Physical Sciences Grade 12 P1 NSC Supplementary 2013 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) FEBRUARY/MARCH/FEBRUARIE/MAART 2013 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 18 pages. Hierdie memorandum bestaan uit 18 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 2 DBE/Feb.–Mar./Feb.–Mrt. 2013 NSC/NSS – Memorandum SECTION A QUESTION 1/VRAAG 1 1.1 X-rays/X-strale (1) 1.2 Momentum  (1) 1.3 Huygens (principle)/Huygens (se beginsel) (1) 1.4 ampere/ampère  (1) 1.5 Isolated/closed  (1) Geïsoleerde/geslote [5] QUESTION 2/VRAAG 2 2.1 B  (2) 2.2 A  (2) 2.3 A  (2) 2.4 D  (2) 2.5 A  (2) 2.6 D  (2) 2.7 B  (2) 2.8 A  (2) 2.9 B  (2) 2.10 C  (2) [20] TOTAL SECTION A/TOTAAL AFDELING A: 25 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 3 DBE/Feb.–Mar./Feb.–Mrt. 2013 NSC/NSS – Memorandum SECTION B/AFDELING B QUESTION 3/VRAAG 3 3.1 50 N downwards/afwaarts (2) 3.2 3.2.1 OPTION 1/OPSIE 1 Notes/Aantekeninge Downward positive: Afwaarts positief: Accept/Aanvaar: g or/of a Accept/Aanvaar: v f = v i + 2a∆y  2 2 v 2f = v i2 + 2a∆x ∴ v f = 0 + 2(9,8)(0,8)  2 2 v 2 = u 2 + 2as ∴ v f = 3,96 m∙s-1  downward /afwaarts Downward negative: Afwaarts negatief: v f = v i + 2a∆y  2 2 ∴ v f = 02 + 2(-9,8)(-0,8)  2 ∴ v f = -3,96 m∙s-1 ∴v f = 3,96 m∙s-1  downward /afwaarts OPTION 2/OPSIE 2 Notes/Aantekeninge (E p + E k ) top/bo = (E p + Ek) bottom/onder  mgh + 0 = 0 + ½mv2 Accept/Aanvaar: (9,8)(0,8) = ½ v2 (U + K) top/bo = (U + K) bottom/onder v = 3,96 m∙s-1  downward/afwaarts (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/Feb.–Mar./Feb.–Mrt. 2013 NSC/NSS – Memorandum 3.2.2 POSITIVE MARKING FROM QUESTION 3.2.1 Notes/Aantekeninge: POSITIEWE NASIEN VAN VRAAG 3.2.1 Substitution: F app and v f must OPTION 1/OPSIE 1 have the same sign. Downward positive/Afwaarts positief: Substitusie: F app en v f moet F net Δt = Δp OR F net Δt = m(v f – v i )  dieselfde tekens hê. (F app + mg)Δt = Δp (-50 + (0,2)(9,8)Δt  = 0,2(- 3,43 – 3,96)  ∴ Δt = 0,03 s  (3 x 10-2 s) Downward negative/Afwaarts negatief: F net Δt = Δp OR F net Δt = m(v f – v i )  (F app + mg)Δt = Δp (50 - (0,2)(9,8)Δt = 0,2[3,43 – (-3,96)]  ∴ Δt = 0,03 s  (3 x 10-2 s) OPTION 2/OPSIE 2 Notes/Aantekeninge Downward positive/Afwaarts positief: Accept/Aanvaar: F net = ma v = u + at F app + mg = ma v f = v i + a ∆t (-50 + (0,2)(9,8) = 0,2a  both formulae ∴ a = -240,2 m∙s-2 beide formules v f = vi + a ∆t -3,43 = 3,96 + (-240,2)∆t  ∴ Δt = 0,03 s  (3 x 10-2 s) Downward negative/Afwaarts negatief: F net = ma F app + mg = ma (50 – (0,2)(9,8) = 0,2a  both formulae ∴ a = 240,2 m∙s -2 beide formules v f = vi + a ∆t 3,43 = -3,96 + (240,2)∆t  ∴ Δt = 0,03 s  (3 x 10-2 s) (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/Feb.–Mar./Feb.–Mrt. 2013 NSC/NSS – Memorandum 3.3 POSITIVE MARKING FROM QUESTION 3.2. POSITIEWE NASIEN VAN VRAAG 3.2. OPTION/OPSIE 1 Ground as zero reference and downward negative: Grond as nulverwysing en afwaarts negatief: Marks/ Criteria for graph/Kriteria vir grafiek: Punte Correct shape (both curves)  Korrekte vorm (beide krommes) Graph starts at y = 0,8 m at t = 0 s  Grafiek begin by y = 0,8 m at t = 0 s Second maximum height at y = 0,6 m  Tweede maksimum by y = 0,6 m s Contact time shown as space on x axis between two curves. Kontaktyd aangetoon as spasie op x-as tussen twee  krommes. Time at which ball leaves the floor shown as t = 0,434 s.  Tyd wanneer die bal die vloer verlaat getoon as t = 0,434 s. 0,8 Position/Posisie (m) 0,6 0 0,404 0,434 Time/tyd(s) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/Feb.–Mar./Feb.–Mrt. 2013 NSC/NSS – Memorandum OPTION/OPSIE 2 Ground as zero reference and downward positive: Grond as nulverwysing en afwaarts positief: Marks/ Criteria for graph/Kriteria vir grafiek: Punte Correct shape (both curves)  Korrekte vorm (beide krommes) Graph starts at y = -0,8 m at t = 0 s  Grafiek begin by y = -0,8 m at t = 0 s Second maximum height at y = -0,6 m  Tweede maksimum by y = -0,6 m s Contact time shown as space on x axis between two curves. Kontaktyd aangetoon as spasie op x-as tussen twee  krommes. Time at which ball leaves the floor shown as t = 0,434 s.  Tyd wanneer die bal die vloer verlaat getoon as t = 0,434 s. 0,404 0,43 0 4 Position/Posisie (m) Time/tyd(s) -0,6 -0,8 (5) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/Feb.–Mar./Feb.–Mrt. 2013 NSC/NSS – Memorandum QUESTION 4/VRAAG 4 4.1 The total (linear) momentum remains constant/is conserved  in an isolated/a closed system/the absence of external forces.  Die totale lineêre momentum bly konstant/behoue  in 'n geïsoleerde sisteem/geslote sisteem/die afwesigheid van eksterne kragte.  Notes/Aantekeninge: The mark for 'closed/isolated system' is only awarded if used in conjunction with momentum. Die punt vir 'geslote/geïsoleerde sisteem' word slegs toegeken indien saam met momentum gebruik. Accept: The total momentum before a collision equals the total momentum after a collision in a closed system. Aanvaar: Die totale momentum voor 'n botsing is gelyk aan die totale momentum na 'n botsing in 'n geslote sisteem. (2) 4.2 To the right as positive/Na regs as positief: Σ p before/voor = Σ p after/na  (0,01)(300) + (1,99)(0) = (0,01 + 1,99)v f2  ∴ v f2 = 1,5 m·s-1  To the right as negative/Na regs as negatief: Σ p before/voor = Σ p after/na  (0,01)(-300) + (1,99)(0) = (0,01 + 1,99)v f2  ∴v f2 = -1,5 m·s-1 ∴v f2 = 1,5 m·s-1  Other formulae/Ander formules: Notes/Aantekeninge: m 1 v i1 + m 2 v i2 = m 1 v f1 + m 2 v f2 3 or/of If no formula/principle – Max. 4 m 1u 1 + m 2u 2 = m 1v1 + m 2v2 3 or/of Indien geen formule/beginsel – Maks. m 1 v i1 + m 2 v i2 = (m 1 + m 2 )v f2 4 (4) 4.3 Inelastic/Onelasties  Kinetic energy is not conserved./Kinetiese energie bly nie behoue nie. OR/OF Inelastic/Onelasties Objects stick together/Voorwerpe heg aan mekaar.  OR/OF Inelastic/Onelasties Structural damage to the block./Strukturele skade aan blok.  OR/OF Inelastic/Onelasties There is deformation to the block/bullet./Daar is vervorming van die blok. OR/OF Inelastic/Onelasties  Energy converted to other forms such as sound and heat./Energie word omgeskakel na ander vorms soos klank en hitte. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 8 DBE/Feb.–Mar./Feb.–Mrt. 2013 NSC/NSS – Memorandum 4.4 POSITIVE MARKING FROM QUESTION 4.2. POSITIEF NASIEN VAN VRAAG 4.2. Option 1/Opsie 1: Notes/Aantekeninge: W net = ΔK  Accept/Aanvaar: OR /OF Ek 1 1 F net ∆xcos θ = mv 2f − mv i2 2 2 1 (8)Δxcos180°  = (2)(02 − 1,52 )  2 ∴ ∆x = 0,28 m  Option 2/Opsie 2: Notes/Aantekeninge: F net = ma Accept/Aanvaar: ∴ (-8) = 2a  v f = v i + 2a∆y 2 2 ∴ a = - 4 m·s-2 Both v 2 = u 2 + 2as formulae v f = v i + 2a∆x 2 2 Beide 02 = (1,5)2 + 2(-4) ∆x  formules ∴ ∆x = 0,28 m  (4) [12] QUESTION 5/VRAAG 5 5.1 The total mechanical energy remains constant/is conserved in an isolated/closed system.  Die totale meganiese energie bly konstant/bly behoue in 'n geïsoleerde/geslote sisteem OR/OF The sum of the potential and kinetic energy remains constant  in an isolated/closed system.  Die som van die potensiële en kinetiese energies bly konstant in 'n geïsoleerde/geslote sisteem Notes/Aantekeninge: The mark for 'closed/isolated system' is only awarded if used in conjunction with energy. Die punt vir 'geslote/geïsoleerde sisteem' word slegs toegeken indien saam met energie gebruik. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 9 DBE/Feb.–Mar./Feb.–Mrt. 2013 NSC/NSS – Memorandum 5.2 5.2.1 Free-body diagram Accept/Aanvaar: Force diagram/kragtediagram Vrye kragtediagram   w w Accepted labels/Aanvaarde benoemings F g /F w /force of Earth on boy/weight/392 N/mg/gravitational force w F g /F w /krag van Aarde op seun/gewig/392 N/mg/gravitasiekrag (1) 5.2.2 Free-body diagram Accept: Force diagram Vrye kragtediagram Aanvaar: Kragtediagram f N f N   w w OR/OF OR/OF f f N N   w// w//  w⊥  w⊥ Accepted labels/Aanvaarde benoemings F g /F w /force of Earth on girl/weight/215,6 N/mg/gravitational force w F g /F w /krag van Aarde op meisie/gewig/215,6 N/mg/gravitasiekrag F friction /F f /friction f F wrywing /F f /F w /wrywing N F N /F normal /normal force/force of slide or surface on girl (3) 5.3 1 1 Notes/Aantekeninge: K i + U i = K f + U f  OR mv i2 + mgh i = mv 2f + mgh f Accept/Aanvaar 2 2 Ep & Ek 1 0 + (40)(9,8)(1,5)  = ( 40)v 2f + 0  2 ∴ v f = 5,42 m·s-1  (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 10 DBE/Feb.–Mar./Feb.–Mrt. 2013 NSC/NSS – Memorandum 5.4 Option 1/Opsie 1 Accept/ W net = ΔK  Aanvaar: 1 1 ∆y/∆x w∆xcos θ + f∆xcos θ = mv 2f − mv i2 2 2 1 1 mg∆xcos θ + f∆xcos θ = mv 2f − mv i2 2 2 1 (22)(9,8)(3)cos60°  + (1,9)(3)cos180° = (22)( v 2f − 02 )  2 ∴ v f = 5,37 m·s-1  Option 2/Opsie 2 Accept/ W net = ΔK  Aanvaar: w//∆xcos θ + f∆xcos θ = ½ mv f 2 – ½ mv i 2 ∆y/∆x 1 1 mgsin θ xcos θ + f∆xcos θ = mv 2f − mv i2 2 2 1 (22)(9,8)sin30°(3)cos0°  + (1,9)(3)cos180° = (22)( v 2f − 02 )  2 ∴ v f = 5,37 m·s-1  Option 3/Opsie 3 Accept/ W net = ΔK  Aanvaar 1 1 h/∆y/∆x mghcos θ + f∆xcos θ = mv 2f − mv i2 2 2 1 (22)(9,8)(1,5)cos0°  + (1,9)(3)cos180° = (22)( v 2f − 02 )  2 ∴v f = 5,37 m·s-1  Option 4/Opsie 4 Accept/ W net = ΔK  Aanvaar - ∆U + W f = ∆K h/∆y/∆x - (mgh f – mgh i ) + W f = ∆K 1 - (0 – (22)(9,8)(1,5)  + (1,9)(3)cos180° = (22)( v 2f − 02 )  2 ∴ v f = 5,37 m·s-1  (5) 5.5 Equal to/Gelyk aan (1) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 11 DBE/Feb.–Mar./Feb.–Mrt. 2013 NSC/NSS – Memorandum QUESTION 6/VRAAG 6 6.1 The approaching observer (higher f) passes the source at t = 6 s and moves away (lower f) from the source.  Die naderende waarnemer (hoër f) beweeg verby die bron by t = 6 s en beweeg weg (laer f) van die bron af. (1) 6.2 6.2.1 OPTION 1/OPSIE 1 Notes: Approaching observer: • Any other Doppler formula, e.g. Naderende waarnemer: v − vL fL = - Max. 3 v ± vL v + vL v − vs 4 fL = f s OR/OF fL = fs  v ± vs v Aantekeninge: 340 + v L ∴ 900  = (850)  • Enige ander Dopplerformule, bv. 340 v − vL ∴ v L = 20 m∙s-1  fL = - Maks. 3 v − vs 4 OPTION 2 / OPSIE 2 Observer moving away: Waarnemer beweeg weg: v ± vL v − vL fL = f s OR/OF fL = fs  v ± vs v 340 − v L ∴800  = (850)  340 ∴ v L = 20 m∙s-1  (5) 6.2.2 POSITIVE MARKING FROM QUESTION 6.2.1 POSITIEWE NASIEN van VRAAG 6.2.1 Option 1/Opsie 1: Notes/Aantekeninge Δx = v i Δt + 2 a∆t  1 2 Accept/Aanvaar: s = ut / s = vt = (20)(6) + ½(0) ∆ t ∴ ∆ x = 120 m  s = ut + 1 at 2 2 Δy = v i Δt + 1 a∆t 2 Option 2/Opsie 2: 2 Δx = vΔ t  = (20)(6)  ∴ ∆ x = 120 m  (3) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 12 DBE/Feb.–Mar./Feb.–Mrt. 2013 NSC/NSS – Memorandum QUESTION 7/VRAAG 7 7.1 B (1) 7.2 Constructive  interference  Konstruktiewe interferensie (2) 7.3 7.3.1 c = fλ  Notes/Aantekeninge 3 x 108 = 4,54 x 1014 λ  Accept/Aanvaar v = f λ ∴ λ = 6,61 x 10-7 m  (3) 7.3.2 POSITIVE MARKING FROM QUESTION 7.3.1 POSITIEWE NASIEN VAN VRAAG 7.3.1 OPTION 1/OPSIE 1 Notes/Aantekeninge 1 central band Accept final answer in range: tan θ = 2 9,94 x10-6 to 9,95 x10-6 m screen dis tan ce Aanvaar finale antwoorde in 1 (0,2)  die gebied: 9,94 x10-6 to 9,95 x10-6 m ∴ tan θ = 2 1,5  IF distance is not half of ∴ θ = 3,81° 4 mλ central band: Max sin θ =  6 a INDIEN afstand nie helfte van (1)(6,61× 10 −7 ) 4 sin3,81° =  sentrale band is nie: Maks a 6 ∴ a = 9,95 x 10-6 m  OPTION 2/OPSIE 2 Notes/Aantekeninge 1 central band Accept final answer in range: tan θ = 2 9,94 x10-6 to 9,95 x10-6 m screen dis tan ce Aanvaar finale antwoorde in 1 (0,2)  die gebied: 9,94 x10-6 to 9,95 x10-6 m ∴ tan θ = 2 1,5  IF distance is not half of ∴ θ = 3,81° 4 central band: Max mλ 6 sin θ =  a INDIEN afstand nie helfte van ( −1)(6,61× 10 −7 ) sentrale band is nie: Maks 4 sin(-3,81°) =  6 a ∴ a = 9,95 x 10-6 m  (6) 7.4 Decreases/Verminder  (1) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 13 DBE/Feb.–Mar./Feb.–Mrt. 2013 NSC/NSS – Memorandum QUESTION 8/VRAAG 8 8.1 Stores (electric) charge/energy.  Stoor (elektriese) lading/energie. OR/OF Releases (stored) charge instantly/very fast.  Stel (gestoorde) lading onmiddellik vry/baie vinnig vry. (1) 8.2 The brightness of the bulb decreases (gradually)  until it stops glowing/dies. Die helderheid van die gloeilamp verminder (geleidelik). OR/OF The bulb glows dimmer  until it stops glowing/dies. Die gloeilamp gloei flouer totdat dit ophou gloei/uitbrand. (1) 8.3 8.3.1 0 (V)  (1) 8.3.2 12 V  (1) 8.4 POSITIVE MARKING FROM QUESTION 8.3.2. POSITIEWE NASIEN VAN VRAAG 8.3.2. V 8.4.1 E=  d 12 =  5,4 × 10 − 3 = 2,22 x 103 V·m-1  (2 222,22 V·m-1) (3) 8.4.2 POSITIVE MARKING FROM QUESTION 8.4.1. POSITIEWE NASIEN VAN VRAAG 8.4.1. F E=  q F ∴ 2,22 x 103 =  1,6 × 10 −19 ∴ F = 3,56 x 10-16 N  (3) 8.5 8.5.1 5,4 mm – 3,8 mm = 1,6 mm  (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 14 DBE/Feb.–Mar./Feb.–Mrt. 2013 NSC/NSS – Memorandum 8.5.2 POSITIVE MARKING FROM QUESTION 8.4.2 & 8.5.1. POSITIEWE NASIEN VAN VRAAG 8.4.2 & 8.5.1. W = FΔxcosθ  = (3,56 x 10-16)(1,6 x 10-3)cos 0° = 5,69 x 10-19 J  (4) [15] QUESTION 9/VRAAG 9 9.1 9.1.1 Potential difference/Potensiaalverskil  (1) 9.1.2 Temperature/Temperatuur  Resistance/Weerstand (1) 9.1.3 Current is directly proportional to potential difference.  Stroom is direk eweredig aan potensiaalverskil OR/OF The ratio of potential difference to current is constant.  Die verhouding van potensiaalverskil tot stroom is konstant. IF/INDIEN: Current is proportional to potential difference.  Stroom is eweredig aan potensiaalverskil. (2) 9.1.4 0,18 − 0  Notes/Aantekeninge: Gradient/m = = 0,36 Accept any set of correct values 0,5 − 0  from the graph. 1 Aanvaar enige stel waardes vanaf R= = 2,78 Ω  0,36 die grafiek. (4) 9.2 9.2.1 1 1 1 = +  R p R1 R 2 1 1 = +  6 10 ∴ R p = 3,75 Ω  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 15 DBE/Feb.–Mar./Feb.–Mrt. 2013 NSC/NSS – Memorandum 9.2.2 POSITIVE MARKING FROM QUESTION 9.2.1. POSITIEWE NASIEN VAN VRAAG 9.2.1. OPTION 2 / OPSIE 2 Notes/Aantekeninge V p = I 10 Ω R Do not penalise for = 0,6 x 10  subscripts. = 6V  Any one Moenie penaliseer indien Enige een onderskrifte weggelaat is Vp Ip = nie. Rp 6 =  3,75 = 1,6 A  OPTION 1/OPSIE 2 Notes/Aantekeninge V 10 Ω = I 10 Ω R Do not penalise for = 0,6 x 10  subscripts. = 6V  Any one Moenie penaliseer indien Enige een onderskrifte weggelaat is V I 6Ω = nie. R 6 =  6 =1A I tot = 1 + 0,6 = 1,6 A  (4) 9.2.3 POSITIVE MARKING FROM QUESTION 9.2.1. POSITIEWE NASIEN VAN VRAAG 9.2.1. E = I(R + r)  30  = 1,6(3,75 + 5 + 8 + r)  ∴r=2Ω (4) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 16 DBE/Feb.–Mar./Feb.–Mrt. 2013 NSC/NSS – Memorandum QUESTION 10/VRAAG 10 10.1 Q/split ring commutator/commutator  Q/splitringkommutator/kommutator (1) 10.2 Replace Q/split ring commutator with slip rings.  Vervang Q/splitringkommutator met sleepringe. (1) 10.3 AC can be stepped-up at power stations/WS kan by die kragstasie verhoog word  to reduce energy loss during transmission./om energieverlies tydens transmissie te verminder.  (2) 10.4 Notes/Aantekeninge 10.4.1 I If subscripts omitted: no mark for I rms/wgk = max/maks  formula 2 Indien onderskifte weggelaat is: 0,35 geen punt vir formule =  2 ∴ I rms/wgk = 0,25 A  (3) 10.4.2 OPTION 1/OPSIE 1 Notes/Aantekeninge 2 Vrms/wgk Do not penalise if subscripts are omitted. Pave / gemid =  Moenie penaliseer indien onderskrifte R weggelaat is nie 240 2 60 =  R ∴ R = 960 Ω  OPTION 2/OPSIE 2 P ave/gemid = I2 rms/wgk R  60 = (0,25)2R  ∴ R = 960 Ω  OPTION 3/OPSIE 3 Notes/Aantekeninge Vrms/wgk Accept/Aanvaar: R=  V Irms/wgk R= as formula/formule I 240 =  0,25 = 960 Ω  (3) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 17 DBE/Feb.–Mar./Feb.–Mrt. 2013 NSC/NSS – Memorandum QUESTION 11/VRAAG 11 11.1 Quantum/packet of energy/Kwantum/pakkie energie found in light/In lig gevind  (2) 11.2 11.2.1 OPTION 1/OPSIE 1 hc E=  λ  -19 (6,63 × 10 −34 )(3 × 108 ) 6,9 x 10  =  λ ∴ λ = 2,9 x 10-7 m  OPTION 2 / OPSIE 2 Notes/Aantekeninge E = hf Accept/Aanvaar: v = f λ 6,9 x 10-19  = 6,63 x 10-34f   Both formulae ∴ f = 1,04 x 1015 Hz Beide formules c = fλ 3 x 108 = 1,04 x 1015 λ  ∴ λ = 2,88 x 10-7 m  (5) 11.2.2 E = Wo + Ek  6,9 x 10-19 = 6,4 x 10-19 + E k  ∴ E k = 5 x 10-20 J  (3) 11.3 11.3.1 Increases/Vermeerder  • More photons (packets of energy) strike the surface of the metal per unit time./Meer fotone (pakkies energie) tref die oppervlakte van die metaal per eenheid tyd.  • More (photo)electrons ejected per unit time./Meer (foto)elektrone vrygestel per eenheid tyd.  (3) 11.3.2 Increases/Vermeerder  • (Photo)electrons are emitted with higher kinetic energy/move faster./ (Foto)elektrone word vrygestel met hoër kinetiese energie/beweeg vinniger.  • Increase in rate of flow of charge./Same number of charges pass a point in a shorter time./Toename in tempo van vloei van lading/dieselfde aantal lading beweeg verby 'n punt in 'n korter tyd.  (3) [16] TOTAL SECTION B/TOTAAL AFDELING B: 125 GRAND TOTAL/GROOTTOTAAL: 150 Copyright reserved/Kopiereg voorbehou