NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12

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1 NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 1 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) NOVEMBER/NOVEMBER 009(1) MEMORANDUM PUNTE: 150 This memorandum consists o 1 pages. Hierdie memorandum bestaan uit 1 bladsye.

2 Physical Sciences (P1)/Fisiese Wetenskappe (V1) - - DoE/November 009(1) Learning Outcomes and Assessment Standards Leeruitkomste en Assesseringstandaarde LO1/LU1 LO/LU LO3/LU3 AS1.1.1: Design, plan and conduct a scientiic inquiry to collect data systematically with regard to accuracy, reliability and the need to control variables. Ontwerp, beplan en voer 'n wetenskaplike ondersoek uit om data te versamel ten opsigte van akkuraatheid, betroubaarheid en die kontroleer van veranderlikes. AS1.1.: Seek patterns and trends, represent them in dierent orms, explain the trends, use scientiic reasoning to draw and evaluate conclusions, and ormulate generalisations. Soek patrone en tendense, stel dit in verskillende vorms voor, verduidelik die tendense, gebruik wetenskaplike beredenering om gevolgtrekkings te maak en te evalueer, en ormuleer veralgemenings. AS1.1.3: Select and use appropriate problem-solving strategies to solve (unseen) problems. Kies en gebruik geskikte probleemoplossingstrategieë om (ongesiene) probleme op te los. AS1.1.4: Communicate and deend scientiic arguments with clarity and precision. Kommunikeer en verdedig wetenskaplike argumente duidelik en presies. AS1..1: Deine, discuss and explain prescribed scientiic knowledge. Deinieer, bespreek en verduidelik voorgeskrewe wetenskaplike kennis. AS1.. Express and explain prescribed scientiic principles, theories, models and laws by indicating the relationship between dierent acts and concepts in the candidate's own words. Verduidelik en druk voorgeskrewe wetenskaplike beginsels, teorieë, modelle en wette uit deur die verwantskap tussen verskillende eite en konsepte in die kandidaat se eie woorde aan te dui. AS1..3: Apply scientiic knowledge in everyday-lie contexts. Pas wetenskaplike kennis in kontekste van die alledaagse lewe toe. AS1.3.1 Research, discuss, compare and evaluate scientiic and indigenous knowledge systems and knowledge claims by indicating the correlation among them, and explain the acceptance o dierent claims. Doen navorsing, bespreek, vergelyk en evalueer wetenskaplike en inheemse kennissisteme en kennisaansprake deur die ooreenkoms tussen hulle aan te dui, en verduidelik die aanvaarding van verskillende aansprake. AS1.3.: Research case studies and present ethical and moral arguments rom dierent perspectives to indicate the impact (pros and cons) o dierent scientiic and technological applications. Vors gevallestudies na en bied etiese en morele argumente aan uit verskillende perspektiewe om die impak (voordele en nadele) van verskillende wetenskaplike en tegnologiese toepassings aan te dui. AS1.3.3: Evaluate the impact o scientiic and technological research and indicate the contribution to the management, utilisation and development o resources to ensure sustainability continentally and globally. Evalueer die impak van wetenskaplike en tegnologiese navorsing en dui die bydrae tot bestuur, benutting en ontwikkeling van bronne aan om volhoubaarheid kontinentaal en globaal te verseker.

3 Physical Sciences (P1)/Fisiese Wetenskappe (V1) DoE/November 009(1) SECTION A/AFDELING A QUESTION 1/VRAAG (gravitational) potential (energy) (gravitasionele) potensiële (energie) [1..1] (1) 1. watt / W [1..1] (1) 1.3 Coherent / coherence Koherent [1..1] (1) 1.4 gamma rays /-strale γ rays / γ -strale [1..1] (1) 1.5 Photons / otone Quanta / kwanta [1..1] (1) [5] QUESTION /VRAAG.1 is 9,8 m s - at its maximum height. is 9,8 m s - by sy maksimum hoogte. the velocity is zero at its maximum height./ die snelheid is nul by sy maksimum hoogte. [1..1] (). the magnitude o the MOMENTUM o the bullet is equal (opposite in direction) to the magnitude o the momentum o the gun. die grootte van die MOMENTUM van die koeël is gelyk (teenoorgesteld in rigting) aan die grootte van die momentum van die geweer....the MOMENTUM o the bullet is opposite in direction to the momentum o the gun... die MOMENTUM van die koeël is teenoorgesteld in rigting aan die momentum van die geweer the MOMENTUM o the bullet is equal to, but opposite in direction to the momentum o the gun. die MOMENTUM van die koeël is gelyk aan, maar teenoorgesteld in rigting aan die momentum van die geweer.

4 Physical Sciences (P1)/Fisiese Wetenskappe (V1) DoE/November 009(1). the CHANGE in MOMENTUM (impulse) o the bullet is equal in magnitude (opposite in direction) to the change in momentum o the gun. die verandering in MOMENTUM van die koeël is gelyk in grootte (teenoorgesteld in rigting) aan die verandering in momentum van die geweer.. the CHANGE in MOMENTUM (impulse) o the bullet is equal to, but opposite in direction to the change in momentum o the gun. die VERANDERING in MOMENTUM (impuls) van die koeël is gelyk aan, maar teenoorgesteld in rigting aan die verandering in momentum van die geweer.... the magnitude o the FORCE that the bullet exerts on the gun is equal (opposite in direction) to the orce that the gun exerts on the bullet. die grootte van die KRAG wat die koeël op die geweer uitoeen is gelyk (teenoorgesteld in rigting) aan die krag wat die geweer op die koeël uitoeen.... the FORCE that the bullet exerts on the gun is equal to, but opposite in direction to the orce that the gun exerts on the bullet. die KRAG wat die koeël op die geweer uitoeen is gelyk aan, maar teenoorgesteld in rigting, aan die krag wat die geweer op die koeël uitoeen. [1..3] ().3 result o diraction / intererence gevolg van diraksie / intererensie by a triangular prism deur n driehoekige prisma [1..3] ().4 dierent potential dierences verskillende potensiaalverskille Identical resistors Identiese resistors Identical resistors connected in parallel Identiese resistors in parallel geskakel Non-identical resistors...in parallel...dierent current Nie identiese resistors... in parallel... verskillende stroom [1..] ()

5 Physical Sciences (P1)/Fisiese Wetenskappe (V1) DoE/November 009(1).5 A line absorption spectrum n Lynabsorpsiespektrum when electrons move rom higher to lower energy levels. wanneer elektrone van hoër na laer energievlakke beweeg. [1..1] () [10] QUESTION 3/VRAAG C [1.1.] () 3. A [1..3] () 3.3 B [1.1.] () 3.4 C [1..3] () 3.5 A [1..1] () [10] TOTAL SECTION A/TOTAAL AFDELING A: 5

6 Physical Sciences (P1)/Fisiese Wetenskappe (V1) DoE/November 009(1) SECTION B/AFDELING B QUESTION 4/VRAAG Option 1/Opsie 1 Statements not correct (or no) / Stellings nie reg nie (o nee) The bricks will experience the same (gravitational) acceleration / ree all and thus reach the ground at the same time. Die bakstene ondervind dieselde (gravitasie) versnelling /vryval en bereik dus die grond gelyktydig. Option /Opsie Pete is correct or Alex is wrong / Pete is reg o Alex is verkeerd The smaller brick experiences less air resistance, thus larger acceleration and reaches the ground irst. Die kleiner baksteen ondervind minder lugweerstand, dus groter versnelling en tre die grond eerste. Option 3/Opsie 3 Alex is correct or Pete is wrong / Alex is reg o Pete is verkeerd In the presence o air resistance, the larger brick, with larger mass, experiences a larger net orce downwards, thus largest acceleration and reaches the ground irst In die aanwesigheid van lugweerstand, ondervind die groter baksteen met groter massa n groter netto awaartse krag, dus grootste versnelling en tre die grond eerste. Option 4/Opsie 4 Both are correct / Beide is reg Pete correct: The smaller brick experiences less air resistance, thus larger acceleration and reaches the ground irst. Die kleiner baksteen ondervind minder lugweerstand, dus groter versnelling en tre die grond eerste Alex correct: In the presence o air resistance, the larger brick, with larger mass, experiences a larger net orce downwards, thus largest acceleration and reaches the ground irst In die aanwesigheid van lugweerstand, ondervind die groter baksteen met groter massa, n groter netto awaartse krag, dus grootste versnelling en tre grond eerste. [1..3] (3)

7 Physical Sciences (P1)/Fisiese Wetenskappe (V1) DoE/November 009(1) 4..1 Any two / Enige twee: Ensure that both bricks are dropped rom same height Maak seker dat beide bakstene vana dieselde hoogte laat val word Ensure that both bricks are dropped at the same time Maak seker dat beide bakstene gelyktydig laat val word Ensure that the stopwatch starts at instant that each brick is released and stopped at the instant that each brick reaches the ground Maak seker dat die stophorlosie begin die oomblik as elk van die bakstene gelos word, en gestop word die oomblik as elke baksteen die grond bereik Repeat the experiment several times and use the average o the results Herhaal die eksperiment verskeie kere en gebruik die gemiddelde van die resultate Make sure that v i = 0 or both bricks Maak seker dat v i = 0 vir beide bakstene Make sure that there is no strong wind Maak seker dat daar geen sterk wind is nie Use bricks made o the same material / o same density Gebruik bakstene gemaak van dieselde materiaal / met dieselde digtheid [1.1.1] () 4.. External orce(s) may be present e.g. riction/air resistance / strong wind blowing Eksterne krag(te) kan teenwoordig wees bv. wrywing / lugweerstand / sterk wind wat waai [1.1.1] (1)

8 Physical Sciences (P1)/Fisiese Wetenskappe (V1) DoE/November 009(1) 4.3 Option 1/Opsie 1: Downward direction positive / Awaartse rigting positie: A: Δy = v i Δt + ½aΔt 8 = (0)Δt + ½(9,8)Δt Δt = 1,8 s B: Δy = v i Δt + ½aΔt 8 = v ib (1,8 0,6) + ½(9,8)(1,8 0,6) v ib = 8,43 m s -1 (8,43 to/tot 8,48 m s -1 ) Downward motion negative / Awaartse beweging negatie: A: Δy = v i Δt + ½aΔt - 8 = (0)Δt + ½(- 9,8)Δt Δt = 1,8 s B: Δy = v i Δt + ½aΔt - 8 = v ib (1,8 0,6) + ½(- 9,8)(1,8 0,6) v ib = - 8,43 m s -1 v ib = 8,43 m s -1 ( 8,43 to/tot 8,48 m s -1 ) Option /Opsie : Downward direction positive / Awaartse rigting positie: A: v = v a Δy = 0 + (9,8)(8) v = 1,5 m s -1 + i v = v i +aδt 1,5 = 0 +(9,8)Δt Δt = 1,8 s B: Δy = v i Δt + ½aΔt 8 = v ib (1,8 0,6) + ½(9,8)(1,8 0,6) v ib = 8,43 m s -1 (8,43 to/tot 8,48 m s -1 ) Downward direction negative / Awaartse rigting negatie: A: v = v + a Δy i = 0 + (- 9,8)(- 8) v = - 1,5 m s -1 v = v i +aδt - 1,5 = 0 +(- 9,8)Δt Δt = 1,8 s B: Δy = v i Δt + ½aΔt - 8 = v ib (1,8 0,6) + ½(- 9,8)(1,8 0,6) v ib = - 8,43 m s -1 v ib = 8,43 m s -1 (8,43 to/tot 8,48 m s -1 )

9 Physical Sciences (P1)/Fisiese Wetenskappe (V1) DoE/November 009(1) Option 3/Opsie 3: Downward direction positive / Awaartse rigting positie: A: v = v a Δy = 0 + (9,8)(8) v = 1,5 m s -1 + i v + vi (0 + 1,5) Δy = Δt - 8 = Δt Δt = 1,8 s B: Δy = v i Δt + ½aΔt 8 = v ib (1,8 0,6) + ½(9,8)(1,8 0,6) v ib = 8,43 m s -1 (8,43 to/tot 8,48 m s -1 ) Downward direction negative / Awaartse rigting negatie: A: v = v + a Δy = 0 + (- 9,8)(- 8) v i = - 1,5 m s -1 v + vi (0 1,5) Δ y = Δt 8 = Δt Δt = 1,8 s B: Δy = v i Δt + ½aΔt 8 = v ib (1,8 0,6) + ½(- 9,8)(1,8 0,6) v ib = - 8,43 m s -1 v ib = 8,43 m s -1 (8,43 to/tot 8,48 m s -1 ) Option 4/Opsie 4: A: W net = Δ K mgh = 1 mv - 1 mv i m(9,8)(8) = 1 m v - 0 v = 1,5 m s -1 A: (U + K) i = (U + K) mgh + 0 = mv m(9,8)(8) + 0 = m v v = 1,5 m s -1 v = v i +aδt 1,5 = 0 +(9,8)Δt Δt = 1,8 s B: Δy = v i Δt + ½aΔt 8 = v ib (1,8 0,6) + ½(9,8)(1,8 0,6) v ib = 8,43 m s -1 (8,43 to/tot 8,48 m s -1 ) [1.1.3] (7) [13]

10 Physical Sciences (P1)/Fisiese Wetenskappe (V1) DoE/November 009(1) QUESTION 5/VRAAG Option 1/Opsie 1: Direction o motion as positive / Rigting van beweging as positie: F net = ma -30 = (3)a a = - 10 m s - v = v + a Δx i = (7) + (-10)() v = 3 m s -1 Option /Opsie : W net = Δ K or/o Δ E k F Δ xcos θ = 1 mv - 1 mv (30)()cos180 = 1 (3) v - 1 (3)(7) - 60 = 1,5 v - 73,5 v = 3 m s -1 i Option 3/Opsie 3: W appl = Δ U + Δ K - W i Max./Maks.: 0 = 0 + ( 1 mv - 1 mv ) - F Δ xcos θ = (3) v - 1 (3)(7) - (30)()cos180 0 = 1,5 v - 73, v = 3 m s -1 Option 4 / Opsie 4: (U + K) i + W = (U + K) Max./Maks.: (0 + i 1 mv ) + F Δ xcosθ = mv (3)(7) + 30()cos180 = (3) v 73,5 60 = 1,5 v v = 3 m s -1 W appl = Δ U + Δ K + W (U + K) i W = (U + K) Option 5 / Opsie 5: (U + K) i = (U + K) - W Max./Maks.: 4 5 (0 + 1 mv ) = mv - F Δ xcos θ i (3)(7) = (3) v 73,5 = 1,5 v + 60 v = 3 m s -1-30()cos180 (U + K) i = (U + K) +W 0 5 [1.1.3] (5)

11 Physical Sciences (P1)/Fisiese Wetenskappe (V1) DoE/November 009(1) 5. w // Any one o the ollowing labels / Enige een van volgende benoemings: w parallel or/o w // F g(parallel) or/o F g// mgsin0 Component o weight parallel to incline / komponent van gewig parallel aan skuinste Checklist / kontrolelys Free-body diagram / vrye kragtediagram Direction o orce indicated as parallel to and down incline (not needed to show incline) Rigting van krag getoon as parallel aan en awaarts teen skuinste (skuinste hoe nie getoon te word nie) Correct label / korrekte benoeming [1.1.] () 5.3 Option 1/Opsie 1: (U + K) i = (U + K) mv i = mgh (3)(3) = (3)(9,8)h + 0 h = 0,46 m As single step/as een stap: (U + K) i = (U + K) (3)(3) = (3)(9,8)h (3)(3) = (3)(9,8) dsin0 d = 1,34 m sin 0 = d h = Option /Opsie : 0,46 d = 1,34 m d W net = Δ K (or/o Δ Ek) F g// Δ xcos θ = 1 mv - 1 mv i (3)(9,8)sin0 (d)cos180 = 0-1 (3)(3) - 10,06d = - 13,5 d = 1,34 m Option 3/Opsie 3: W net = Δ K (or/o Δ Ek) W gravity = K K i mghcos180 o = 1 mv - 1 mv i (3)(9,8)h(-1) = 0-1 (3)(3) h = 0,46 m sin0 = d h = 0,46 d = 1,34 m d

12 Physical Sciences (P1)/Fisiese Wetenskappe (V1) DoE/November 009(1) Option 4 / Opsie 4: Direction o motion as positive / Rigting van beweging as positie: F net = ma mgsin0 = ma - (3)(9,8)sin0 = 3a a = - 3,35 m s - v = v + a Δx i 0 = (3) + (-3,35)(d) d = 1,34 m QUESTION 6 / VRAAG m m v im + m b v bi = (m m +m b )v [1.1.3] (5) [1] + (87)v im + 0 = (87 + )(,4) v im = 3,01 m s -1 [1..3] (4) 6. Option 1/Opsie 1: K(beore/voor) = ½mv = ½(87)(3,01) + 0 = 394,11 J = (391,5 i 3 m s -1 ) K(ater/na) = ½mv = ½(109)(,4) = 313,9 J Collision is inelastic / No Botsing is nie-elasties / Nee Option /Opsie : Δ K = K(ater/na) - K(beore/voor) = ½mv (ater/na) - ½mv (beore/voor) = ½(109)(,4) (½(87)(3,01) + 0) = 313,9-394,11 = - 80,19 J Collision is inelastic / No Botsing is nie-elasties / Nee [1..3] (6)

13 Physical Sciences (P1)/Fisiese Wetenskappe (V1) DoE/November 009(1) 6.3 Option 1 / Opsie 1: W net = ΔE k F net Δxcos θ = ½m( v v i ) F net ()(-1) = ½(87 + )(0 -,4 ) F net = 156,96 N I/indien 156,96 N: minus 1 Option / Opsie : v = vi + aδx 0 =,4 + a() a = - 1,44 m s - F net = ma = (87 + )( - 1,44) = - 156,96 N F net = 156,96 N Option 3 / Opsie 3: v + v i 0 +,4 Δx = Δt = ( ) Δ t Δ t = 1,67 s F Δt = Δp = mv mv net i F net (1,67) = (87 + )(0,4) F net = - 156,65 N F net = 156,65 N Option 4 / Opsie 4: v + v i 0 +,4 Δx = Δt = ( ) Δ t Δ t = 1,67 s v = v i +a Δ t 0 =,4 + a(1,67) a = - 1,44 m s - F net = ma = (87 + )(-1,44) = - 156,96 N F net = 156,96 N [1..3] (5) [15]

14 Physical Sciences (P1)/Fisiese Wetenskappe (V1) DoE/November 009(1) QUESTION 7/VRAAG Towards the person / Na die persoon toe [1.1.] (1) 7. When the source moves towards a stationary observer waves in ront o the source is compressed resulting in a shorter wavelength, resulting in a higher requency (speed o sound constant) Wanneer die bron n stilstaande waarnemer nader, word golwe voor die bron saamgepers wat 'n korter gollengte tot gevolg het wat n hoër rekwensie tot gevolg het (spoed van klank konstant) [1..] () 7.3 Formulae accepted / Formules aanvaar: v ± v L L = s v ± v s = ( ) s s = 43,53 Hz [1..3] (4) [7] QUESTION 8/VRAAG A (chemical) substance that (selectively) absorb light o certain requencies / colours and (selectively) transmits / relects others. n (Chemiese) sto wat (selektie) lig van sekere rekwensies / kleure absorbeer en ander (selektie) deurlaat / weerkaats. [1..1] () 8. The manuacturing o pigments made all colours aordable or all people./ Vervaardiging van pigmente het alle kleure vir alle mense bekostigbaar gemaak. At the same time people, e.g. the Mexicans, could have lost their jobs and only income. / Terseldertyd het mense, bv. die Meksikane, hulle werk en enigste inkomste verloor. [1.3.1] () 8.3 Subtractive / Subtraktie [1..1] (1) 8.4 A: magenta + yellow / geel B: magenta + cyan / siaan [1..3] ()

15 Physical Sciences (P1)/Fisiese Wetenskappe (V1) DoE/November 009(1) 8.5 Option 1/Opsie 1: Yellow light / Geel lig Cyan paint relects blue and green light (Yellow light contains green light and red light) Only green light (in yellow light) will be relected and it appears green Siaanver weerkaats blou en groen lig (Geel lig bevat groen en rooi lig) Slegs groen lig (in geel lig) sal weerkaats word en dit kom groen voor Option /Opsie : Green light / Groen lig Cyan paint relects blue and green light When green light shines onto it, only green light will be relected and it appears green Siaanver weerkaats blou en groen lig Wanneer groen lig daarop skyn, word slegs groen lig weerkaats en dit kom groen voor. [1..3] (3) [10] QUESTION 9 / VRAAG Diraction / Diraksie [1.1.] (1) 9..1 Each point on a wave ront acts as a source o (spherical) secondary wave ronts / wavelets (that propagates in the orward direction). Elke punt op n golront dien as n bron van (seriese) sekondêre golronte / golies (wat in n voorwaartse rigting voortplant). [1..1] () 9.. Dark bands orm where wave ronts / wavelets interere destructively. Red/bright bands orm where wave ronts / wavelets interere constructively. Donker bande vorm waar golronte / golwe destruktiewe intererensie ondergaan. Rooi/helder bande vorm waar golronte / golwe konstruktiewe intererensie ondergaan. [1.1.4] ()

16 Physical Sciences (P1)/Fisiese Wetenskappe (V1) DoE/November 009(1) 9..3 Diraction is inversely proportional to the slit width / Diraction Diraksie is omgekeerd eweredig aan die spleetwydte / 1 Diraksie α a α 1 a The degree o diraction / Angle at which minima occurs increases with decreasing slit width Mate van diraksie / Hoek waar minima voorkom neem toe met aname in spleetwydte [1.1.4] () 9..4 White light consists o dierent colours with dierent wavelengths Amount o diraction diers or dierent colours / dierent wavelengths. Wit lig bestaan uit verskillende kleure met verskillende gollengtes. Mate van diraksie verskil vir verskillende kleure / gollengtes. [1.1.4] () [9] QUESTION 10 / VRAAG kq1q F = r 9 6 (9 10 )(4 10 )(6 10 F = (0,4) F = 1,35 N 6 ) kq1q F = r (9 10 )(4 10 )(-6 10 ) F = (0,4) F = -1,35 N Magnitude o / Grootte van F = 1,35 N [1..3] (4) 10. our / vier (4) [1..] (1)

17 Physical Sciences (P1)/Fisiese Wetenskappe (V1) DoE/November 009(1) 10.3 kq E(6 μ C) = r 9 (9 10 )(6 10 = (0,) 6 = 1,35 x 10 6 N C -1 to the let/na links ) kq E(4 μ C) = r 9 (9 10 )(4 10 = (0,6) 6 = 1 x 10 5 N C -1 (to the right / na regs) To the right as positive/na regs as positie: E net/netto = - 1,35 x x 10 5 = - 1,5 x 10 6 N C -1 ) E net/netto = 1,5 x 10 6 N C -1 to the let / na links kq E net = = 9 x (4 10 ) ( + ) r (0,) (0,6) =- 1,35 x x 10 5 = - 1,5 x 10 6 E net/netto = 1,5 x 10 6 N C -1 in the direction o the ield o the 6 μ C charge/in die rigting van veld van die 6 μ C lading [1.1.3] (6) ( ) + ( 6 10 ) New charge/nuwe lading = = -1 x 10-6 C or/o 1 μ C kq1q U = r (9 10 )( 1 10 )( 1 10 ) = (0,4) U =,5 x 10 - J (0,0 J) [1.1.3] (5) [16] QUESTION 11 / VRAAG V Potential dierence measured when: switch is open / no current lows / circuit is open/no work done is in external circuit Potensiaalverskil gemeet wanneer: die skakelaar oop is / geen stroom vloei nie / stroombaan oop is / geen arbeid verrig word in die eksterne stroombaan nie [1..] ()

18 Physical Sciences (P1)/Fisiese Wetenskappe (V1) DoE/November 009(1) 11. Option 1 / Opsie 1: Em = IR + Ir 9 = 3(3R) + 3(0,3) R = R 1 = 9 0,9 = 0,9 Ω 9 Option / Opsie : Em = IR + Ir 9 = V ext + (3)(0,3) V ext = 8,1 V V ext = I(R 1 + R ) 8,1 = 3(3R) R 1 = 0,9 Ω Option 3 / Opsie 3: em = IR + Ir 9 = V ext + (3)(0,3) V ext = 8,1 V 8, 1 V 1 = =,7 V (R1 + R = 3R) 3 Option 4 / Opsie 4: R t = I V = 3 9 = 3 Ω R + R 1 = 3 0,3 =,7 Ω = 3R R 1 = V 1 = I,7 3 = 0,9 Ω, 7 R 1 = R = = 0,9 Ω 3 Option 5 / Opsie 5: V int = Ir = (3)(0,3) = 0,9 V V ext = 9 0,9 = = 8,1 V V 1 = IR 1 V 1 = 3R V R = IR V R = 3(R) = 6R. V 1 + V R = 3R + 6R = 9R 8,1 = 9R R = 0,9 Ω [1.1.3] (5) Decreases / Verminder [1..] (1) Increases / Vermeerder Resistance decreases / Weerstand verminder Current increases / Stroom vermeerder Ir increases / Ir vermeerder Increases / Vermeerder Current passes through wire QN / wire QN shorts the parallel combination o resistors R and R 3 All the current passes through R 1 and also through battery, thus Ir increases Die stroom gaan deur draad QN / draad QN veroorsaak n kortsluiting van die parallelle kombinasie resistors R en R 3 Al die stroom gaan deur R 1 en deur die battery, dus verhoog Ir [1..] (3) [11]

19 Physical Sciences (P1)/Fisiese Wetenskappe (V1) DoE/November 009(1) QUESTION 1/VRAAG Vrms 36 Irms = = = 3 A R 1 V 4 Ω + V 8 Ω = 36 V and/en V 4 Ω :V 8 Ω = 1: V 4 Ω = 1 V V 1 I 4 Ω = = = 3 A R 4 [1...3] (3) 1.1. Imax Irms = 3 = I max Imax = 4,4 A [1..3] (3) P ave = I rmsr = (3) (4) = 36 W V rms (speaker/luidspreker 1) = I rms R = (3)(4) = 1 V P ave = V rms I rms = (1)(3) = 36 W V rms (speaker/luidspreker 1) = I rms R = (3)(4) = 1 V V 1 rms P ave = = = 36 W R 4 [1..3] (3) 1. P 4 Ω = ½ P 8 or P 8 Ω = P 4 Ω or /o Smaller / Kleiner (P 4 Ω < P 8 Ω ) or/o P 8 Ω > P 4 Ω ) P ave = I rmsr, but since I rms is constant / omdat I wgk konstant is P α R P 4 Ω = ½ P 8 or P 8 Ω = P 4 Ω or /o Smaller / Kleiner P ave = V rms I rms, since I rms is constant / omdat I wgk konstant is α V P ave [1..] (3) [1]

20 Physical Sciences (P1)/Fisiese Wetenskappe (V1) DoE/November 009(1) QUESTION 13 / VRAAG Electromagnetic induction / Faraday's law Elektromagnetiese induksie / Faraday se wet [1..1] (1) 13. Provides a (sliding) contact (between coil and conducting wires) / Ensures ree rotation Verska n (glyende) kontak (tussen die spoel en die geleidende drade )/ Verseker dat spoel vrylik roteer [1..1] (1) 13.3 Em/Emk (V) B A C A t(s) D Checklist / Kontrolelys Criteria or graph / Kriteria vir graiek Correct shape with ull cycle (ignore i more than one cycle shown / Korrekte vorm met volle siklus (ignoreer indien meer as een siklus getoon word) Points A, B, C and D correctly indicated/punte A, B, C en D korrek aangedui, Marks/ Punte [1.1.] (3) 13.4 Increase the speed at which the coil rotates Verhoog die spoed waarteen die spoel roteer [1..3] (1) 13.5 (Splitring) commutator (Splitring)kommutator [1..3] (1) [7] QUESTION 14/VRAAG Photoelectric eect / Foto-elektriese eek [1..1] (1) 14. The minimum energy o light needed to emit (photo)electrons rom a metal Die minimum energie benodig deur lig om (oto-)elektrone uit n metaal vry te stel [1..1] ()

21 Physical Sciences (P1)/Fisiese Wetenskappe (V1) DoE/November 009(1) 14.3 Option 1 / Opsie 1: 34 8 hc (6,63 10 )(3x10 ) E/h = = λ = 3,58 x J h 1 + W 0 = 3,58 x J = W0 mv 3,58 x = W Option / Opsie : 1 h = W0 + mv hc 1 = W 0 + mv λ 34 8 (6,63 10 )(3x10 ) W 0 = W 0 = 3,58 x J + 0 Option 3 / Opsie 3: 8 c 3x10 = = = 5,41 x Hz λ E = h = (6,63 x )(5,41 x ) = 3,59 x J h 1 + W 0 = 3,59 x J = W0 mv 3,59 x = W Option 4 / Opsie 4: 34 8 (6,63 10 )(3x10 ) W 0 = h 0 = = 3,58 x J [1.1.3] (6) 14.4 Increases / Vermeerder With light o higher intensity more photons strikes the metal surace per second / Met lig van hoër intensiteit tre meer otone die metaaloppervlak per sekonde Thus more (photo)electrons are emitted per second, resulting in a bigger current./ Dus word meer (oto-)elektrone per sekonde vrygestel wat n hoër stroom tot gevolg het. [1..] (3) 14.5 Decreases / Verminder [1..] (1) [13] TOTAL SECTION B / TOTAAL AFDELING B: 15 GRAND TOTAL / GROOTTOTAAL: 150

NATIONAL SENIOR CERTIFICATE GRADE 12

NATIONAL SENIOR CERTIFICATE GRADE 1 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (Vr1) SEPTEMBER 015 MEMORANDUM MARKS: 150 TIME: 3 hours This memorandum consists of 10 pages. Physical Sciences/P1