐹𝑛𝑒𝑡 ∆𝑡 = ∆𝑝 formula refers to proportionality – allocate marks. If the change in momentum remains constant, the net force is inversely proportional to the time of contact. Thus, if the time increases the net force will decrease.  AFNEEM ✓ 𝐹𝑛𝑒𝑡 ∆𝑡 = ∆𝑝 Indien die verandering in momentum konstant bly, is die netto krag omgekeerd eweredig aan die kontaktyd. ✓ Dus, indien die tyd vermeerder sal die netto krag verminder. ✓ (3) [24] 9

Downloaded from hlayiso.com 10841/22 QUESTION/VRAAG 4 4.1 A force for which the work done in moving an object between two points is independent of the path taken. ✓✓ ʼn Krag waarvoor die arbeid verrig om die voorwerp te beweeg tussen twee punte onafhanklik is van die pad wat geneem word. Marking criteria/Nasienriglyne If any of the underlined words/phrases in the correct context is omitted, deduct 1 mark. If the word work is omitted, zero marks. Indien enige van die onderstreepte woorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. Indien die woord arbeid weggelaat is, geen (2) punte. 4.2 Force of gravity OR Gravitational force ✓/ Gravitasiekrag OF swaartekrag (1) 4.3 The net work done on an object by a net force is equal to the change in the object’s kinetic energy.✓✓ Die netto arbeid verrig op ʼn voorwerp deur ʼn netto krag is gelyk aan die verandering in die voorwerp se kinetiese energie. OR/OF The work done on an object by a net force is equal to the change in the object’s kinetic energy. ✓✓/ Die arbeid verrig op ʼn voorwerp deur ʼn netto krag is gelyk aan die verandering in kinetiese energie van die voorwerp. Marking criteria/Nasienriglyne If any of the underlined words/phrases in the correct context is omitted, deduct 1 mark. Indien enige van die onderstreepte woorde/frases in die korrekte konteks uitegelaat is, trek een punt af. (2) 4.4 OPTION/OPSIE 1 Change substitution. K together. Formula Left right answer Wnc = ΔK + ΔU  𝑓𝑘 . 𝛥𝑥. 𝑐𝑜𝑠180 = (½ mv2- ½ mv2) + (mgh - mgh) 𝑓𝑘 𝑥 (−8) = ½(2)6,922 - ½(2) 22 + 0 – 2(9,8)(8)sin 30   𝑓𝑘 𝑥 (−8) = -34,5136 𝑓𝑘 = 4,3142 𝑁 𝑓𝑘 = 4,31 N  (4) 10

Downloaded from hlayiso.com 10841/22 OPTION/OPSIE 2 ℎ = 8 sin 30 = 4𝑚 𝑊𝑛𝑐 = ∆𝐸𝑘 + ∆𝐸𝑝  ∆𝐸𝑘 = ∆𝐸𝑘𝑓 - ∆𝐸𝑘𝑖 = ½ (2) 6,922 – ½ (2) 22  this mark goes for both substitutions = 43,886 J ∆𝐸𝑝 = ∆𝐸𝑝𝑓 - ∆𝐸𝑝𝑖 = 2(9,8)0 - 2(9,8)4 = -78,4 J 𝑊𝑛𝑐 = ∆𝐸𝑘 + ∆𝐸𝑝 = 43,886 – 78,4 = -34,514 J 𝑓∆𝑥 cos 180 = −34,514 𝑓 = 4,31 N  OPTION/OPSIE 3 𝑊𝑛𝑒𝑡 = 𝐹𝑛𝑒𝑡 cos 𝜃 = ∆ 𝐸𝑘  either or 2 Fnet (8)cos 0 = ½ mvf - ½ m vi 2 8F = ½ (2)6,922 - ½ (2) 22  Fnet = 5,486 N Fnet = + Fg// - f 5,486 = ( 2 )(9,8)sin 30 - f  𝑓 = 4,31 N  4.5 OPTION/OPSIE 1 𝑊𝑛𝑒𝑡 = 𝑓𝑘 . 𝑥. 𝑐𝑜𝑠𝜃 = ∆𝐸𝑘  (0,35(2)9,8)cos180o = 0 - ½ (2)(6,922)  - 6,86 x = - 47,8864 x = 6,981 m  OPTION/OPSIE 2 𝑓∆𝑥 cos 𝜃 = ∆ 𝐸𝑘  𝜇𝐾 𝑥 𝑐𝑜𝑠𝜃 = 𝐸𝑘𝑓 – 𝐸𝑖 0,35(2)(9,8) ∆ 𝑥 cos 180   = ½ (2)02 - ½ (2)(6,922) 𝑥 = 6,98 𝑚  (4) [13] 11

Downloaded from hlayiso.com 10841/22 QUESTION/VRAAG 5 5.1 5.1.1 Change in frequency observed or frequency received/The speed of the race car  Frequency of listener as observed by the phone. Verandering in frekwensie waargeneem of frekwensie ontvang/Die spoed van die renmotor Frekwensie van luisteraar soos waargeneem deur die foon. (2) 5.1.2 Speed of sound in air /stationary listener/siren set at 200 Hz Spoed van klank in lug/ stilstaande luisteraar/sirene gestel op 200 Hz (2) 5.1.3 What is the relationship between the change in the frequency that will be heard relative to the speed of the vehicle?  Wat is die verwantskap tussen die verandering in die frekwensie wat gehoor sal word relatief tot die spoed van die voertuig? OR/OF How can the speed of the race car be determined using the change in the frequency and the relative motion between the observer and the source?  Hoe kan die spoed van die renmotor bepaal word deur gebruik te maak van die verandering in die frekwensie en die relatiewe beweging tussen die waarnemer en die bron? Marking criteria/Nasienriglyne Refer to change in frequency and relative motion/Doppler effect. Verwys na verandering in frekwensie en relatiewe beweging/Doppler effek. In context with Doppler and scenario given. Starting with Is/Do/Does/Will/Can one mark if answer to question is yes or no. Begin sin met Sal/Is/Kan een punt indien antwoord vir vraag ja of nee is. (2) 5.2 OPTION/OPSIE 1 𝑣±𝑣 Marking guidelines/ 𝑓𝐿 = 𝑣 ±𝑣 𝐿 𝑓𝑆  𝑆 Nasienriglyne: TOWARDS/NA: 340 ✓ Formula/Formule 𝑓𝐿 =  𝑥 200 ✓ Substitution towards/ 340−𝑣𝑠 Either or AWAY/WEG: Invervanging na 𝑓𝐿 = 340 𝑥 200 ✓ Substitution away/ 340+ 𝑣𝑠 Invervanging weg ∆𝑓 = 𝑓𝑓 – 𝑓𝑖 ✓ Substitution change 340 340 -67,15  = 340+𝑣 𝑥 200 − 340−𝑣 𝑥 200 in/Invervanging vir 𝑠 𝑠 340 340 verandering -67,15 = 200 340+𝑣 − 340−𝑣 𝑠 𝑠 ✓ Answer/Antwoord v = 57,02 m.s-1.  Range for answers. (55- 57) (5) 12

Downloaded from hlayiso.com 10841/22 OPTION/OPSIE 2 𝑣±𝑣 𝑓𝐿 = 𝑣 ±𝑣 𝐿 𝑓𝑆  𝑆 ∆𝑓 = 𝑓𝑓 – 𝑓𝑖 𝑣±𝑣 𝑣±𝑣 ∆𝑓 = 𝑣 ±𝑣 𝐿 𝑓𝑆 - 𝑣 ±𝑣 𝐿 𝑓𝑆 𝑆 𝑆 340 340 (-)67,15  = 340+𝑣 𝑥 200  - 340−340 𝑥200  𝑠 v = 57,02 m.s-1.  Ignore negative in front of the 67,15 during marking. OPTION 3 If learners add or subtract the 67,15 Hz from 200 Hz. They will only do one sum. Give mark allocation as follows: ✓ formula ✓ ✓ ✓ substitution ✓ answer 𝑣±𝑣 𝑓𝐿 = 𝑣 ±𝑣 𝐿 x𝑓𝑆  𝑆 TOWARDS/NA: 340 267,15 ✓ = 340−𝑣𝑠 ✓𝑥 200 vs = 85,46 m.s-1✓ OR/OF if sum done as AWAY/WEG: 340 132,85 = 𝑥 200 340 + 𝑣𝑠 vs = 171,86 m.s-1 5.3 AWAY  The spectrum of the distant star moved towards the longer wavelengths/ lower frequencies, thus stretching out the waves proving that it is moving away.  WEG ✓ Die spektrum van die verafgeleë ster beweeg na die langer golflengtes/laer frekwensies, dus die uitrekking van die golwe bewys dat dit besig is om weg te beweeg. (3) [14] 13

Downloaded from hlayiso.com 10841/22 QUESTION/VRAAG 6 6.1 The electric field at a point is the electrostatic force experienced per unit positive charge placed at that point.  Die elektriese veld by ʼn punt in die elektrostatiese krag ondervind per eenheid positiewe lading geplaas by daardie punt. 2 marks or zero. (2) 6.2 Marking criteria/Nasienriglyne ✓ Shape/form/ all lines need to be curved / Vorm / alle lyne moet ‘n kurwe hê. ✓ Direction of arrows towards sphere/Rigting van pyle na die sfere ✓ All other field rules applied. / Alle ander veldreëls toegepas. Do not penalise if different amounts of lines around the charges. Ignore if different number of lines are drawn around each charge. / Ignoreer indien die verskillende aantal lyne getrek is rondom elke lading. (3) 6.3 OPTION/OPSIE 1 𝐹𝑛𝑒𝑡 𝑜𝑛/𝑜𝑝 𝐵 = 𝐹𝐴 𝑜𝑛/𝑜𝑝 𝐵 + 𝐹𝐶 𝑜𝑛/𝑜𝑝 𝐵 𝑘𝑄𝐴 𝑄𝐵 𝑘𝑄𝐶 𝑄𝐵 𝐹𝑛𝑒𝑡 𝑜𝑛/𝑜𝑝 𝐵 = +  𝑟2 𝑟2 9 𝑥 109 𝑥 5,6 𝑥 10−9 𝑥 2,34 𝑥 10−9 9 𝑥 109 𝑥 7,46 𝑥 10−9 𝑥 2,34 𝑥 10−9 0,004078 ✓ = +  0,0062 𝑟2 r = 0,01399 m (accept/aanvaar 0,014 m)  (5) 14

Downloaded from hlayiso.com 10841/22 OPTION/OPSIE 2 𝑘𝑄𝐴 𝑄𝐵 𝐹𝐴 𝑜𝑛/𝑜𝑝 𝐵 = ✓ 𝑟2 9 𝑥 109 𝑥 5,6 𝑥 10−9 𝑥 2,34 𝑥 10−9 = ✓ 0,0062 = 0,003276 N 𝑘𝑄𝐶 𝑄𝐵 𝐹𝐶 𝑜𝑛/𝑜𝑝 𝐵 = 𝑟2 9 𝑥 109 𝑥 7,46 𝑥 10−9 𝑥 2,34 𝑥 10−9 = ✓ 𝑟2 1,57 𝑥 10−7 𝐹𝐶 𝑜𝑛/𝑜𝑝 𝐵 = 𝑟2 𝐹𝑛𝑒𝑡 𝑜𝑛/𝑜𝑝 𝐵 = 𝐹𝐴 𝑜𝑛/𝑜𝑝 𝐵 + 𝐹𝐶 𝑜𝑛/𝑜𝑝 𝐵 1,57 𝑥 10−7 0,004078 ✓= 0,003272 + 𝑟2 r = 0,01399 m (accept/aanvaar 0,014 m)  accept 0,01 m rounding off 6.4 6.4.1 𝑄𝐴 + 𝑄𝐵 𝑄𝑛𝑒𝑤/𝑛𝑢𝑤𝑒 = 2 = +5,6 + (-2,34)  2 = 1,63 nC = 1,63 x 10-9 C  accept either or (2) 6.4.2 B is positively charged and C negative. Attractive pattern.  The electric field pattern has changed. It is now originating from B and ending at C. This is because B is now positively charged whilst C remains negatively charged. B is positief gelaai en C is negatief gelaai. Aantrekkende patroon.  Die elektriese veldpatroon het verander. Dit begin nou vanaf B en eindig by C . Dit is omdat B nou positief gelaai is terwyl C negatief gelaai bly. (3) [15] 15

Downloaded from hlayiso.com 10841/22 QUESTION/VRAAG 7 7.1 7.1.1 OPTION/OPSIE 1 1 1 1 Marking guidelines:/ = +  𝑅𝑝 𝑅1 𝑅2 Nasienriglyne: 1 1 1 ✓ Formula = +  RT/Formule RT 𝑅𝑝 30 10 ✓ Substitution/ 15 Invervanging R1 = 2 ✓ Formula RT/ = 7,5  Formule RT RT = Rs + Rp + r ✓ Substitution/ = 7,5 + 0,6 + 3,5  or (4,1) Invervanging = 11,6  ✓ Answer/Antwoord OPTION/OPSIE 2 RT = Rs + Rp + r  10 𝑥 30 = 3,5 ✓ + ✓ + 0,6  10+30 = 11,6  (5) 7.1.2 OPTION/OPSIE 1 VP = I x RP Marking guidelines:/ = 2,07 x 7,5 ✓ Nasienriglyne: = 15,53 V 𝐼2 = 𝑉𝑃 ✓ ✓Substitution/ 𝑅 Invervanging = 15,53 ✓ ✓ Answer/ 10 Antwoord = 1,55 A ✓ OPTION/OPSIE 2 RATIO 1 : 3   ¾ :¼  2,07 x ¾  = 1,55 A  OPTION/OPSIE 3 𝑉// = 24 – 𝑉𝑟 - 𝑉𝑒𝑥 = 24 – (2,06897x0,6)– (2,06897 x 3,5) ✓ = 15,517 V 𝑉// 𝐼𝐴2 = R heater/𝑣𝑒𝑟𝑤𝑎𝑟𝑚𝑒𝑟 15,517 = ✓ 10 = 1,55 A ✓ (3) 16

Downloaded from hlayiso.com 10841/22 7.2 DECREASES ✓ Increasing R increases the total resistance of the circuit, hence reduces the overall circuit current, ✓ This will reduce the reading on A2 and from P = I2R, heat dissipated reduces. ✓ (relationship between current and power) VERLAAG ✓ Toename van R lei tot toename in totale weerstand van die stroom, daarom verminder die totale stroom van die stroombaan. Dit sal die lesing op A 2 verminder en vanaf P=I2r, word hitte gelewer verminder. OR/OF I decreases or V decreases ✓I I verminder of V verminder 𝑉2 P = I2 R or P = ✓ 𝑅 ∴ Less heat dissipated. ✓/Minder hitte word versprei. (3) [11] 17

Downloaded from hlayiso.com 10841/22 QUESTION/VRAAG 8 8.1 1,6 1,4 Potential difference (V) 1,2 1,0 Potensiaalverskil (V) 0,8 0,6 0,4 0,2 50 100 150 200 250 300 350 400 450 500 550 600 650 700 Current (mA) Stroom (mA) Marking Guidelines/Nasienriglyne: ✓ Calibration on both axes/Intervalle op beide asse ✓ 5 points plotted correctly/5 punte korrek geplot ✓ Close fit line through most of the points/Beste paslyn deur die meeste punte ✓ Negative gradient/Negatiewe gradient Do not penalise for A in place of mA (4) 8.2 8.2.1 1,5 V  (1) 8.2.2 ∆𝑦 0,5−1,5 gradient = ∆𝑥 = 652 𝑥 10−3 −0 if learners swop final and initial do not penalise = - 1,53  r = 1,53 Ω  (can use any of the values from the table.)/(enige waardes vanuit die tabel kan gebruik word) must have a positive as the answer (3) 8.3 INCREASES  As the resistance decreases the current will increase.  Resistance is inversely proportional to the current. As the current increases, the temperature of the battery will increase due to an increase in total resistance since internal resistance stays the same. (relationship between current and energy) W = I2RΔt VERHOOG ✓ Soos die weerstand verminder sal die stroom verhoog. ✓ Weerstand is omgekeerd eweredig aan die stroom. Soos die stroom verhoog, verhoog die temperatuur van die battery as gevolg van die groter totale weerstand en die interne weerstand bly konstant. ✓ W = I2RΔt (3) [11] 18

Downloaded from hlayiso.com 10841/22 QUESTION/VRAAG 9 9.1 Electrical energy to mechanical energy/ ✓✓ 2 or zero Elektriese energie na meganiese energie (2) 9.2 Direct current: The current keeps flowing in one direction only. ✓✓ Gelykstroom: Die stroom hou aan om slegs in een rigting te beweeg. OR/OF The current carrying charges move in only one direction. ✓✓ Die stroom wat ladings dra, beweeg net in een rigting. Marking criteria/Nasienriglyne: If any of the underlined words/phrases in the correct context is omitted, deduct 1 mark. Indien enige van die onderstreepte woorde/frases in die korrekte konteks uitgelaat word, trek een punt af. (2) 9.3 FROM Q to P ✓/VANAF Q tot P (1) 9.4 Replace the commutator (split ring) ✓ with slip rings. ✓/ Vervang die kommutator (splitring) met ʼn sleepring. (2) ✓ 9.5 𝑃𝑎𝑣𝑒 = 𝑉𝑟𝑚𝑒 𝐼𝑟𝑚𝑠  2000 = 220 𝐼𝑟𝑚𝑠  𝐼𝑟𝑚𝑠 = 9,09 A 𝐼𝑚𝑎𝑥 = 𝐼𝑟𝑚𝑠 √2  = 9,09 x √2 ✓ = 12,86 A  If combined formulae – give the credit. (5) [12] 19

Downloaded from hlayiso.com 10841/22 QUESTION/VRAAG 10 10.1 The minimum frequency of light needed to emit electrons from a certain metal surface  Die minimum frekwensie van lig benodig om elektrone vry te stel vanuit ʼn sekere metaal se oppervlakte OR/OF A certain minimum frequency of incident radiation which will cause photoelectrons to be emitted from the surface of a metal✓✓ ʼn Sekere minimum frekwensie van invallende strale wat sal veroorsaak dat foto- elektrone uit die oppervlak van ʼn metaal vrygestel word Marking criteria/Nasienriglyne: If any of the underlined words/phrases in the correct context is omitted, deduct 1 mark. If learner explains minimum energy then zero marks are allocated. It has to be frequency. Indien enige van die onderstreepte woorde/frases in die korrekte konteks uitgelaat word, trek 1 punt af. Indien leerder minimum energie verduidelik, dan geen punte toegeken nie. Dit moet frekwensie wees. (2) 10.2 10.2.1 𝑊0 = ℎ𝑓0  Marking guidelines/ = 6,63 x 10-34 5,1 x 1014  Nasienriglyne: = 3,38 x 10-19 J  ✓ Formula/Formule ✓ Substitution/Vervanging ✓ Answer/Antwoord (3) 10.2.2 OPTION/OPSIE 1 Marking guidelines/ 𝑐 Nasienriglyne: E= ℎ   ✓ Formula/Formule = 6,63 x 10-34 x 3 x 108  ✓ ✓Substitution/Vervanging 400 x 10-9✓ ✓ Answer/Antwoord = 4,97 x 10-19 J  20

Downloaded from hlayiso.com 10841/22 OPTION 2: OPSIE 2 𝑐 𝑓 =  = 3 x 108 ✓ 400 x 10-9 = 7,50 x 1014 Hz E= ℎ𝑓 = 6,63 x 10-34 x 7,5 x 1014  = 4,97 x 10-19 J  (4) 10.2.3 Positive marking from Marking Question 10.2.1 and 10.2.2 guidelines/Nasienriglyne: Positiewe nasien vanaf Vraag ✓ Formula/Formule 10.2.1 en 10.2.2 ✓ ✓Substitution/Vervanging 𝑐 ✓ Answer/Antwoord E= ℎ  = 𝑊0 + 𝐸𝑘max/𝑚𝑎𝑘𝑠  4,97 x 10-19 ✓ = 3,38 x 10-19 + Notes/Notas: Ekmax/maks✓ • If max. is omitted from Ek Ekmax/maks = 1,59 x 10-19 J  formula, marks are lost for the formula. The rest can however be marked./ • Indien maks. uitgelaat word van formule met Ek, word punte verbeur vir die formule. Die res kan wel steeds punte kry. (4) 10.3 Not to be marked. Mark moved to 10.2.2 [13] TOTAL/TOTAAL: 150 21