# Physical Sciences Grade 12 P1 NSC Supplementary 2012 Memo > Grade 12 Physical Sciences Paper 1 memorandum from the NSC Supplementary 2012, 15 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/physical-sciences-p1-feb-march-2012-memo-afr-eng/1775cde4-925f-4643-b3d3-8f103e8853f9 - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2012 - Exam period: NSC Supplementary - Paper: 1 - Curriculum: CAPS - Pages: 15 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Physical Sciences](https://www.hlayiso.com/library/grade-12/physical-sciences) › [Physical Sciences Grade 12 P1 NSC Supplementary 2012 Memo](https://www.hlayiso.com/library/physical-sciences-p1-feb-march-2012-memo-afr-eng/1775cde4-925f-4643-b3d3-8f103e8853f9) ## About this document Physical Sciences · Grade 12 · NSC Supplementary · 2012. Memorandum, 15 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 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 2012 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 15 pages. Hierdie memorandum bestaan uit 15 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 2 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 2 DBE/Feb.–Mar./Mrt. 2012 NSC/NSS – Memorandum Learning Outcomes and Assessment Standards Leeruitkomste en Assesseringstandaarde LO/LU 1 LO/LU 2 LO/LU 3 AS 12.1.1: AS 12.2.1: AS 12.3.1: Design, plan and conduct a Define, discuss and explain Research, discuss, compare and scientific inquiry to collect data prescribed scientific knowledge. evaluate scientific and systematically with regard to Definieer, bespreek en indigenous knowledge systems accuracy, reliability and the need verduidelik voorgeskrewe and knowledge claims by to control variables. wetenskaplike kennis. indicating the correlation among Ontwerp, beplan en voer ʼn them, and explain the wetenskaplike ondersoek uit om acceptance of different claims. data te versamel ten opsigte van Doen navorsing, bespreek, akkuraatheid, betroubaarheid en vergelyk en evalueer die kontroleer van veranderlikes. wetenskaplike en inheemse kennissisteme en kennisaansprake deur die ooreenkoms aan te dui en verduidelik die aanvaarding van verskillende aansprake. AS 12.1.2: AS 12.2.2: AS 12.3.2: Seek patterns and trends, Express and explain prescribed Research case studies and represent them in different forms, scientific principles, theories, present ethical and moral explain the trends, use scientific models and laws by indicating arguments from different reasoning to draw and evaluate the relationship between different perspectives to indicate the conclusions, and formulate facts and concepts in own words. impact (pros and cons) of generalisations. Verduidelik en druk different scientific and Soek patrone en tendense, stel voorgeskrewe wetenskaplike technological applications. dit in verskillende vorms voor, beginsels, teorieë, modelle en Vors gevallestudies na en lewer verduidelik tendense, gebruik wette uit deur die verwantskap etiese en morele argumente uit wetenskaplike beredenering om tussen verskillende feite en verskillende perspektiewe om die gevolgtrekkings te maak en te konsepte in eie woorde aan te impak (voordele en nadele) van evalueer, en formuleer dui. verskillende wetenskaplike en veralgemenings. tegnologiese toepassings aan te dui. AS 12.1.3: AS 12.2.3: AS 12.3.3: Select and use appropriate Apply scientific knowledge in Evaluate the impact of scientific problem-solving strategies to everyday life contexts. and technological research and solve (unseen) problems. Pas wetenskaplike kennis in indicate the contribution to the Kies en gebruik geskikte kontekste van die alledaagse management, utilisation and probleemoplossingstrategieë om lewe toe. development of resources to (ongesiene) probleme op te los. ensure sustainability continentally and globally. Evalueer die impak van wetenskaplike en tegnologiese navorsing en dui die bydrae tot bestuur, benutting en ontwikkeling van bronne om volhoubaarheid kontinentaal en globaal te verseker. AS 12.1.4: Communicate and defend scientific arguments with clarity and precision. Kommunikeer en verdedig wetenskaplike argumente duidelik en presies. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 3 DBE/Feb.–Mar./Mrt. 2012 NSC/NSS – Memorandum SECTION A/AFDELING A QUESTION 1/VRAAG 1 1.1 Kinetic energy/Kinetiese energie  (1) 1.2 Interference/Interferensie  (1) 1.3 Ohm  (1) 1.4 Electromagnetic induction/Elektromagnetiese induksie  OR/OF Faraday's law/Faraday se wet (1) 1.5 (Line) emission (spectrum)  (1) (Lyn)emissie(spektrum) [5] QUESTION 2/VRAAG 2 2.1 B  (2) 2.2 B  (2) 2.3 C  (2) 2.4 B  (2) 2.5 C  (2) 2.6 A  (2) 2.7 D  (2) 2.8 D  (2) 2.9 C  (2) 2.10 A  (2) [20] TOTAL SECTION A /TOTAAL AFDELING A: 25 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/Feb.–Mar./Mrt. 2012 NSC/NSS – Memorandum SECTION B/AFDELING B QUESTION 3/VRAAG 3 3.1 Accepted Labels/Aanvaarde benoemings Fg / Fw / force of Earth on stone/weight/mg/gravitational force w Fg / Fw / krag van Aarde op klip/gewig/mg/gravitasiekrag  w  (1) 3.2.1 Option 1/Opsie 1: Upward positive/Opwaarts positief: Upward negative/Opwaarts negatief: v f = vi + a ∆t  v f = vi + a ∆t  0 = 10 +(-9,8) ∆ t  0 = -10 + 9,8 ∆ t  ∴ ∆ t = 1,02 s  ∴ ∆ t = 1,02 s  Option 2/Opsie 2: Upward positive/Opwaarts positief: Upward negative/Opwaarts negatief: v f = v i + 2a∆y v f = v i + 2a∆y 2 2 2 2 02 = 102 + 2(-9,8) ∆ y  Both 02 = (-10)2 + 2(9,8) ∆ y   Both formulae/ formulae/ ∴ y = 5,1 m ∴ y = -5,1 m Beide Beide formules formules  v + vf   v + vf  Δy =  i  Δt Δy =  i  Δt  2   2  10 + 0 − 10 + 0 5,1 = ( )∆t  -5,1 = ( )∆t  2 2 ∴ ∆ t = 1,02 s  ∴ ∆ t = 1,02 s  (4) 3.2.2 POSITIVE MARKING FROM QUESTION 3.2.1 TO QUESTION 3.2.2 POSITIEWE NASIEN VAN VRAAG 3.2.1 NA VRAAG 3.2.2 Option 1/Opsie 1: Upward positive/Opwaarts positief: Upward negative/Opwaarts v f = v i + 2a∆y  negatief: 2 2 v f = v i + 2a∆y  2 2 02 = 102 + 2(-9,8) ∆ y  ∴ ∆ y = 5,1 m 02 = (-10)2 + 2(9,8) ∆ y  ∴ ∆ y = -5,1 m Height/Hoogte = 50 +  5,1 = 55,1 m  Height/Hoogte = 50 +  5,1 = 55,1 m  Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/Feb.–Mar./Mrt. 2012 NSC/NSS – Memorandum Option 2/Opsie 2: Option 3/Opsie 3: Upward positive/Opwaarts positief: Consider downward motion/  v + vf  Beskou afwaartse beweging: Δy =  i  Δt   2   v + vf  Δy =  i  Δt  10 + 0  2  ∆y = ( )1 ,02  2 − 10 + 0 ∆y = ( )1,02  ∴ = 5,1 m 2 ∴ = -5,1 m Height = 50 +  5,1 = 55,1 m  Height = 50 +  5,1 = 55,1 m  Upward negative/Opwaarts negatief: Upward negative/Opwaarts  v + vf  negatief: Δy =  i  Δt   2   v + vf  Δy =  i  Δt  − 10 + 0  2  ∆y = ( )1,02  2 − 10 + 0 ∆y = ( )1,02  ∴ ∴ ∆ y = - 5,1 m 2 ∴ = - 5,1 m Height/Hoogte = 50 +  5,1 Height/Hoogte = 50 + 5,1 = 55,1 m  = 55,1 m (4) 3.3 Marks/ Criteria for graph/Kriteria vir grafiek Punte Correct shape/Korrekte vorm  Final position lower than initial position.  Graph ends on x axis./Grafiek eindig op x-as.  Upward positive/Opwaarts positief Upward negative/Opwaarts negatief position/posisie (m) 0 position/posisie (m) 50 time/tyd(s) -50 0 time/tyd(s) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/Feb.–Mar./Mrt. 2012 NSC/NSS – Memorandum 3.4 Option 1/Opsie 1 1 ∆y = v i ∆t + a∆t 2  2 1 1,5 = vi (0,1) + (9,8)(0,1)2  2 ∴ vi = 14,51 m∙s-1 From maximum height/Van maksimum hoogte: v f = v i + 2a∆y  2 2 14,512 = (0)2 + 2(9,8) ∆ y  ∴ ∆ y = 10,74 m Height/Hoogte = 55,1 –10,74 = 44,36 m  Option 2/Opsie 2 Option 3/Opsie 3 1 1 ∆y = v i ∆t + a∆t 2  ∆y = v i ∆t + a∆t 2  2 2 1 1 1,5 = vi (0,1) + (9,8)(0,1)2  1,5 = vi (0,1) + (9,8)(0,1)2  2 2 ∴ vi = 14,51 m∙s-1 ∴ vi = 14,51 m∙s-1 Downwards from top of tower to top From original point of projection:/Van of window:/Afwaarts van bopunt van oorspronklike punt van projeksie toring tot bopunt van venster v f = v i + 2a∆y  v f = v i + 2a∆y  2 2 2 2 14,512  = (10)2 + 2(9,8) ∆ y  14,512 = (-10)2 + 2(9,8) ∆ y  ∴ ∆ y = 5,64 m ∴ ∆ y = 5,64 m Height/Hoogte = 50 – 5,64 Height/Hoogte = 50 – 5,64 = 44,36 m  = 44,36 m  Option 4/Opsie 4 ∆y 1,5 v = = = 15 m∙s-1 ∆t 0,1 v + vf v = i = 15 2 ∴ vi + vf = 30 m∙s-1 ∴ vf = 30 – vi v f = vi + a ∆t  30 - vi = vi + 9,8(0,1)  ∴ vi = 14,51 m∙s-1 v f = v i + 2a∆y  2 2 14,512  = (0)2 + 2(9,8) ∆ y  ∴ ∆ y = 10,74 m Height/Hoogte = 55,1 – 10,74 = 44,36 m  (7) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/Feb.–Mar./Mrt. 2012 NSC/NSS – Memorandum QUESTION 4/VRAAG 4 4.1 Impulse is the product of the (net/average) force and the time during which the force acts.  Impuls is die produk van die (netto/gemiddelde) krag en die tyd waartydens die krag inwerk.  OR/OF Impulse is the change in momentum.  Impuls is gelyk aan verandering in momentum.  (2) 4.2 Option 1/Opsie 1: Option 2/Opsie 2: Upward positive:/Opwaarts positief Upward negative:/Opwaarts negatief Fnet ∆t = ∆p  Fnet ∆t = ∆p  = m(vf – vi) = m(vf – vi) = 0,15(3,62 - (-6,2))  = 0,15[(-3,62 - (6.2))  = 1,473 N∙s / kg∙m∙s-1 = -1,473 N∙s /kg∙m∙s-1 upward/opwaarts Fnet ∆t = 1,473 N∙s /kg∙m∙s-1  upward/opwaarts Option 3/Opsie 3: Option 4/Opsie 4: Upward positive: /Opwaarts positief Upward negative: /Opwaarts negatief Fnet ∆t = ∆p  Fnet ∆t = ∆p  = mvf – mvi = mvf – mvi) = (0,15)(3,62) – (0,15)(-6,2)  = (0,15)(-3,62) – (0,15)(6,2)  = 1,473 N∙s / kg∙m∙s-1 = -1,473 N∙s /kg∙m∙s-1 upward/opwaarts Fnet ∆t = 1,473 N∙s /kg∙m∙s-1 upward/opwaarts (3) 4.3 (U + K)top/bo = (U + K)bottom/onder K(bottom/onder) = U(top/bo) mghf + ½ m v 2f = mghi + ½ m v i2 Max.: 0 4 (0,15)(9,8)h + 0  = 0 + ½(0,15)(6,2)2  ∴ h = 1,96 m  Other formulae/Ander formules: Emech(A) = Emech(B) / Emech(i) = Emech(f) 1,96 Emech(top) = Emech(bottom) = 0,65 m (Ep + Ek)A = (Ep + Ek)B 3 (Ep + Ek)bottom = (Ep + Ek)top Yes/Meets requirements  Ep + Ek)i = (Ep + Ek)f (U + K)i = (U + K)f Ja/Voldoen aan vereistes.  (U + K)A = (U + K)B mghB + 21 mv i2 = mghB + 21 mv 2f (5) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 8 DBE/Feb.–Mar./Mrt. 2012 NSC/NSS – Memorandum QUESTION 5/VRAAG 5 5.1 The energy of an object due to its position  above the surface of the earth.  Die energie van 'n voorwerp as gevolg sy posisie  bokant die oppervlak van die aarde.  (2) 5.2 Option 1/Opsie 1: Wnet = ∆K  mg∆ycosθ + W f = ½m v 2f – ½m v i2 (2)(9,8)(2)cos0° + Wf = ½(2)(5)2– 0  ∴ W f = -14,2 J  Option 2/Opsie 2: Wnet = ∆K  -∆U + W f = ½m v 2f – ½m v i2 mgh + Wf = ½m v 2f – ½m v i2 (2)(9,8)(2) + W f = ½(2)(5)2– 0  ∴ W f = -14,2 J  (6) 5.3 No/Nee  Friction is present/Wrywing is aanwesig.  (2) 5.4.1 Σ pbefore = Σ pafter  Notes/Aantekeninge: (2)(5) + (9)(0) = 2vf2 + (9)(1)  Other formulae/Ander formules: m1vi1 + m2vi2 = m1vf1 + m2vf2 ∴ vf2 = 0,5 m·s-1  m1u1 + m2u2 = m1v1 + m2v2 (4) 5.4.2 K(total after/total na) = ½m1 v 2f + ½m2 v 2f  = ½(2)(0,5)2 + ½(9)(1)2  = 4,75 J  K(total before) ≠ K(total after)  ∴ inelastic K(totaal na) ≠ K(totaal voor)  ∴ onelasties (5) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 9 DBE/Feb.–Mar./Mrt. 2012 NSC/NSS – Memorandum QUESTION 6/VRAAG 6 6.1 v ± vL v fL = f s OR fL = fs  v ± vs v − vs 340 + 0 ∴ 1 050 = (980)  340 − v s ∴ vs = 22,67 m∙s-1  (4) 6.2 Waves in front of the moving source are compressed. The observed wavelength decreases.  For the same speed of sound, a higher frequency will be observed. Golwe voor die bewegende bron word saamgepers. Die waargenome golflengte verminder.  Vir dieselfde spoed van klank sal 'n hoër frekwensie waargeneem word. (2) 6.3 Any ONE/Enige EEN: • Determine whether arteries are clogged/narrowed  so that precautions can be taken in advance/to prevent heart attack /stroke.  Bepaal of are verstop/vernou is,  sodat voorsorg getref kan word/om hartaanvalle/beroerte te voorkom.  • Determine heartbeat of foetus to assure that child is alive/does not have a heart defect. Bepaal die hartklop van 'n fetus om seker te maak of baba leef/geen hartdefekte het nie. (2) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 10 DBE/Feb.–Mar./Mrt. 2012 NSC/NSS – Memorandum QUESTION 7/VRAAG 7 7.1 Criteria for investigative question/Kriteria vir ondersoekende Mark/Punt vraag: The dependent and independent variables are stated.  Die afhanklike en onafhanklike veranderlikes is genoem. - Asks a question about the relationship between dependent and independent variables.  Vra 'n vraag oor die verwantskap tussen die afhanklike en onafhanklike veranderlikes. Dependent variable: Afhanklike veranderlike: Broadness of central (bright) band/degree of diffraction Breedte van sentrale (helder) band/mate van diffraksie Independent variable: Onafhanklike veranderlike: Wavelength (of light)/Golflengte (van lig) Example/Voorbeeld: How will the width of the central band change/differ when the wavelength (of the light) changes/is increased/is decreased? Hoe sal die breedte van die sentrale helder band verander wanneer die golflengte (van die lig) toeneem/afneem? (2) 7.2 Slit width/Spleetwydte  Distance between slit and screen/Afstand tussen spleet en skerm.  (2) 7.3 0,033 tan θ =  ∴ θ = 4,19(4)° 0,45 mλ sin θ =  a (1)λ sin4,19°  =  5,6 × 10 −7 ∴ λ = 4,1 x 10-8 m  (5) 7.4 Greater than/Groter as  Red light has a longer wavelength (and is diffracted more.)  Rooilig het 'n langer golflengte (en word meer diffrakteer.) OR/OF (2) Diffraction/Diffraksie α λ  [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 11 DBE/Feb.–Mar./Mrt. 2012 NSC/NSS – Memorandum QUESTION 8/VRAAG 8 8.1 The (electrostatic) force experienced at a point  per unit charge at that point.  Die elektrostatiese krag ondervind by 'n punt  per eenheidslading by daardie punt.  OR/OF The (electrostatic) force experienced  by a charge placed at that point divided by the charge itself.  Die (elektrostatiese) krag ondervind  deur 'n lading geplaas by daardie punt gedeel deur die lading self.  (2) 8.2 Marks/ Criteria for sketch/Kriteria vir skets Punte Correct shape as shown.  Korrekte vorm soos getoon Direction from positive to negative.  Rigting van positief na negatief. Field lines start on spheres and do not cross. Veldlyne begin op elke sfeer en kruis nie.  (3) 8.3 kQ EP =  r2 (9 × 10 9 )(5 × 10 −9 ) = (30 × 10 −3 )2  = 5 x 104 N∙C-1 to the right/na regs kQ EQ = r2 (9 × 10 9 )(5 × 10 −9 ) = (10 × 10 −3 )2  = 4,5 x 105 N∙C-1 to the right/na regs Enet = 5 x 104 + 4,5 x 105 = 5 x 105 N∙C-1 to the right/na regs  (6) 8.4 POSITIVE MARKING FROM QUESTION 8.3 TO QUESTION 8.4/ POSITIEWE NASIEN VAN VRAAG 8.3 NA VRAAG 8.4 F E=  q F 5 x 105 =  1,6 × 10 −19 F = 8 x 10-14 N  (3) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 12 DBE/Feb.–Mar./Mrt. 2012 NSC/NSS – Memorandum QUESTION 9/VRAAG 9 9.1 9.1.1 rheostat/reostaat A resistor V Criteria for circuit diagram/Kriteria vir stroombaandiagram Mark/Punt Battery connected to the resistor as shown – correct symbols used.  Battery aan resistor geskakel soos getoon – korrekte simbole is gebruik. Rheostat connected in series with resistor – correct symbols used.  Reostaat in serie geskakel met resistor – korrekte simbole is gebruik. Ammeter connected in series so that it measures the current through resistor – correct symbols used.  Ammeter in serie geskakel sodat dit die stroom deur die resistor meet – korrekte simbole is gebruik. Voltmeter connected in parallel across resistor – correct symbols used.  Voltmeter in parallel geskakel oor resistor – korrekte simbole is gebruik. (4) 9.1.2 Temperature/Temperatuur  (1) 9.1.3 B V The ratio is greater than that of A.  I B V Die verhouding is groter as die van A.  I OR/OF B I The ratio is smaller than that of A.  V B I Die verhouding is kleiner as die van A.  V (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 13 DBE/Feb.–Mar./Mrt. 2012 NSC/NSS – Memorandum 9.2 9.2.1 1 1 1 1 1 = + = +  R r1 r2 4 16 ∴ R = 3,2 Ω Reffective/effektief = 3,2 Ω + 2 Ω + 0,8 Ω =6Ω (4) 9.2.2 Option 1/Opsie 1: Option 2/Opsie 2: V = IR  emf = I(R + r)  12 = I(6)  12 = I(5,2 + 0,8)  I=2A I=2A (3) 9.2.3 Option 1/Opsie 1: Option 2/Opsie 2: R Vparallel = IR  Vp = p x V  =(2)(3,2)  R = 6,4 V 3,2 =  x 12  = 6,4 V 6 6,4 ∴ V8Ω = 3,2 V  V8Ω = = 3,2 V  2 Option 3/Opsie 3: Option 4/Opsie 4: 4 emf = I(R + r)  I8Ω = (2)  20 12 = IR2 Ω + Vp + Ir = 0,4 A 12 = (2)(2) + Vp + (2)(0,8)  Vp = 6,4 V V8Ω = IR  = (0,4)(8)  6,4 = 3,2 V  V8Ω = = 3,2 V  2 (4) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 14 DBE/Feb.–Mar./Mrt. 2012 NSC/NSS – Memorandum QUESTION 10/VRAAG 10 10.1 10.1.1 (a) Reverses the direction of the current in the coil each half cycle.  Keer die stroomrigting in die spoel elke halwe siklus.  OR/OF Maintains constant direction of rotation of the coil. Onderhou die konstante rigting van rotasie van die spoel. (1) 10.1.1 Makes electrical contact (with the commutator).  (b) Maak elektriese kontak (met kommutator).  OR/OF Allows split-ring commutator to rotate freely. Laat splitringkommutator toe om vry te roteer. OR/OF Allows charges to flow/current in and out of the coil. Laat vloei van lading/stroom in en uit spoel toe. (1) 10.1.2 B to/na A  (1) 10.1.3 Maximum/Maksimum  (1) 10.1.4 Any ONE/Enige EEN: • Increase the current in the coil.  Verhoog die stroom in die spoel.  • Increase the magnitude of the magnetic field./Use a stronger magnet. Vergroot die grootte van die magneetveld./Gebruik 'n sterker magneet. • Increase the number of turns in the coil. Verhoog die aantal windinge in die spoel. • Use a soft iron core as the core of the coil. Gebruik 'n sagte ysterkern in die spoel. (1) 10.2 10.2.1 Any ONE/Enige EEN: • Can be transmitted over long distances without major energy loss.  Kan oor groot afstande oorgedra word sonder groot energieverlies.  • The potential difference can be increased or decreased. Die potensiaalverskil kan verhoog of verlaag word. (1) 10.2.2 (a) Vrms/wgk = Vmax/ maks  2 Vmax/ maks 230 =  2 Vmax/maks = 325, 27 V  (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 15 DBE/Feb.–Mar./Mrt. 2012 NSC/NSS – Memorandum 10.2.2 (b) Pave = Vrms/wgkIrms/wgk  2 000 = (230)Irms/wgk  Irms/wgk = 8,70 A  (8,696 A) (3) [12] QUESTION 11/VRAAG 11 11.1 Photoelectric effect/Foto-elektriese effek  (1) 11.2 11.2.1 E = hf  = (6,63 x 10-34)(6,16 x 1014)  = 4,08 x 10-19 J  (3) 11.2.2 E = W0 + K  4,08 x 10-19  = (6,63 x 10-34)f0 + 5,6 x 10-20  f0 = 5,31 x 1014 Hz  (5) 11.3 11.3.1 Increases  More photoelectrons emitted per second  Vermeerder  Meer foto-elektrone vrygestel per sekonde  (2) 11.3.2 Remains the same  Intensity does not affect energy.  Bly dieselfde  Intensiteit het geen effek op energie nie.  OR/OF Remains the same  The frequency of light remains the same.  Bly dieselfde  Die frekwensie van die lig bly dieselfde.  (2) [13] TOTAL SECTION B/TOTAAL AFDELING B: 125 GRAND TOTAL/GROOTTOTAAL: 150 Copyright reserved/Kopiereg voorbehou ## Related documents - [Physical Science P1 (SG) (English) June 2012 QP](https://www.hlayiso.com/library/physical-science-p1-sg-english-june-2012-qp/0c084a64-a915-451d-b115-ecf0fdc0bc4b) — Physical Sciences · QP · 2012 · Eng - [Physical Sciences P1 Nov 2012 Eng](https://www.hlayiso.com/library/physical-sciences-p1-nov-2012-eng/f3e1efe1-d287-4610-8583-1bb6eb811406) — Physical Sciences · QP · 2012 · Eng - [Physical Science P2 (SG) (English and Afrikaans) June 2012 Memo](https://www.hlayiso.com/library/physical-science-p2-sg-english-and-afrikaans-june-2012-memo/230c1a41-a0da-4842-92a1-abb8df5e31ac) — Physical Sciences · Memo · 2012 · Afr - [Gr 12 Physical Sciences P1 (English and Afrikaans) June 2024 Possible Answers](https://www.hlayiso.com/library/gr-12-physical-sciences-p1-english-and-afrikaans-june-2024-possible-answers/f78ca708-f7f0-4bc5-8ff9-36594b1b1052) — Physical Sciences · Memo · 2024 · Afr - [Physical Sciences Physics P1 MEMO May June 2024 KZN](https://www.hlayiso.com/library/physical-sciences-physics-p1-memo-may-june-2024-kzn/49134363-7da2-4c98-ab54-c6d186d63bdc) — Physical Sciences · Memo · 2024 - [2024 MCED Sept Gr 12 P.Sciences P1 MEMO](https://www.hlayiso.com/library/2024-mced-sept-gr-12-p-sciences-p1-memo/10a9100b-159b-46b7-bf75-b22879248ca6) — Physical Sciences · Memo · 2024 ## More from Grade 12 Physical Sciences - [PHYSICAL SCIENCE P1 QP JCT 2025 AFR D watermark](https://www.hlayiso.com/library/physical-science-p1-qp-jct-2025-afr-d-watermark/b9429ac5-acbb-4fae-989d-af9e1478a96b) — QP · 2025 · Afr - [PHYSICAL SC P2 QP GR12 SEPTEMBER 2025 AFR1 watermark](https://www.hlayiso.com/library/physical-sc-p2-qp-gr12-september-2025-afr1-watermark/37cff2db-7c1a-4ec8-8ac5-07ec5bcf87a9) — QP · 2025 - [Physics Grade 12 NSC P2 QP September 2025 Gauteng](https://www.hlayiso.com/library/physics-grade-12-nsc-p2-qp-september-2025-gauteng/4d3cbea6-10ec-47ff-98ee-50646165ae9f) — QP · 2025 - [Physics Grade 12 NSC P1 QP September 2025 KZN](https://www.hlayiso.com/library/physics-grade-12-nsc-p1-qp-september-2025-kzn/6d225dc8-24c6-4c28-a181-df906d5a2c0e) — QP · 2025 - [Physical Sciences Physics Grade 12 PS01 QP March 2025 Limpopo](https://www.hlayiso.com/library/physical-sciences-physics-grade-12-ps01-qp-march-2025-limpopo/482ee9af-e88a-4818-9120-11ab9e045a1c) — QP · 2025 - [Physics Grade 12 NSC P1 QP September 2025 Limpopo](https://www.hlayiso.com/library/physics-grade-12-nsc-p1-qp-september-2025-limpopo/b65e9832-a6be-4312-8f5a-57425d647121) — QP · 2025 ## Collection - [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers) --- Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. 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