# Physical Sciences Grade 12 P1 NSC Supplementary 2015 Memo > Grade 12 Physical Sciences Paper 1 memorandum from the NSC Supplementary Exam 2015, 17 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/physical-sciences-p1-feb-march-2015-memo-afr-eng/12e143e4-5d6d-42a2-a7d6-a8a4ec637dbd - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2015 - Exam period: NSC Supplementary Exam - Paper: 1 - Curriculum: CAPS - Pages: 17 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 2015 Memo](https://www.hlayiso.com/library/physical-sciences-p1-feb-march-2015-memo-afr-eng/12e143e4-5d6d-42a2-a7d6-a8a4ec637dbd) ## About this document Physical Sciences · Grade 12 · NSC Supplementary Exam · 2015. Memorandum, 17 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 2015 FEBRUARIE/MAART 2015 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 17 pages. Hierdie memorandum bestaan uit 17 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./Feb.–Mrt. 2015 NSC/NSS – Memorandum QUESTION 1/VRAAG 1 1.1 A  (2) 1.2 C (2) 1.3 D  (2) 1.4 B (2) 1.5 D (2) 1.6 A (2) 1.7 A (2) 1.8 C (2) 1.9 A (2) 1.10 B (2) [20] 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./Feb.–Mrt. 2015 NSC/NSS – Memorandum QUESTION 2/VRAAG 2 2.1 When one body exerts a force on a second body, the second body exerts a force of equal magnitude in the opposite direction on the first body. Wanneer een liggaam 'n krag op 'n tweede liggaam uitoefen, oefen die tweede liggaam 'n krag van gelyke grootte in die teenoorgestelde rigting op die eerste liggaam. OR/OF: When body A exerts a force on body B, body B will exert a force of equal magnitude but opposite in direction on body A. Indien liggaam A 'n krag uitoefen op liggaam B, sal B 'n krag van gelyke grootte maar teenoorgesteld in rigting op liggaam A uitoefen. ACCEPT/AANVAAR (for 1 mark only/vir slegs 1 punt) Action and reaction are equal and opposite. Aksie en reaksie is gelyk en teenoorgesteld (2) 2.2 Accept/Aanvaar Notes/Aantekeninge FA Force diagram Do not penalise for N Kragtediagram length of arrows Moenie vir die lengte van f FA die pyltjies penaliseer nie N If w is not shown but F // T and F⊥ are shown give 1 w f mark for both. Indien w nie aangetoon T is nie maar F // en F⊥ is w getoon, ken 1 punt toe vir beide. Accept the following symbols/Aanvaar die volgende simbole. N F N ; Normal;/Normaal  FA 40 N f F f ,f k  w F G Weight/Gewig; Gravitational force/Gravitasiekrag T Tension/Spanning; F T ;  (5) 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./Feb.–Mrt. 2015 NSC/NSS – Memorandum 2.3.1 OPTION 1/OPSIE 1 For the 1 kg block/Vir die 1 kg blok; f k = µkN = µ k mgcosθ = 0,29 (1 x 9,8 cos 30o)  = 2,46 N OPTION 2/OPSIE 2 BY PROPORTION:/DEUR EWEREDIGHEID The smaller mass = ¼ of the larger mass Die kleiner massa = ¼ die groter massa ∴frictional force/wrywingskrag = ¼ (10)  = 2,5 N (3) 2.3.2 POSITIVE MARKING FROM QUESTION 2.2 Notes/Aantekeninge POSITIEWE NASIEN VANAF VRAAG 2.2 Learners need not show how (1) and (2) were combined OPTION 1/OPSIE 1 Leerders hoef nie aan te toon F net = ma hoe (1) en (2) gekombineer is nie. For 1 kg block/Vir 1 kg blok The first correct substitution F A – {(T+f k ) + mgsinθ} = ma for equation (1) should carry 2 40 – {T + 2,46 + 1(9,8)(sin30o)}= (1 x) a  marks. 40 – T – 7,36 = a The second substitution must 32,64 – T =a........(1) carry 1 mark. For 4 kg block/Vir 4 kg blok Die eerste korrekte vervanging T – (mg sinθ + f k ) = 4a vir vergelyking (1) moet 2 T – (4 x 9,8 sin30o + 10) = 4a punte tel. T- 29,6 = 4a.........(2) Die tweede vervanging tel 1 From (1) and (2)/Vanaf (1) en (2) punt. a = 0,61 m∙s-2 T = 29,6 + (4(0,61)  T = 32,04 N OPTION 2/OPSIE 2 Consider the blocks as a single system. Beskou die blokke as 'n enkele sisteem. F A – [ (f tot ) - {(4+1) gsin30o}]= (4+1)a 40 – (10 -2,46) –(5(9,8)sin30o) = 5a ∴a = 0,61 m∙s-2 For 1 kg block/Vir 1 kg blok F net = ma F A – {(T+f k ) + mgsinθ} = ma 40 – {T + 2,46 + 1(9,8)(sin30o)} = (1 x) a  40 – T – 7,36 = a 32,64 – T = 0,61 T = 32,04 N 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./Feb.–Mrt. 2015 NSC/NSS – Memorandum OR/OF For 4 kg block/Vir 4 kg blok Fnet - ma T – (mg sinθ + f k ) = 4a T – (4 x 9,8 sin30o + 10) = 4a T- 29,6 = 4a T = 29,6 + (4)(0,61)  (6) = 32,04 N [16] QUESTION 3/VRAAG 3 3.1 Free fall/Vrye val ACCEPT/AANVAAR Vertically accelerated motion/projectile motion. Vertikale versnelde beweging /projektielbeweging (1) 3.2.1 Downward motion as positive Notes / Aantekeninge Afwaartse beweging as positief Accept/Aanvaar g or/of a ∆y = v i ∆t + ½ a∆t2 ∆x = v i ∆t + ½ a∆t2 s = ut + ½ at2 30 = v i (1,5) + ½ (9,8)(1,5)2  v i = 12,65 m∙s-1 Upward motion as positive Opwaartse beweging as positief ∆y = v i ∆t + ½ a∆t2 -30 = v i (1,5) + ½ (-9,8)(1,5)2  v i = 12,65 m∙s-1 (4) 3.2.2 OPTION 1/OPSIE 1 Notes / Aantekeninge Positive marking from QUESTION 3.2.1 For/Vir XB Positiewe nasien vanaf VRAAG 3.2.1 Accept/Aanvaar g or/of a Downward motion as positive v2 = u2 + 2as Afwaartse beweging as positief The height must be written down in v f 2 = v i 2 + 2a∆y order to score the final mark. 12,652 = 0 + 2(9,8) ∆y Die hoogte moet neergeskryf word ∆y = 8,16 m om die finale punt te kry. Height/Hoogte XC = XB + BC (30 + 8,16) = 38,16 m Height is/Hoogte is 38,16 m 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./Feb.–Mrt. 2015 NSC/NSS – Memorandum Upward motion as positive Opwaartse beweging as positief v f 2 = v i 2 + 2a∆y (-12,65)2 = 0 + 2(-9,8) ∆y ∆y = - 8,16 m Height/Hoogte XC = XB + BC (-30) +(- 8,16) = -38,16 m Height is /Hoogte is 38,16 m)  (5) OPTION /OPSIE 2 Notes / Aantekeninge Positive marking from QUESTION 3.2.1 Start with time for XB Positiewe nasien vanaf VRAAG 3.2.1 Begin met tyd vir XB Accept/Aanvaar Downward motion as positive g or/of a v=u+at Afwaartse beweging as positief v2 = u2 + 2as v B = v X + a∆t s = ut + ½ at2 12,65 = 0 + 9,8∆t ∆t = 1,29 s ∆y = v i ∆t + ½ a∆t2 = 0 + ½ (9,8(1,29)2 ∆y = 8,15 m Height/Hoogte XC = XB + BC (30 + 8,15) = 38,15 m Upward motion as positive Opwaartse beweging as positief v B = v X + a∆t -12,65 = 0 + (-)9,8∆t ∆t = 1,29 s ∆y = v i ∆t + ½ a∆t2 = 0 + ½ (-9,8(1,29)2 ∆y = -8,15 m Height/Hoogte XC = XB + BC (-30) + (-8,15) = 38,15 m (5) 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./Feb.–Mrt. 2015 NSC/NSS – Memorandum OPTION 3/OPSIE 3 Notes / Aantekeninge Positive marking from QUESTION 3.2.1 start with velocity at C Positiewe nasien vanaf VRAAG 3.2.1 Accept/Aanvaar g or/of a v = u+at Downward motion as positive v2 = u2 + 2as Afwaartse beweging as positief The height must be written down in v C = v B + a∆t order to score the final mark. = 12,65 + 9,8 (1,5)  Die hoogte moet neergeskryf word = 27,35 m∙s-1 om die finale punt te kry. v C = v X 2 + 2a∆y 2 (27,35)2 = 0 + 2(9,8) ∆y ∴ ∆y = 38,16 m Height is /Hoogte is 38,16 m Upward motion as positive Opwaartse beweging as positief v C = v B + a∆t = -12,65 + (-9,8) (1,5)  = -27,35 m∙s-1 v C 2 = v X 2 + 2a∆y (-27,35)2 = 0 + 2(-9,8) ∆y ∴ ∆y = -38,16 m Height/Hoogte = 38,16m (5) OPTION 4/OPSIE 4 Notes / Aantekeninge Positive marking from QUESTION 3.2.1 Accept/Aanvaar Positiewe nasien vanaf VRAAG 3.2.1 mgh i + ½ mv i 2 = mgh f + ½ mv f 2 ∆U + ∆K =0 Take point B as the zero position (mgh + 0)  = 0 + (½ m(12,65)2) and XH = h Neem punt B is nul posisie en h = 8,16 m XH = h XC = h + 30 = (30 + 8,16) = 38,16 m (5) 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./Feb.–Mrt. 2015 NSC/NSS – Memorandum 3.3 OPTION 1/OPSIE 1 Notes / Aantekeninge Criteria/Kriteria Mark/Punt For each line correctly drawn as shown  Vir elke lyn korrek geteken soos getoon Acceleration/Versnelling Both axes correctly labelled  Beide asse korrek benoem (m∙s-2) 0 Accept/Aanvaar t(s) Only 2 marks for this answer/Slegs 2 punte vir hierdie antwoord Acceleration/Versnelling (m∙s-2) 0 t(s) 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./Feb.–Mrt. 2015 NSC/NSS – Memorandum OPTION 2/OPSIE 2 Notes / Aantekeninge Criteria/Kriteria Mark/Punt For each line correctly drawn as shown  Acceleration/Versnelling (m∙s-2) Vir elke lyn korrek geteken soos getoon Both axes correctly labelled  Beide asse korrek benoem 0 Only 2 marks for this answer/Slegs 2 t(s) punte vir hierdie antwoord Acceleration/Versnelling (m∙s-2) 0 t(s) (3) [13] 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./Feb.–Mrt. 2015 NSC/NSS – Memorandum QUESTION 4/VRAAG 4 4.1 W net = ΔK  Notes / Aantekeninge W net = ½ (M + m)(v f 2 –v i 2) 1 mark for either of the formulae indicated W fr = fΔxcosθ =½ (M + m)(v f 2 –v i 2) 1 punt vir enige van die formule 10 x 2 cos180= ½ (7,02)(0-v2)  aangedui v bb = 2,39 m∙s-1 (2,387) m∙s-1 Accept/Aanvaar W nc = ΔK + ΔU with ΔU =0 (5) 4.2 The total linear momentum of an (isolated) closed Notes/Aantekeninge system remains constant. 2 or/of 0 Die totale lineêre momentum in 'n geslote (geïsoleerde) sisteem bly konstant ACCEPT/AANVAAR In an isolated system the total momentum before collision equals the total momentum after collision. In 'n (geïsoleerde) geslote sisteem is die totale momentum voor botsing gelyk aan die totale momentum na botsing. (2) 4.3. Positive marking from QUESTION 4.1 Notes / Aantekeninge Positiewe nasien vanaf VRAAG 4.1 Accept/Aanvaar Σp i = Σp f  (m 1 + m 2 )v i = m 1 v 1f + m 2 v 2f m 1 v 1i + m 2 v 2i = (m 1 +m 2 )v f 0,02v i + (7)(0) = (7,02)(2,39) 837,84 m∙s-1 (for learners working with 2,387 m∙s-1/vir leerders wat met 0,02v i  = 7,02 (2,39) 2,387 m∙s-1 werk) v i = 838,89 m∙s-1  (4) [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./Feb.–Mrt. 2015 NSC/NSS – Memorandum QUESTION 5/VRAAG 5 5.1 ∆U +∆K = 0 Notes / Aantekeninge (5)(9,8)( 5) + 0 + (0 + ½ (5v f 2) = 0 Accept/Aanvaar v f = 2 × 9,8 × 5 Mgh i + ½ Mv i 2 = Mgh f + ½ Mv f 2 = 9,90 m∙s-1 (9,899 m∙s-1) (4) 5.2 No friction/zero resultant force so there is no loss in energy./Only conservative forces present./Mechanical energy is conserved Geen wrywing/nul resulterende krag dus is daar geen verlies in energie nie/ Slegs konserwatiewe kragte is teenwoordig./Meganiese energie bly behoue (2) 5.3 A force for which the work done is path Notes / Aantekeninge dependent. Accept/Aanvaar 'n Krag waarvoor arbeid verrig afhanklik A force which does not conserve van die pad gevolg is mechanical energy./'n Krag wat nie meganiese energie behoue laat bly nie. (2) 5.4 OPTION 1/OPSIE 1 W nc = ∆U +∆K  F ∆x cosθ = ∆U +∆K (18 ∆x cos 180) = (5 ) ( 9,8) (3 -0) + ½ (5) (0 – 9,902) ∆x = 5,4458m 3 θ =sin-1  5,4458 θ =33,43o OPTION 2/OPSIE 2 W net = W f + W G  W net =f ∆x cosθ + mgsinθΔxcosθ 3 =[(18 ) ∆x cos 180o) + 5 (9,8) ( Δx ) cos180o] Δx = -18∆x - 147 W net = ∆K ∆K = ½ ( 5) (0 – 9,902) = -245,025 -18∆x – 147 = -245,025 ∆x = 5,4458 m 3 θ =sin-1  5,4458 θ =39,43o (7) [15] 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./Feb.–Mrt. 2015 NSC/NSS – Memorandum QUESTION 6/VRAAG 6 6.1.1 v = fλ 340 λ= 520 = 0,65 m (2) 6.1.2 v ± vL Notes / Aantekeninge fL = fs  Accept/Aanvaar v ± vs v 340  fL = f fL = (520 )  v -vs s (340 - 15)  f L = 544 Hz v = fλ 340  λ= 544 = 0,63m (6) 6.2 The wavelength in QUESTION 6.1.2 is shorter because the waves are compressed as they approach the observer.   Die golflengte in VRAAG 6.1.2 is korter omdat die golwe saamgedruk word soos hulle die waarnemer nader. (2) 6.3 The red shift occurs when the spectrum of a distant star moving away from the earth is shifted toward the red end of the spectrum.  Rooi verskuiwings vind plaas wanneer die spektrum van 'n vêr afgeleë ster wat vanaf die aarde wegbeweeg na die rooi ent van die spektrum skuif. (2) [12] 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./Feb.–Mrt. 2015 NSC/NSS – Memorandum QUESTION 7/VRAAG 7 7.1 The net electrostatic force on a charged particle due to the presence of another charged particle is directly proportional to the product of the chargesand inversely proportional to the square of the distance between them (their centres)  Die netto elektrostatiese krag op 'n gelaaide deeltjies as gevolg van die teenwoordigheid van 'n ander gelaaide deeltjie is direk eweredig aan die produk van die ladings en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle (hul middelpunte) OR/OF The force of attraction or repulsion between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.  Die aantrekkings- of afstotingskrag tussen twee puntladings is direk eweredig aan die produk van die ladings en omgekeerd eweredig aan die kwadraat van die afstand tussen hulle. OR/OF Any two charged particles will exert an electrostatic force on each other where the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between the charged particles. (their centres) Enige twee gelaaide deeltjies sal 'n elektrostatiese krag op mekaar uitoefen waar die krag direk eweredig is aan die produk van die ladings en omgekeerd eweredig is aan die kwadraat van die afstand tussen hulle (tussen hul middelpunte) (2) 7.2 KQ 1Q 2 7.2.1 F=  r2 -1 (9 × 10 9 )Q 2  1,44 × 10 = (0,5)2 Q = 2 x10-6 C (4) 7.2.2 Positive marking from QUESTION 7.2.1 Positiewe nasien vanaf VRAAG 7.2.1 Q = ne 2 x10-6 = n(1,6x10-19)  n = 1,25 x1013 electrons/elektrone (3) 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./Feb.–Mrt. 2015 NSC/NSS – Memorandum 7.3 7.3.1 Left /Links (west/wes)  (1) 7.3.2 Take right as positive/Neem regs as positief Notes / Aantekeninge E net = E A + E B    No. electrons should be 4 (9 × 10 9 )(2 × 10 -6 ) (9 × 10 9 )Q final (3 × 10 ) = - + removed = n f – n i (1,5)2 (1)2 allocate the 1 mark for the Q final = 4,22 x 10-6 C subtraction Aantal elektrone wat Q = ne verwyder moet word = n f – n i 4,22 x10-6 = n(1,6x10-19)  Ken 1 punt toe vir aftrekking n f = 2,64 x1013 electrons/elektrone electrons removed/elektrone verwyder = (2,64 x1013 +1,25 x1013)  (8) = 3,89 x1013 electrons/elektrone [18] QUESTION 8/VRAAG 8 8.1.1 Ensure that the wires have:/Maak seker dat die drade The same length/dieselfde lengte het. The same thickness/cross-sectional area/dieselfde dikte/deursnit-area/ oppervlakte het (2) 8.1.2 Wire A (Resistor A)/Draad A  ΔV R=  ΔI Accept any correct coordinates chosen from the graph 4,4 Aanvaar enige korrekte koördinate van die grafiek - RA =  = 11 Ω gekies. 0,4 2,2 RB =  = 5,5 Ω 0,4 E = I2 R∆t For the same time and current, the heating in A will be higher because its resistance is higher than that of B.  Vir dieselfde tyd en stroom, sal die verwarming in A hoër wees omdat sy weerstand groter is as die van B. ACCEPT/AANVAAR: P = I2R For the same current, the heat produced per unit time in A will be higher because its resistance is higher than that of B.  Vir dieselfde stroom, sal die hitte vrygestel per eenheidstyd in A hoër wees omdat sy weerstand groter is as die van B. (8) 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./Feb.–Mrt. 2015 NSC/NSS – Memorandum 8.2.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 I 5,5Ω : I 11Ω V = IR 2:1 V 11 Ω = 0,2 x 11 I 5,5Ω = (0,2)(2)  =2,2 V = 0,4 A V 5,5 = V 11 = 2,2 V 2,2 I 5,5 = 5,5 = 0,4 A (3) 8.2.2 OPTION 1/OPSIE 1 Notes / Aantekeninge V = IR Accept/Aanvaar I tot = (0,4 + 0,2)  R1R 2 RP =  = 0,6 A R1 + R 2 1 1 1 = + + ... R p R1 R 2 11× 5,5 =  1 1 1 11+ 5,5 = +  Rp 11 5,5 = 3,67 Ω R P = 3,67 Ω RT = RP + RA = 3,67 + 11 =14,67Ω ε = I(R + r)  9 = 0,6(14,67 +r)  r = 0,33 Ω  (7) OPTION 2/OPSIE 2 I tot = (0,4 + 0,2)  = 0,6 A V ext = V 11 Ω + V //  = [I tot (R 11 ) + 2,2] = 0,6 (11)  + 2,2 = 8,8V ε = V ext + I tot (r)  9 = 8,8 + 0,6r r = 0,33 Ω (7) 8.2.3 Decrease/Afneem  The total resistance increases / Die totale weerstand neem toe (2) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 16 DBE/ Feb.–Mar./Feb.–Mrt. 2015 NSC/NSS – Memorandum QUESTION 9/VRAAG 9 9.1.1 OPTION 1/OPSIE 1 Notes / Aantekeninge V 2 rms Assume correct formula for V rms and give a P av =  mark if the substitution is correct1 mark R Aanvaar die korrekte formule vir V wgk en ken 'n 2  340   punt toe indien die vervanging korrek is.   100 =  2  R  R = 578 Ω OPTION 2/OPSIE 2 Vmax 340 V rms = = = 240,04 2 2 V 2 rms P ave =  R 240,04 2 100 =  R R = 578 Ω  (5) 9.1.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 P av = I rms V rms  V rms = Irms R 340 340 100 = I rms  = I rms (578)  2 2 100 I rms = 0,417 A I rms = 340 2 = 0,417 A  (3) 9.2 Can be stepped up or down/ can be transmitted with less power loss. Kan verhoog of verlaag word/ kan versend word met minder energie verlies. (1) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 17 Downloaded from hlayiso.com Physical Sciences P1/Fisiese Wetenskappe V1 17 DBE/Feb.–Mar./Feb.–Mrt. 2015 NSC/NSS – Memorandum QUESTION 10/VRAAG 10 10.1 The minimum energy needed to emit an electron  from (the surface of) a metal.  Die minimum energie benodig om 'n elektron uit die (oppervlak van) 'n metaal vry te stel. (2) 10.2 1 2 E = W0 + mv max 2 Any ONE OF/ENIGE EEN van c 1 2 h = W0 + mv max λ 2 (6,63 × 10 −34 )(3 × 10 8 )  = (3,36 × 10 −19 ) + 2,32 × 10 −19  (λ ) λ = 3,50 x 10-7 m (4) 10.3 POSITIVE MARKING FROM QUESTION 10.2 POSITIEWE NASIEN VANAF VRAAG 10.2 1 E = W0 + mv max 2 2 OR/OF c 1 h = W0 + mv max 2 λ 2   (6,63 × 10 -34 )(3 × 10 8 ) = (3,65 × 10 -19 ) + E k (3,50 × 10 )-7  -19 E = 2,03 x 10 J (4) 10.4.1 Increasing the intensity does not change the energy/ frequency/wavelength of the incident photons/The energy of a photon remains unchanged (for the same frequency). Verhoging van die intensiteit, verander nie die energie/frekwensie/golflengte van die invallende fotone nie/Die energie van die foton bly onveranderd (vir dieselfde frekwensie). (1) 10.4.2 Increases./Neem toe (1) - 10.4.3 More photons (packets of energy) strike the surface of the metal per unit time hence more (photo) electrons ejected per unit time (leading to increased current). Meer fotone (energie pakkies) tref die oppervlakte van die metaal per eenheidstyd, gevolglik word meer (foto)elektrone per eenheidstyd vrygestel (wat (2) lei tot 'n verhoogde stroom). 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