Physical Sciences Grade 12 Paper 1 NSC November 2023 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) NOVEMBER 2023 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 28 pages. Hierdie nasienriglyne bestaan uit 28 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 2 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 1/VRAAG 1 1.1 B  (2) 1.2 A  (2) 1.3 D  (2) 1.4 B (2) 1.5 B  (2) 1.6 A  (2) 1.7 D  (2) 1.8 B  (2) 1.9 C  (2) 1.10 A  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 3 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 2/VRAAG 2 2.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark/ Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. When a resultant/net force acts on an object, the object will accelerate in the direction of the force. The acceleration is directly proportional to the resultant/net force and inversely proportional to the mass of the object.  Wanneer 'n resulterende/netto krag op 'n voorwerp inwerk, sal die voorwerp in die rigting van die krag versnel. Die versnelling is direk eweredig aan die netto/resulterende krag en omgekeerd eweredig aan die massa van die voorwerp. OR/OF The resultant/net force acting on an object is equal to the rate of change of momentum of the object in the direction of the resultant/net force. (2 or 0) Die resulterende/netto krag wat op 'n voorwerp inwerk is gelyk aan die tempo van verandering van momentum in die rigting van die resulterende/netto krag. (2 of 0) (2) 2.2 N T f w Accepted labels/Aanvaarde benoemings N FN/Normal/Fnormal/Fnormaal/Normaal f (kinetic) friction/5,88 N /Ff/fk/(kinetiese) wrywing /Fw w Fg/Fw/weight/mg/39,2 N/gravitational force Fg/Fw/gewig/mg/39,2 N/gravitasiekrag T FT/Fstring/tension/spanning/Ftou Notes/Aantekeninge  Mark is awarded for label and arrow./Punt word toegeken vir byskrif en pyltjie.  Do not penalise for length of arrows./Moenie vir die lengte van die pyltjies penaliseer nie.  If w is not shown but wǁ and w┴ are shown, give 1 mark for both./ Indien w nie getoon is nie maar wǁ en w┴ is getoon, ken 1 punt toe vir beide.  If arrows do not touch the dot/Indien pyle nie die kolletjie raak nie: Max/Maks 3 4  Any other additional force(s)/Enige ander addisionele krag(te): Max/Maks 3 4  If everything correct, but no arrows/Indien alles korrek, maar geen pyltjies: Max/Maks 3 4 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne 2.3.1 For block A/Vir blok A: UP THE INCLINE AS POSITIVE/ TEEN DIE SKUINSTE OP AS POSITIEF Fnet = ma T – fk – wǁ = ma Any one /Enige een T – fk – mgsinθ = ma T – 5,88 – 4(9,8)sin35° = 4(2)  T = 36,36 N  DOWN THE INCLINE AS POSITIVE/ TEEN DIE SKUINSTE AF AS POSITIEF Fnet = ma – T + fk + wǁ = ma Any one /Enige een – T + fk + mgsinθ = ma – T + 5,88 + 4(9,8)sin35° = 4(–2)  T = 36,36 N  (4) 2.3.2 POSITIVE MARKING FROM QUESTION 2.3.1 POSITIEWE NASIEN VANAF VRAAG 2.3.1 NOTE: If systems approach is used, learner gets mark for the answer only. NOTA: Indien sisteem benadering gebruik word, kry leerder slegs ‘n punt vir die antwoord. For block B/Vir blok B: UP THE INCLINE AS POSITIVE/ TEEN DIE SKUINSTE OP AS POSITIEF Fnet = ma F –T – fk – wǁ = ma F – T – fk – mgsinθ = ma F – 36,36 – 13,23 – 9(9,8)sin35° = (9)(2)  F = 118,18 N  DOWN THE INCLINE AS POSITIVE/ TEEN DIE SKUINSTE AF AS POSITIEF Fnet = ma – F +T + fk + wǁ = ma – F + T + fk + mgsinθ = ma –F + 36,36 + 13,23 + 9(9,8)sin35° = (9)( –2)  F = 118,18 N  (3) 2.4.1 INCREASES/TOENEEM  (1) 2.4.2  Since and m are constant, as decreases, normal force/w⟘/mgcos will increase.  Aangesien en m konstant is, soos afneem sal normaalkrag/w⟘/mgcos toeneem. OR/OF N = mgcos  Frictional force is directly proportional to normal force/  Wrywing is direk eweredig aan normaalkrag OR/OF f N /f w⟘ / f N / f = N (2) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 3/VRAAG 3 3.1 Motion under the influence of gravitational force only.  Accept weight/gravity. Beweging slegs onder die invloed van gravitasiekrag. Aanvaar swaartekrag/gewig/gravitasie. (2 or/of 0) OR/OF Motion in which the only force acting is gravitational force. Accept weight/gravity. Beweging waar die enigste krag wat inwerk, gravitasiekrag is. Aanvaar swaartekrag/gewig/gravitasie.(2 or/of 0) (2) NOTE: If projectile is defined: ⁄ NOTA: Indien projektiel gedefinieer is: ⁄ 3.2 Marking criteria/Nasienkriteria  Correct formula for vi./Korrekte formule vir vi.  Correct substitution into formula./Korrekte vervanging in formule.   Correct final answer/Korrekte finale antwoord: 10,74 m·s-1  Note: If energy principles are used, max: 1/3 for answer. B Nota: Indien energiebeginsels gebruik word, maks: 1/3 vir antwoord. OPTION 1/OPSIE 1 A-B: UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF A C v v y v v y 2 2 0 = v + 2( 9,8)(5,89)  0 = v + 2(9,8)( 5,89)  vi = 10,74 m·s-1  vi = 10,74 m·s-1  OPTION 2/OPSIE 2 B-C: UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF G v v y v v y v = 0 + 2( 9,8)( 5,89)  v = 0 + 2(9,8)(5,89)  vi = 10,74 m·s-1  vi = 10,74 m·s-1  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne 3.3.1 POSITIVE MARKING FROM QUESTION 3.2. POSITIEWE NASIEN VANAF VRAAG 3.2 Marking criteria/Nasienkriteria  Any one of the correct formulae leading to the velocity at which the ball strikes the ground./Enige een van die korrekte formules wat lei tot die snelheid waarmee die bal die grond tref.   Correct substitution leading to the velocity at which the ball strikes the ground (values of vi and vf can be swopped)./Korrekte vervanging wat lei tot die snelheid waarmee die bal die grond tref (waardes van vi en vf kan omgeruil word).   Correct formula or ∆Ek/Ek/Ek lost./Korrekte formule vir ∆Ek/Ek/Ek verlore.   Correct substitution into ∆Ek/Ek lost formula (values of vi and vf can be swopped, ignore negative vi or vf values)./ Korrekte vervanging in ∆Ek/Ek verlore formule (waardes van vi en vf kan omgeruil word, ignoreer negatiewe vi of vf waardes).   Correct final answer/Korrekte finale antwoord: +/- 68,31 J  Range/Gebied: (67,91 – 69,34J) Note: Accept if downwards is taken as positive. Nota: Aanvaar indien afwaarts as positief geneem is. OPTION 1/OPSIE 1 A-G: v v y  2 v = (10,74) + 2(–9,8)(–15,3)  vf = – 0,38 m∙s-1 OPTION 2/OPSIE 2 B-G: DURING COLLISION/ v v y  2 TYDENS BOTSING v = 0 + 2(–9,8)( –21,19)  ∆Ek = Ekf – Eki B vf = – 0,38 m∙s-1 OPTION 3/OPSIE 3 ∆Ek = ½mv – ½mv Any one/ Enige een C-G: ∆Ek = ½m v – v ) v v y  ∆Ek = ½(0,5)[(11,92)2 – (20,38)2]  2 v = (–10,74) + 2(–9,8)(–15,3)  ∆Ek/Ek lost = – 68,31 J  A C vf = – 0,38 m∙s-1 OPTION 4/OPSIE 4 OR/OF A-G: y v t t ∆Ek = ½(0,5)[(20,38)2 – (11,92)2]  – ,3 0, t – ,8 t ∆Ek/Ek lost = 68,31 J  t = 3,18 s G v v ∆y ( )∆t  0, v –15,3 = ( )3,18  vf = – 0,38 m∙s-1 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 5/OPSIE 5 B-G: y v t t – , 0 t – ,8 t t = 2,08 s v v ∆y ( )∆t  0 v –21,19 = ( )2,08  vf = – 0,38 m∙s-1 OPTION 6/OPSIE 6 C-G: y v t t – ,3 – 0, t – ,8 t t = 0,98 s v v DURING COLLISION/ ∆y ( )∆t  TYDENS BOTSING B ∆Ek = Ekf – Eki – 0, v Any one/ –15,3 = ( )0,98  ∆Ek = ½mv – ½mv Enige een ∆Ek = ½m v – v ) vf = – 20,48 m∙s-1 OPTION 7/OPSIE 7 ∆Ek = ½(0,5)[(11,92)2 – (20,38)2]  A C A-G: ∆Ek/Ek lost = – 68,31 J  y v t t – ,3 0, t – ,8 t OR/OF t = 3,18 s ∆Ek = ½(0,5)[(20,38)2 – (11,92)2]  vf = vi + a t  ∆Ek/Ek lost = 68,31 J  vf = 10,74 + (–9,8)(3,18)  G v = –20,42 m∙s-1 f OPTION 8/OPSIE 8 B-G: y v t t – , 0 t – ,8 t t = 2,08 s vf = vi t vf = 0 + (–9,8)(2,08)  vf = – 0,38 m∙s-1 OPTION 9/OPSIE 9 C-G: y v t t – ,3 – 0, t – ,8 t t = 0,98 s vf = vi t vf = – 0, + (–9,8)(0,98)  vf = –20,34 m∙s-1 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 8 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 9 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 10/OPSIE 10 A-G: W net Ek Any one/ w ycosθ ½mv – ½mv Enige een (0,5)(9,8)(15,3)cos0° = ½(0,5)v – ½(0,5)(10,74)2  vf = 20,38 m·s-1 OPTION 11/OPSIE 11 B-G: W net = Ek B Any one/ w ycosθ mv – ½mv Enige een (0,5)(9,8)(21,19)cos0° = ½(0,5)vf2 – 0  vf = 20,38 m·s-1 OPTION 12/OPSIE 12 C-G: W net Ek Any one/ A C w ycosθ mv – ½mv Enige een (0,5)(9,8)(15,3)cos0° = ½(0,5) v – ½ (0,5)(10,74)2 vf = 20,38 m·s-1 OPTION 13/OPSIE 13 A-G OR/OF C-G: Wnc = K + U Any one/ G Wnc = [ ½ m(v – v ) ]+ [ mg(hf - hi) ] Enige een 0 = [ ½mv - ½mv ]+ [ mghf - mghi] DURING COLLISION/TYDENS 0 = [ ½(0,5) v – (10,74)2] + [0 –(0,5)(9,8)(15,3)]  BOTSING vf = 20,38 m·s-1 ∆Ek = Ekf – Eki OPTION 14/OPSIE 14 ∆Ek = ½mv – mv Any one/ Enige een B-G: ∆Ek = ½m v – v ) Wnc = K + U ∆Ek = ½(0,5)[(11,92)2 – (20,38)2]  Any one/ Wnc =[ ½ m(v – v )] + [ mg(hf - hi) ] Enige een ∆Ek/Ek lost = – 68,31 J  0 = [ ½mvf2 - ½mvi2 ] + [ mghf - mghi ] 0 = [ ½(0,5) v – 0 ] + [ 0 – (0,5)(9,8)(21,19) ]  OR/OF vf = 20,38 m·s-1 OPTION 15/OPSIE 15 ∆Ek= ½(0,5)[(20,38)2 – (11,92)2]  ∆Ek/Ek lost = 68,31 J  y v t t – ,3 0, t – ,8 t t = 3,18 s Fnet ∆t ∆p Any one/Enige een mg∆t m vf – vi) (-9,8)(3,18) = vf – 10,74  -1 vi = – 20,38 m·s  OPTION 16/OPSIE 16 (Ek + EP)top = (Ek + EP)bottom Any one/ (0 + mgh)top = ( mv + 0)bottom Enige een 2 0 + (0,5)(9,8)(21,19) = (0,5)v + 0  -1 vi = 20,3795 m·s  (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 10 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne 3.3.2 Marking criteria/Nasienkriteria  Correct formula to c lcul te ∆t./Korrekte formule om ∆t te bereken.   Correct subst tut on to c lcul te ∆t./Korrekte vervanging om ∆t te bereken.   Correct final answer./Korrekte finale antwoord: 1,22 s  OPTION 1/OPSIE 1 G-P: G-P: UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF vf = vi t vf = vi t 0 = 11,92 + (– ,8 t  0 = (– , ,8 t  P t = 1,22 s  t = 1,22 s  OPTION 2/OPSIE 2 P-G: P-G: UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF vf = vi t vf = vi t , 0 ,8 t  G G – 11,92 = 0 + (– ,8 t  t = 1,22 s  t = 1,22 s  OPTION 3/OPSIE 3 G - G: G - G: UPWARDS AS POSITIVE/ DOWNWARDS AS POSITIVE/ OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF vf = vi t vf = vi t –11,92 = 11,92 + (– ,8 t  11,92 = – , ,8 t  t = 2,43 s t = 2,43 s , 3 , 3 t to reach hmax/t tot hmaks = t to reach hmax/t tot hmaks = = 1,22 s  = 1,22 s  OPTION 4/OPSIE 4 y v t t  G-P: , , t – ,8 t  UPWARDS AS POSITIVE/ t , s OPWAARTS AS POSITIEF v v y OR/OF 2 0 = (11,92) 2(– ,8) y v v y = 7,25 m ∆y ( )∆t  , 0 7,25 = ( )∆t  t , s Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 11 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne G-P: v t y t  P DOWNWARDS AS POSITIVE/ – , – , t ,8 t  AFWAARTS AS POSITIEF t , s v v y 0 = (–11,92)2 2( ,8) y OR/OF y = –7,25 m v v ∆y ( )∆t  G G – , 0 –7,25 = ( )∆t  t , s OPTION 5/OPSIE 5 y v t t  P-G: – , – ,8 t  UPWARDS AS POSITIVE/ t , s OPWAARTS AS POSITIEF v v y OR/OF 2 (–11,92) = 0 2(– ,8) y v v y = –7,25 m ∆y ( )∆t  0– , –7,25 = ( )∆t  t , s P-G: y v t t  DOWNWARDS AS POSITIVE/ , , t ,8 t  AFWAARTS AS POSITIEF t , s v v y 2 (11,92) = 0 2( ,8) y OR/OF y = 7,25 m v v ∆y ( )∆t  0 , 7,25 = ( )∆t  t , s Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 12 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 6/OPSIE 6 UPWARDS AS POSITIVE/ W net Ek OPWAARTS AS POSITIEF w ycosθ mv - ½mv 0, ,8 ∆ycos 80 = ½(0,5)(0 – (11,92)2 º y v t t  ∆y , m , , t – ,8 t  t , s OR/OF OR/OF v v ∆y ( )∆t  Wnc = K + U , 0 Wnc =[ ½ m(v – v ) ] + [ mg(hf – hi) ] 7,25 = ( )∆t  0 = [ ½mv – ½mv ] + [ mghf – mghi ] t , s 0 = ½(0,5)[(0 – (11,92)2] + (0,5)(9,8)h h = 7,25 m DOWNWARDS AS POSITIVE/ AFWAARTS AS POSITIEF OR/OF y v t t  – , – , t ,8 t  t , s ∑EMi ∑EMf 2 ½mvi + mghi = ½mv + mghf OR/OF ½ (0,5)(11,92)2 + 0 = 0 + (0,5)(9,8)(hf) v v h = 7,25 m ∆y ( )∆t  – , 0 –7,25 = ( )∆t  t , s OPTION 7/OPSIE 7 DOWNWARDS AS POSITIVE/ UPWARDS AS POSITIVE/ AFWAARTS AS POSITIEF OPWAARTS AS POSITIEF Fnet ∆t ∆p m vf – vi)  Fnet ∆t ∆p m vf – vi)  mg∆t m vf – vi) mg∆t m vf – vi) 0, ,8 ∆t 0, [0 – (11,92)]  0, ,8 ∆t 0, [0 – (–11,92)]  ∆t = 1,22 s  ∆t = 1,22 s  (3) 3.4 POSITIVE MARKING FROM QUESTIONS 3.2 AND 3.3.2 POSITIEWE NASIEN VANAF VRAE 3.2 EN 3.3.2 3.4.1 11,92 (m·s-1)  (1) 3.4.2 10,74 (m·s-1)  (1) 3.4.3 1,22 (s)  (1) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 13 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 4/VRAAG 4 4.1 591 N to the right/original direction of bullet/Accept East  591 N na regs/in oorspronklike rigting van koeël/AanvaarOos (1) 4.2 Marking criteria/Nasienkriteria  Any correct formula for conservation of momentum./Enige korrekte formule vir behoud van momentum.   Correct substitutions. /Korrekte vervangings.   Correct final answer./Korrekte finale antwoord: 395,58 m·s-1 Range/Gebied: (394 to 395,58 m·s-1) Σpi Σpf Any one/ mbvbi + mtvti = {mb + mt)vf Enige een (0,03)vbi+(2,7)(-3)=(0,03+2,7)(1,38)  (0,03)vbi = 11,86 vi = 395,58 m·s-1 OPTION 1/OPSIE 1  Fnet ∆t ∆p m vf – vi ) OR/OF (591)(0,02) = 2,7[vf – (–3)] ∆p(bullet) = – ∆p(trolley)  vf = 1,38 m·s-1 m(vf – vi ) = – (591)(0,02)  (0,03)(1,38 – vi) = –11,82 1,38 – vi = –394 vi = 395,38 m·s-1 (4) 4.3 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The total (linear) momentum in an isolated system is conserved/remains constant.  Die totale (lineêre) momentum in 'n geïsoleerde sisteem bly behoue/konstant. Accept for 1 mark/Aanvaar vir 1 punt In a isolated system the total momentum before a collision is equal to the total momentum after a collision. In ‘n geïsoleerde sisteem is die totale momentum voor ‘n botsing gelyk aan die totale momentum na ‘n botsing. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 14 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne 4.4 POSITIVE MARKING FROM QUESTION 4.2. POSITIEWE NASIEN VANAF VRAAG 4.2. OPTION 1/OPSIE 1 RIGHT AS POSITIVE/REGS AS POSITIEF ∑ pi ∑ pf mxvix + myviy = mxvfx + myvfy Any one/Enige een (0,03)(395,58) + (2,7)( –3) = vf (0,03 + 2,7) vf = 1,38 m·s-1  Range/Gebied: (1,36 – 1,38 m·s-1) LEFT AS POSITIVE/LINKS AS POSITIEF ∑ pi ∑ pf mxvix + myviy = mxvfx + myvfy Any one/Enige een (0,03)( –395,58) + (2,7)(3) = vf (0,03 + 2,7) vf = –1,38 m·s-1 vf = 1,38 m·s-1  Range/Gebied: (1,36 –1,38 m·s-1) OPTION 2/OPSIE 2 Fnet ∆t ∆p m vf – vi)  (591)(0,02) = 2,7[vf – (–3)]  -1 vf = 1,38 m·s  Range/Gebied: (1,36 – 1,38 m·s-1) OPTION 3/OPSIE 3 RIGHT AS POSITIVE/REGS AS POSITIEF ∆p(bullet) ∆p(trolley) = 0 Any one/Enige een m1(v1f – v1i) + m2 (v2f – v2i) = 0 (0,03)(vf – 395,58)  + (2,7)(vf + 3)  = 0 -1 vf = 1,38 m·s Range/Gebied:(1,36 – 1,38 m·s-1) LEFT AS POSITIVE/LINKS AS NEGATIEF ∆p(bullet) ∆p(trolley) = 0 Any one/Enige een m1(v1f – v1i) + m2 (v2f – v2i) = 0 (0,03)(vf + 395,58)  + (2,7)(vf – 3)  = 0 -1 vf = –1,38 m·s -1 (4) vf = 1,38 m·s  Range/Gebied:(1,36 – 1,38 m·s-1) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 15 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 5/VRAAG 5 5.1 N N ACCEPT/AANVAAR f f w w Accepted labels/Aanvaarde benoemings Fw /Fg /FEarth on trolley/FAarde op trollie /mg /gravitational force/ w gravitasiekrag/weight/gewig f Ff /fk/(kinetic) Friction/(kinetiese) wrywing/Fw N FN /Fsurface on trolley /Foppervlak op trollies /Normal/Normaal Notes/Aantekeninge:  Mark awarded for label and arrow./Punt toegeken vir benoeming en pyltjie.  Do not penalise for length of arrows./Moenie vir die lengte van die pyltjies penaliseer nie.  Any other additional force(s)/Enige ander addisionele krag(te): Max/Maks ⅔  If everything is correct, but no arrows/Indien alles korrek is, maar geen pyltjies: Max/Maks ⅔  If force(s) do not make contact with the dot /Indien krag(te) nie met die kolletjie kontak maak nie: Max/Maks ⅔ (3) 5.2 Initial kinetic energy/Aanvanklike kinetiese energie/EkA /Eki/Ki (1) 5.3 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark/Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The net/total work done (on an object) is equal to the change in the object's kinetic energy.  Die netto/totale arbeid wat (op 'n voorwerp) verrig is, is gelyk aan die verandering in die voorwerp se kinetiese energie. OR/OF The work done on an object by a resultant/net force is equal to the change in the object's kinetic energy.  Die arbeid verrig op in voorwerp deur die resultante/netto krag is gelyk aan die verandering in die voorwerp se kinetiese energie. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 16 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne 5.4 Marking criteria/Nasienkriteria OPTION 1/OPSIE 1  Relating frictional force to gradient./Verband tussen wrywingskrag en helling.   Correct subtitution of two values or ratio from the graph./Korrekte vervanging van twee waardes of verhouding vanaf die grafiek.   Formula to calculate mass./Formule om massa te bereken.   Correct substitution of µ and 9,8./Korrekte vervanging van µ en 9,8.   Correct final answer/Korrekte finale antwoord: 2,27 kg  y ∆x gradient = = =  x E , 3 ,  OR/OF OR/OF =  8 f =4N fk k FN  4 = (0,18)(m)(9,8)  m = 2,27 kg  Marking criteria/Nasienkriteria OPTIONS 2 TO 5/OPSIES 2 TOT 5  Correct formula for work./Korrekte formule vir arbeid.   Correct substitution of two co-ordinate values from the graph. Correct negative ∆Ek value (any two correct co-ordinates used).  Korrekte vervanging van twee koördinaat waardes vanaf die grafiek Korrekte negatiewe ∆Ek waarde (enige twee korrekte koördinate gebruik).  Formula to calculate mass./Formule om massa te bereken.   Correct substitution of µ and 9,8./Korrekte invervanging van µ en 9,8.   Correct final answer/Korrekte finale antwoord: 2,27 kg  OPTION 2/OPSIE 2 W nc = K + U W net = Ek W net = Ekf – Eki Any one/ Wf = Ek Enige een Wf = Ekf – Eki ∆xcos180° = Ekf – Eki – f(1,5)  = 0 – 6  OR/OF fk = k FN  – f(3) = 0 – 12 OR/OF 4 = (0,18)(m)(9,8)  – f(4,5) = 0 – 18 m = 2,27 kg  f=4N OPTION 3/OPSIE 3 W net = Fnet xcos θ  – 6 = f(1,5)cos180º  OR/OF – 12 = f(3)cos180º OR/OF – 18 = f(4,5)cos180º f=4N Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 17 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 4/OPSIE 4 fk = µkN  = 0,18(9,8)(m)  = 1,76m W net ∆Ek Fnet∆xcosθ = Ekf Eki fk∆xcosθ = Ekf Eki Any one/Enige een µkN∆xcosθ = Ekf Eki µkN∆xcos 80o = Ekf Eki (1,76m)(4,5)  = 0 18  OR/OF (1,76m)(3) = 0 12 OR/OF (1,76m)(1,5) = 0 6 m = 2,27 kg  OPTION 5/OPSIE 5 Wnet ∆Ek   µkmg∆xcosθ Ekf Eki   (0,18)(m)(9,8) (4,5)cos180o = 0 18  m = 2,27 kg  OPTION 6/OPSIE 6 fk = µkN  = 0,18(9,8)(m)  = 1,76m Fnet = ma -1,76m = ma -2 a = -1,764 m·s v v y 2 0 = vi 2(-1,764)(4,5)  -1 vi = 3,98 m·s ∆Ek = ½m∆v2 -18 = ½m(0-3,982)  m = 2,27 kg  (6) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 18 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 6/VRAAG 6 6.1.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The change in frequency (pitch) of the sound detected by a listener because the sound source and the listener have different velocities relative to the medium of sound propagation.  Die verandering in frekwensie (toonhoogte) van die klank waargeneem deur 'n luisteraar omdat die klankbron en die luisteraar verskillende snelhede relatief tot die medium waarin die klank voortgeplant word, het. OR/OF An (apparent) change in observed/detected frequency (pitch), as a result of the relative motion between a source and an observer (listener). 'n (Skynbare) verandering in waargenome frekwensie (toonhoogte), as gevolg van die relatiewe beweging tussen die bron en 'n waarnemer (luisteraar). (2) 6.1.2 v v v fL = fS OR/OF fL = fS  v vs v vs v  512,64 = ( ) (550)  v v = 343,04 m∙s-1  Range/Gebied: (332,14 - 343,04 m∙s-1) (5) 6.1.3 a) Remains the same/Bly dieselfde  (1) b) Remains the same/Bly dieselfde  (1) c) Decreases/Afneem  (1) 6.2.1 AWAY FROM/WEG VAN  (1) 6.2.2 Marking criteria/Nasienkriteria: Second mark is only awarded if red is linked to lower frequency/longer wavelength./ Tweede punt word slegs toegeken indien vergelyking getref word tussen rooi en laer frekwensie en langer golflengte.  A lower frequency/longer wavelength  is detected.  The spectral lines are shifted to the red end of the spectrum.   'n Laer frekwensie/langer golflengte word waargeneem.  Die spektrale lyne word geskuif na die rooi end van die spektrum. (2) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 19 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 7/VRAAG 7 7.1. Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. Electric field is a region/space in which an electric charge experiences a force. 'n Gebied/ruimte waarin 'n elektriese lading 'n krag ondervind. NOTE: If electric field at a point is defined: ⁄ NOTA: Indien elektiese veld by 'n punt gedefinieer is: ⁄ (2) 7.2 Criteria for sketch/Kriteria vir skets Marks/Punte Correct direction of field lines./Korrekte rigting van veldlyne.  Correct shape of the electric field lines between charges and on the outside of the charges./Korrekte vorm van elektrieseveld tussen  ladings en die buitekant van die ladings. No field lines crossing each other. Field lines must touch the charge, but not go inside the charge./Geen veldlyne wat mekaar kruis nie.  Veldlyne moet die lading raak, maar nie die lading binnegaan nie. Note: If learner draws field pattern of two opposite charges: ⁄ If only one charge is drawn, max: ⁄ for direction. Nota: Indien leerder elektrieseveld van twee teenoorgestelde ladings teken: ⁄ . Indien slegs een lading geteken is, maks: ⁄ vir rigting. (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 20 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne 7.3 Marking criteria/Nasienkriteria kQ  Formula/Formule: E = 2 .  r  Correct substitution for either A or B /Korrekte vervanging vir of A of B.  Substitution of 27 or –27 for Enet. /Vervanging van 27 of –27 vir Enet.  Subtraction of/Aftrek van (EA – EB OF/OR EB – EA )   Correct final answer/Korrekte finale antwoord: 0,87 (m)  OPTION 1/OPSIE 1 kQ E= 2  r ( x 0 )( 3 x 0- ) EA = r Any one/Enige een ( x 0 )( 3 x 0- ) EB = r Enet = EA – EB ( x 0 )( 3 x 0- )  ( x 0 )( 3 x 0- ) 27 = r r r = 0,87 (m)  OR/OF kQ E=  r2 ( x 0 )( 3 x 0- ) EA = Any one/Enige een r ( x 0 )( 3 x 0- ) EB = r Enet = EB – EA ( x 0 )( 3 x 0- ) ( x 0 )( 3 x 0- ) –27 =  r r r = 0,87 (m ) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 21 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 2/OPSIE 2 Marking criteria/Nasienkriteria:  Formula for Coulomb's law./Formule vir Coulomb se wet.   Correct substitution in Coulomb's formula for either FA or FB./  Korrekte vervanging in Coulomb se formule vir of FA of FB.  Substitution of 27q or –27q for Fnet. /Vervanging van 27q of –27q vir Fnet.  Subtraction of/Aftrek van (FA – FB OF/OR FB – FA)   Correct final answer/Korrekte finale antwoord: 0,87 (m) F=  r ( x 0 )( 3 x 0- ) q FA = r Any one/Enige een ( x 0 )( 3 x 0- ) q FB = r Fnet = FA – FB ( x 0 )( 3 x 0- ) q ( x 0 )( 3 x 0- ) q 27q = r r r = 0,87 (m)  OR/OF F=  r ( x 0 )( 3 x 0- ) q FA = r Any one/Enige een ( x 0 )( 3 x 0- ) q FB = r –Fnet = FB – FA ( x 0 )( 3 x 0- ) q  ( x 0 )( 3 x 0- ) q –27q = – r r r = 0,87 (m)  (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 22 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne 7.4 OPTION 1/OPSIE 1 F = Eq  = (27)(1,6 x 10-19)  = 4,32 x 10-18 N  Note: Do not penalize for -1,6 x 10-19, but penalize for negative final answer. Nota: Moenie penaliseer vir -1,6 x 10-19 nie, maar wel vir negatiewe finale antwoord. POSITIVE MARKING FROM QUESTION 7.3. POSITIEWE NASIEN VANAF VRAAG 7.3. OPTION 2/OPSIE 2 F=  r Fnet = FA – FB  ( x 0 )( 3 x 0- )( , x 0- ) ( x 0 )( 3 x 0- )( , x 0- ) Fnet= 0,8 , = 4,28 x 10-18 N  OR/OF F=  r Fnet = FB – FA  ( x 0 )( 3 x 0- )( , x 0- ) ( x 0 )( 3 x 0- )( , x 0- ) Fnet= , 0,8 = 4,28 x 10-18 N = 4,28 x 10-18 N  (3) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 23 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 8/VRAAG 8 8.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The potential difference (voltage) across a conductor is directly proportional to the current in the conductor at constant temperature.  Die potensiaalverskil (spanning) oor 'n geleier is direk eweredig aan die stroom in die geleier by konstante temperatuur. OR/OF The current in a conductor is directly proportional to the potential difference (voltage) across the conductor if temperature is constant.  Die stroom in 'n geleier is direk eweredig aan die potensiaalverskil (spanning) oor die geleier indien die temperatuur konstant is. OR/OF The ratio of potential difference to current is constant provided the temperature remains the same.  Die verhouding van potensiaalverskil tot stroom is konstant indien die temperatuur konstant bly. (2) 8.2.1 Marking criteria/Nasienkriteria  Any correct formula to calculate the effective resistance of any of the two parallel combinations./Enige korrekte formule om die effektiewe weerstand van enige een van die parallel kombinasies te bereken.  Correct substitution in formula to calculate effective resistance of both parallel combinations./Korrekte vervanging in formule om effektiewe weerstand van beide parallel gedeeltes te bereken.   Adding the 10 Ω to the first parallel combination and using this to calculate the external resistance (Rext)/Bymekaartel van 10 Ω en die gebruik daarvan om die eksterne weerstand te bereken   Correct final answer/Korrekte finale antwoord: 7,5 Ω  OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 R12L = RL + ( ) = + OR/OF R12 =( ) p  10  10 0x 0 = 10 +  = +  OR/OF  10  10 0 0 0 0 = 15 Ω R12 = 5 Ω  Any one/ 3 enige een R12L = RL + R12 Rp = ( )  3 = 10 + 5 15  15 Ω  Any one/ Rp =  15  15 enige een Rp= 7,5 Ω  = + p 3 = +  p Rp = 7,5 Ω  (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 24 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne 8.2.2 POSITIVE MARKING FROM QUESTION 8.2.1. POSITIEWE NASIEN VANAF VRAAG 8.2.1. OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 ε = I(R + r)  ε 12 = I (7,5 + 0,5)  R= / RT =  I = 1,5 A  (7,5 + 0,5) =  I = 1,5 A  (3) 8.2.3 POSITIVE MARKING FROM QUESTIONS 8.2.1 AND 8.2.2 POSITIEWE NASIEN VANAF VRAE 8.2.1 EN 8.2.2 Marking criteria/Nasienkriteria  Substitution of the correct current or potential difference for R3./ Vervanging van die korrekte stroom of potensiaalverskil vir R3.  Correct formula for power, leading to the answer./Korrekte formule vir drywing wat lei tot die antwoord.   Correct substitution to calculate power./Korrekte vervanging om drywing te bereken.  Correct final answer/Korrekte finale antwoord: 8,44 W. 1,5 = 2IR3 ext I = 0,75 A Rext = 3 = x Itotal 3 =( ) 3 3 V = (7,5)(1,5) V = 11,25 V , = x 1,5 I=( )(1,5) ext R3 = = 0,75 A = 0,75 A , 15 = I = 0,75 A OPTION 1/ OPTION 2/ OPTION 3/ OPTION 4/ OPSIE 1 OPSIE 2 OPSIE 3 OPSIE 4 P = I2R V = IR V = IR ε = Vext + Ir  = (0,75)215  = (0,75)(15)  = (0,75)(15) 12=Vext + (1,5)(0,5) Vext = 11,25 V = 8,44 W = 11,25 V = 11,25 V P=  P=  P = VI , = (11,25)(0,75)  , =  = 8,44 W =  = 8,44 W = 8,44 W (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 25 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne 8.3.1 INCREASES/NEEM TOE  (1) 8.3.2  Total resistance of the circuit increases and total current in circuit decreases.  Vinternal/internal volts/Vlost decreases and Vexternal/external volts /VRL increases.   Power output increases  therefore brightness increases.  Totale weerstand van die stroombaan neem toe en die totale stroom neem af.  Vintern/interne volts/Vverlore neem af en Vekstern/eksterne volts /VRL neem toe  Drywing neem toe daarom sal die herlderheid toeneem. OR/OF  ε r 12 = I(15 + 0,5) I = 0,77 A   IL has increased/IL het toegeneem   Power output increases  therefore brightness increases. Drywing neem toe daarom sal die herlderheid toeneem. (3) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 26 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 9/VRAAG 9 9.1.1 Split ring/Commutator/Splitring/Kommutator  (1) 9.1.2 Electrical to mechanical/kinetic  Elektries na meganies/kineties  (1) 9.1.3 Clockwise/Kloksgewys  (2) 9.1.4 Any two of the following./Enige twee van die volgende:  Increase the strength of the magnetic field e.g. use stronger magnets/bring magnets closer/use curved magnets./ Toename in die sterkte van die magneetveld bv. gebruik sterker magnete/bring magnete nader aan mekaar/gebruik geboë magnete.  Increase the current./use battery with higher potential difference./more cells in series./Increase EMF./ Verhoog die stroom./gebruik battery met hoër potensiaalverskil./meer selle in serie./verhoog EMK.  Increase the area of the coil./Vergroot die oppervlakte van die spoel.  Increase the number of turns in the coil./Vermeerder die aantal windings in die spoel. (2) 9.2.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. Root-mean-square current is the alternating current (AC) that dissipates the same amount of energy/heating effect as an equivalent direct current (DC) current.  Die wortelgemiddeldekwadraat-stroom is die wisselstroom (WS) wat dieselfde hoeveelheid energie/verhittingseffek as 'n ekwivalente gelykstroom (GS) het. (2) Note: If energy or heating effect is omitted: 0⁄ Nota: Indien energie of verhittingseffek uitgelaat is: ⁄ 9.2.2 m x Irms =  √ 3, =  √ = 2,55 A  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 27 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne 9.2.3 Marking criteria/Nasienkriteria  Formula for W./Formule vir W.   Correct substitution for W./Korrekte vervanging vir W.   Correct final answer./Korrekte finale antwoord: 69 168 J  (6,92 x 103) (Range/Gebied: 69 167,56 J – 69 168,44 J) OPTION 1/OPSIE 1 W = VI∆t  = (220)(2,62)(120)  = 69 168 J  (6,92 x 103J) OPTION 2/OPSIE 2 V = IR 220 = 2,62R 83, Ω W = I2R∆t = (2,62)2(83,97)(120)  W= ∆t  = 69 168,44 J  (6,92 x 103 J) 0 = ( )(120)  83, = 69 167,56 J  (6,92 x 103 J) OPTION 3/OPSIE 3 Pave = VrmsIrms Pave = rms R rms = (220)(2,62) = (2,62)2(83,97) Pave = = 576,4 W = 576,4 W 0 = 83, = 576,4 W W = P∆t  = (576,4)(120)  = 69 168 J  (6,92 x 103 J) OPTION 4/OPSIE 4 q = I∆t q = (2,62)(120) q = 314,4 C W = Vq  W = 220 x 314,4  W = 69168 J  (6,92 x 103 J) (3) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 28 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 10/VRAAG 10 10.1.1 Marking criteria/Nasienkriteria If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark./Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af. The minimum energy (of incident photons) that can eject electrons from a metal/surface.  Die minimum energie (van invallende fotone) wat elektrone kan vrystel vanuit 'n metaal/oppervlak. NOTE: If reference to frequency: ⁄ NOTA: Indien na frekwensie verwys word: ⁄ (2) 10.1.2 OPTION 1/OPSIE 1 E = hf  E = (6,63 x 10-34)(2,8 x 1016)  E = 1,86 x 10-17 (J)  Since/Aangesien E > W0 (or E – W 0 > 0), electrons will be ejected/elektrone sal vrygestel word OPTION 2/OPSIE 2 W 0 = hf0 6,63 x 10-19 = (6,63 x 10-34)f0  f0 = 1 x 1015 (Hz)  Since/Aangesien f > f0 (or f – f0 > 0), electrons will be ejected/elektrone sal vrygestel word OPTION 3/OPSIE 3 c W0 = 0 , 3 x 0-3 3 x 08  both/beide 6,63 x 10-19 = 0  any one/ -7 0 = 3 x 10 (m) enige een v 3 x 108 = 2,8 x 1016  both/beide -8 λ = 1,07 x 10 (m) Since/Aangesien 0 > λ (or 0 – λ > 0) , electrons will be ejected/elektrone sal vrygestel word. OPTION 4/OPSIE 4 E = W 0 + Ek(max)  Any one/Enige een hf = W 0 + Ek(max) (6,63 x 10-34)(2,8 x 1016) = 6,63 x 10-19 + Ek(max)  Ek(max) = 1,79 x 10-17 (J)  Since/Aangesien Ek(max) > 0,  electrons will be ejected/elektrone sal vrygestel word (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 29 DBE/November 2023 NSC/NSS – Marking Guidelines/Nasienriglyne 10.1.3 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 OPTION 3/OPSIE 3 F F=  E= E= r r  q  × 0 , × 0- 0,0  ( x 0 )( , x 0- ) E= =  0,027 = 0, , x 0 - 0, E = 4 8 0 000 N∙C-1 = 5 000 N∙C-1 Q2 = 5,56 x 10-9 C (4,86 x 106) (5 x 103) F E=  E=  q r 0,0 × 0 4860000 =  5000 = B  q 0, q = 5,56 x 10-9 C QB = 5,56 x 10-9 C n= e , × 0- n=  - , x 0 n = 3,47 x 1010 (electrons/elektrone)  number of photons/aantal fotone = n = 3,47 x 1010(3,475 x 1010) (6) 10.2.1 (Line) Absorption/(Lyn) Absorbsie (1) 10.2.2 Continuous spectrum of white light/rainbow of colours with dark/black lines (replacing specific frequencies)./Kontinue spektrum van wit lig/reënboog van kleure met donker/swart lyne (wat spesifieke frekwensies vervang). (2) 10.2.3 Diagram B  (2) [17] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou