# Physical Sciences Grade 12 P1 Eastern Cape June 2021 Memo > Grade 12 Physical Sciences Paper 1 memorandum from the Eastern Cape June 2021 (Afrikaans), 14 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/phys-sc-p1-memo-gr12-june2021-afrik-english/51075955-8380-4def-a361-3f9b02ab715c - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Language: Afrikaans - Document type: Memorandum - Year: 2021 - Exam period: Eastern Cape June - Paper: 1 - Curriculum: CAPS - Pages: 14 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 Eastern Cape June 2021 Memo](https://www.hlayiso.com/library/phys-sc-p1-memo-gr12-june2021-afrik-english/51075955-8380-4def-a361-3f9b02ab715c) ## About this document Physical Sciences · Grade 12 · Eastern Cape June · 2021 · Afrikaans. Memorandum, 14 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 NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 JUNE/JUNIE 2021 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 MARKING GUIDELINE / NASIENRIGLYN (EXEMPLAR/EKSEMPLAAR) MARKS/PUNTE: 150 This marking guideline consists of 14 pages./ Hierdie nasienriglyn bestaan uit 14 bladsye. ### Page 2 2 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE/JUNIE 2021) GENERAL GUIDELINES/ALGEMENE RIGLYNE 1. CALCULATIONS/BEREKENINGE 1.1 Marks will be awarded for: correct formula, correct substitution, correct answer with unit. Punte sal toegeken word vir: korrekte formule, korrekte substitusie, korrekte antwoord met eenheid. 1.2 No marks will be awarded if an incorrect or inappropriate formula is used, even though there are many relevant symbols and applicable substitutions. Geen punte sal toegeken word waar ʼn verkeerde of ontoepaslike formule gebruik word nie, selfs al is daar relevante simbole en relevante substitusies. 1.3 When an error is made during substitution into a correct formula, a mark will be awarded for the correct formula and for the correct substitutions, but no further marks will be given. Wanneer ʼn fout gedurende substitusie in ʼn korrekte formule begaan word, sal ʼn punt vir die korrekte formule en vir korrekte substitusies toegeken word, maar geen verdere punte sal toegeken word nie. 1.4 If no formula is given, but all substitutions are correct, a candidate will forfeit one mark. Indien geen formule gegee is nie, maar al die substitusies is korrek, verloor die kandidaat een punt. 1.5 No penalisation if zero substitutions are omitted in calculations where correct formula/principle is correctly given. Geen penalisering indien nulwaardes nie getoon word nie in berekeninge waar die formule/beginsel korrek gegee is nie. 1.6 Mathematical manipulations and change of subject of appropriate formulae carry no marks, but if a candidate starts off with the correct formula and then changes the subject of the formula incorrectly, marks will be awarded for the formula and correct substitutions. The mark for the incorrect numerical answer is forfeited. Wiskundige manipulasies en verandering van die onderwerp van toepaslike formules tel geen punte nie, maar indien ʼn kandidaat met die korrekte formule begin en dan die onderwerp van die formule verkeerde verander, sal die punte vir die formule en korrekte substitusies toegeken word. Die punt vir die verkeerde numeriese antwoord word verbeur. 1.7 Marks are only awarded for a formula if a calculation has been attempted, i.e. substitutions have been made or a numerical answer given. Punte word slegs vir ʼn formule toegeken indien ʼn poging tot ʼn berekening aangewend is, d.w.s. substitusies is gedoen of ʼn numeriese antwoord is gegee. 1.8 Marks can only be allocated for substitutions when values are substituted into formulae and not when listed before a calculation starts. Punte kan slegs toegeken word vir substitusies wanneer waardes in formule ingestel word en nie vir waardes wat voor ʼn berekening gelys is nie. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 (EC/JUNE/JUNIE 2021) PHYSICAL SCIENCES P1 /FISIESE WETENSKAPPE V1 3 1.9 All calculations, when not specified in the question, must be done to a minimum of two decimal places. Alle berekenings, wanneer nie in die vraag gespesifiseer word nie, moet tot ʼn minimum van twee desimale plekke gedoen word. 1.10 If a final answer to a calculation is correct, full marks will not automatically be awarded. Markers will always ensure that the correct/appropriate formula is used and that workings, including substitutions, are correct. Indien ʼn finale antwoord van ʼn berekening korrek is, sal volpunte nie outomaties toegeken word nie. Nasieners sal altyd verseker dat die korrekte/toepaslike formule gebruik word en dat bewerkings, insluitende substitusies korrek is. 1.11 Questions where a series of calculations have to be made (e.g. a circuit diagram question) do not necessarily always have to follow the same order. FULL MARKS will be awarded provided it is a valid solution to the problem. However, any calculation that will not bring the candidate closer to the answer than the original data, will not count any marks. Vrae waar ʼn reeks berekeninge gedoen moet word (bv. ʼn stroombaan- diagramvraag) hoef nie noodwendig dieselfde volgorde te hê nie. VOLPUNTE sal toegeken word op voorwaarde dat dit ʼn geldige oplossing vir die probleem is. Enige berekening wat egter nie die kandidaat nader aan die antwoord as die oorspronklike data bring nie, sal geen punte tel nie. 2. UNITS/EENHEDE 2.1 Candidates will only be penalised once for the repeated use of an incorrect unit within a question. Kandidate sal slegs een keer gepenaliseer word vir die herhaaldelike gebruik van ʼn verkeerde eenheid in ʼn vraag. 2.2 Units are only required in the final answer to a calculation. Eenhede word slegs in die finale antwoord op ʼn vraag verlang. 2.3 Marks are only awarded for an answer, and not for a unit per se. Candidates will therefore forfeit the mark allocated for the answer in each of the following situations: Correct answer + wrong unit Wrong answer + correct unit Correct answer + no unit Punte sal slegs vir ʼn antwoord en nie vir ʼn eenheid per se toegeken word nie. Kandidate sal die punt vir die antwoord in die volgende gevalle verbeur: Korrekte antwoord + verkeerde eenheid Verkeerde antwoord + korrekte eenheid Korrekte antwoord + geen eenheid 2.4 SI units must be used except in certain cases, e.g. V.m-1 instead of N.C-1, and cm.s-1 or km.h-1 instead of m.s-1 where the question warrants this. SI eenhede moet gebruik word, behalwe in sekere gevalle, bv. V .m-1 in plaas van N.C-1, en cm.s-1 of km.h-1 in plaas van m.s-1 waar die vraag dit regverdig. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 4 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE/JUNIE 2021) 3. GENERAL/ALGEMEEN 3.1 If one answer or calculation is required, but two are given by the candidate, only the first one will be marked, irrespective of which one is correct. If two answers are required, only the first two will be marked, etc. Indien een antwoord of berekening verlang word, maar twee word deur die kandidaat gegee, sal slegs die eerste een nagesien word, ongeag watter een korrek is. Indien twee antwoorde verlang word, sal slegs die eerste twee nagesien word, ens. 3.2 For marking purposes, alternative symbols (s, u, t etc) will also be accepted. Vir nasiendoeleindes sal alternatiewe simbole (s, u, t ens) ook aanvaar word. 3.3 Separate compound units with a multiplication dot, no a full stop, for example, m.s-1. For marking purposes, m.s-1 and m/s will also be accepted. Skei saamgestelde eenhede met ʼn vermenigvuldigingspunt en nie met ʼn punt nie, byvoorbeeld m.s-1. Vir nasiendoeleindes sal m.s-1 en m/s ook aanvaar word. 4. POSITIVE MARKING/POSITIEWE NASIEN Positive marking regarding calculations will be followed in the following cases: Positiewe nasien met betrekking tot berekeninge sal in die volgende gevalle geld: 4.1 Subquestion to subquestion: When a certain variable is calculated in one subquestion (e.g. 3.1) and needs to be substituted in another (3.2 of 3.3), e.g. if the answer for 3.1 is incorrect and is substituted correctly in 3.2 or 3.3, full marks are to be awarded for the subsequent subquestions. Subvraag na subvraag: Wanneer ʼn sekere veranderlike in een subvraag (bv. 3.1) bereken word en dan in ʼn ander vervang moet word (3.2 of 3.3), bv. indien die antwoord vir 3.1 verkeerd is en word korrek in 3.2 of 3.3 vervang, word volpunte vir die daaropvolgende subvraag toegeken. 4.2 A multistep question in a subquestion: If the candidate has to calculate, for example, current in die first step and gets it wrong due to a substitution error, the mark for the substitution and the final answer will be forfeited. ʼn Vraag met veelvuldige stappe in ʼn subvraag: Indien ʼn kandidaat bv. die stroom verkeerd bereken in ʼn eerste stap as gevolg van ʼn substitusiefout, verloor die kandidaat die punt vir die substitusie sowel as die finale antwoord. 5. NEGATIVE MARKING/NEGATIEWE NASIEN Normally an incorrect answer cannot be correctly motivated if based on a conceptual mistake. If the candidate is therefore required to motivate in QUESTION 3.2 the answer given in QUESTION 3.1, and 3.1 is incorrect, no marks can be awarded for QUESTION 3.2. However, if the answer for e.g. 3.1 is based on a calculation, the motivation for the incorrect answer could be considered. ʼn Verkeerde antwoord, indien dit op ʼn konsepsuele fout gebaseer is, kan normaalweg nie korrek gemotiveer word nie. Indien ʼn kandidaat gevra word om in VRAAG 3.2 die antwoord op VRAAG 3.1 te motiveer en 3.1 is verkeerd, kan geen punte vir VRAAG 3.2 toegeken word nie. Indien die antwoord op bv. 3.1 egter op ʼn berekening gebaseer is, kan die motivering vir die verkeerde antwoord in 3.2 oorweeg word. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 (EC/JUNE/JUNIE 2021) PHYSICAL SCIENCES P1 /FISIESE WETENSKAPPE V1 5 QUESTION/VRAAG 1 1.1 B (2) 1.2 C (2) 1.3 C (2) 1.4 D (2) 1.5 A (2) 1.6 C (2) 1.7 B (2) 1.8 A (2) 1.9 D (2) 1.10 B (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 6 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE/JUNIE 2021) QUESTION/VRAAG 2 2.1 When object A exerts a force on object B, object B simultaneously exerts an oppositely directed force of the same magnitude on object A. Wanneer voorwerp A ʼn krag op voorwerp B uitoefen, oefen voorwerp B ʼn krag gelyk in grootte en in teenoorgestelde rigting op voorwerp A uit. (2) 2.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 FA on/op B FN FA on/op B FN Ff Ff Fg Fg for both Fg components/vir beide komponente (4) 2.3 Data: MB = 25 kg fB = 15 N a = 1,5 m.s-2 FA on B = ? FB on A = ? Fnet = ma FA on B + fB + Fg = ma Any one/Enige een FA on B + fB + mgSin35°= ma FA on B + (- 15) + (- 25x9,8x0,574) = 25x1,5 FA on B – 155,63 = 37,50 FA on B = 193,13 N FB on A = - FA on B FB on A = - 193,13 N FB on A = 193,13 N down the slope/ teen die helling af (6) 2.4 Fnet = ma FNceba + FB on A + fA + Fg = ma Any one/Enige een FNceba + FB on A + fA + mgSin35°= ma FNceba + (- 193,13) + (- 4,5) + (- 10x9,8x0,574) = 10x1,5 FNceba – 253,88 = 15 FNceba = 268,88 N (4) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 (EC/JUNE/JUNIE 2021) PHYSICAL SCIENCES P1 /FISIESE WETENSKAPPE V1 7 QUESTION/VRAAG 3 3.1 When a net/resultant force acts on an object of mass m, the object accelerates in the direction of the net force. The accelereation is directly proportional to the net force and inversely proportional to the mass. Wanneer ’n netto krag op ’n voorwerp inwerk, versnel die voorwerp in die rigting van die netto krag. Die versnelling is direk eweredig aan die netto krag en omgekeerd aan die massa van die voorwerp. (2) 3.2 Fnet = ma Fg + FN + FY = ma Any one/Enige een Fg + FN + Fa Sin40° = ma Fg + (-623,3) + (-500Sin40°) = mx0 Fg - 623,3 – 321,39 = 0 Fg = 944,69 N mg = 944,69 mx9,8 = 944,69 m = 96,40 kg (5) 3.3 For 96,4 kg box / vir 96,4 kg houer Fnet = ma FX + fk + T = ma Any one/Enige een Fa Cos40° + f + T = ma 500 Cos40° + (-115) +(- T) = 96,4a 383,02 – 115 – T = 96,4a 268,02 - T = 96,4a T = 268,02 – 96,4a .............. (1) For 15 kg box / vir 15 kg houer Fnet = ma Any one/Enige een Fg + T = ma mg + T = ma -15x9,8 + T = 15a -147 + T = 15a (Substitute/Stel in vir T) -147 + (268,02 – 96,4a) = 15a 121,02 = 111,40a a = 1,09 m.s-2 (7) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 8 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE/JUNIE 2021) QUESTION/VRAAG 4 4.1 It (projectile) is an object upon which the only force acting is the force of gravity. Dit (projektiel) is ’n voorwerp waarop die enigste krag wat daarop inwerk, die gravitasiekrag is. OR/OF Projectile is an object that experiences only gravitational force. Projektiel is ’n voorwerp wat slegs die gravitasiekrag ondervind. (2) 4.2 OPTION 1 /OPSIE 1 (upwards positive) (downwards positive) (opwaarts positief) (afwaarts positief) vf2 = vi2 + 2gΔy vf2 = vi2 + 2gΔy 02 = 52 + 2 (-9,8) Δy 02 = (-5)2 + 2 (9,8) Δy Δy = 1,28 m Δy = -1,28 m Δy = 1,28 m, (upward) OPTION 2/OPSIE 2 (upwards positive) (downwards positive) (opwaarts positief) (afwaarts positief) vf = vi + g∆t vf = vi + g∆t 0 = (5) + (-9,8 ). ∆t 0 = (5) + (-9,8 ). ∆t ∴∆t = 0,51 s ∴∆t = 0,51 s Δy = viΔt + 1 gΔt2 Δy = viΔt + 1 gΔt2 2 2 = (5)(0,51) + 1 (- 9,8) (0,51)2 = (-5)(0,51) + 1 (9,8)(0,51)2 2 2 ∴Δy = 1,28 m Δy = - 1,28 m ∴Δy = 1,28 m, (upwards) (3) 4.3 The maximum height reached by ball A (Δymax) Die maksimum hoogte bereik deur bal A ( Δymax) = 1,28 + 5 = 6,28 m Maximum height reached by ball B is given by: Makisum hoogte bereik deur bal B is: (upwards positive) (downwards positive) (opwaarts positief) (afwaarts positief) vf2 = vi2 + 2gΔy vf2 = vi2 + 2gΔy 02 = 82 + 2 (-9,8) Δy 02 = (-8)2 + 2 (9,8) Δy ΔyB = 3,27 m Δy = -3,27 m ΔyB = 3,27 m, (upward) Since ΔyB < Δymax OR 3,27 m < 6,28 m Since ΔyB < Δymax ∴ Ball B did not reach the maximum OR 3,27 m < 6,28 m height. ∴ Ball B did not reach the maximum ∴ Bal B het nie die maksimum height. hoogte bereik nie. ∴ Bal B het nie die maksimum hoogte bereik nie. (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 (EC/JUNE/JUNIE 2021) PHYSICAL SCIENCES P1 /FISIESE WETENSKAPPE V1 9 4.4 (upwards positive) (downwards positive) (opwaarts positief) (afwaarts positief) vf = vi + g∆t vf = vi + g∆t 0 = (5) + (-9,8 ). ∆t 0 = (5) + (-9,8 ). ∆t ∴∆t = 0,51 s ∴∆t = 0,51 s vf = vi + g∆t vf = vi + g∆t = (8) + (-9,8 ). 0,51 = (-8) + (9,8 ). 0,51 ∴ vf = 3,00 m.s-1 vf = -3,00 m.s-1 ∴ vf = 3,00 m.s-1 (4) 4.5 (Upwards positive) For ball A/ Vir bal A vf = vi + g∆t vf = (5) + (-9,8 ). ∆t ...... (1) For ball B/ Vir bal B vf = vi + g∆t vf = (8) + (-9,8 ). ∆t ...... (2) Equating (1) and (2)/ Stel (1) gelyk aan (2): Since ball A will be moving downwards then/Aangesien bal A afwaarts beweeg dat: -(5 – 9,8.∆t) = 8 – 9,8.∆t 19,6.∆t = 13 ∆t = 0,66 s ∆t = 0,011min (5) [19] QUESTION/VRAAG 5 5.1 A system on which the net external forces acting is zero. Dit is ’n sisteem waarop die netto eksterne krag wat op die sisteem inwerk, nul is. (2) 5.2 Let east be positive/Laat oos positief wees p = mv = 1 140 x 30 = 34 200 kg.m.s-1, (east/oos) (3) 5.3 Σpi = Σpf mvviv + mcvic = mvvfv +mcvfc 1 650 viv+ 1 140 x 30 = 1 650 x - 12 + 1 140x 10 1 650 viv = 11 400 – 34 200 – 19 800 viv = - 25,82 viv = 25,82 m.s-1 , west (5) 5.4 Principle of conservation of linear momentum The total linear momentum of an isolated system remains constant (preserved). Beginsel van behoud van lineêre momentum. Die totale lineêre momentum in ’n geïsoleerde sisteem bly konstant (behoue). (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 10 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE/JUNIE 2021) 5.5 ∆p = m(vf - vi) ∆p = 1 650 (-12 – (-25,82)) ∆p = 22 803 kg.m.s-1 (3) [16] QUESTION/VRAAG 6 6.1 The net/resultant force acting on an object is equal to the rate of change in momentum of the object. Die netto/resulterende krag wat op ’n voorwerp inwerk, is gelyk aan die tempo van verandering van momentum van die voorwerp. (2) 6.2 Total Ek Before = Ek tree + Ek car 1 1 Any one/Enige een = m v 2tree + m 2 v car 2 2 1 1 = (m)(02) + (1 125)(25)2 2 2 = 351 562,5 J (4) 6.3 Impulse of the tree/Impuls van die boom: Let original direction be positive/Laat die oorspronklike rigting positief wees Fnet∆t (car) = m(vf - vi) = 1 125(-6 – 25) = - 34 875 N Fnet∆t (car) = 34 875 N, backwards/terugwaarts Fnet∆t (tree) = 34 875 N , original direction of the car/ oorspronklie rigting van die motor. (4) 6.4 m(v f - v i ) Fnet (car) = Δt 1 125(-6 - 25) Fnet (car) = 3,5 Fnet (car) = - 9 964 N Fnet (car) = 9 964 N, backwards/terugwaarts (3) 6.5 When the airbags deploy during collision, the contact time of the driver with the airbag increases. According to the equation Fnet∆t = ∆p, when contact time increases the net force experienced by the driver decreases. Then, the extent of injury is reduced. Wanneer die lugsak gedurende die botsing ontplooi, sal die kontaktyd van die bestuurder met die lugsak verhoog. Volgens die vergelyking Fnet∆t = ∆p , wanneer die kontaktyd verhoog sal die netto krag wat deur die bestuurder ervaar word verlaag. Dus sal die omvang van die beserings verminder word. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 (EC/JUNE/JUNIE 2021) PHYSICAL SCIENCES P1 /FISIESE WETENSKAPPE V1 11 6.6 Decrease The final velocity will now be zero. Then, according to the equation: ∆p = m(vf – vi), it follows that ∆p will decrease. If ∆p decreases then Fnet decreases as they are directly proportional to each other. Verminder Die finale snelheid sal nou nul wees. Volgens die vergelyking: ∆p = m(vf – vi) dit volg dat ∆p sal verlaag. As Δp afneem dan sal Fnet verlaag aangesien hulle direk eweredig is aanmekaar. (4) [19] QUESTION/VRAAG 7 7.1 Gm1m2 F= r2 (6,67 x10-11 )( 3,4 x103 )( 5,98 x1024 ) F= (2,32 x106 )2 = 251 959,05 = 2,52 x 105 N (4) 7.2 Newton’s Law of Universal Gravitation Any two objects or particles in the universe attract each other with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. Newton se Universele Gravitasiewet Enige twee liggaam of deeltjie in die heelal trek mekaar aan met ’n krag wat direk eweredig aan die produk van hul massas en omgekeerd eweredig aan die kwadraat van die afstand tussen hul middelpunte aan. (3) 7.3 Gm1m2 F1 = =F r2 m Gm1 2 F2 = 2 2 r 2 Gm1m2 F2 = 2 r2 4 Gm1m2 F2 =2 =2F r2 F2 = 2(2,52 x 105 N) F2 = 5,04 x 105 N (3) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 12 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE/JUNIE 2021) QUESTION/VRAAG 8 8.1 8.1.1 The net work done on an object is equal to the change in the kinetic energy of the object. Die netto arbeid deur ‘n voorwerp verrig is gelyk aan die verandering in kinetiese energie van die voorwerp. OR /OF The amount of work done by a net force on an object is equal to the change in the object’s kinetic energy. Die arbeid verrig deur ’n netto krag is gelyk aan die verandering in kinetiese energie van die voorwerp. (2) 8.1.2 OPTION 1/OPSIE 1 Wnet = W F1 + W F2 + W// + W f = F1ΔxCos180° + F2ΔxCos0° + mgSin30°ΔxCos180° + fΔxCos180° = (20)(20)(-1) + (100)(20)(1) + (8)(9,8)(0,5)(20)(-1) + (8,5)(20)(-1) = -400 + 2000 – 784 – 170 = 646 J OPTION 2/OPSIE 2 Fnet = F2 + F1 + Fg + f = F2 + F1 + mgSin30°+ f = 100 + (-20) + {-(8)(9,8)(0,5)} + (-8,5) = 32,3 N Wnet = FnetΔxCos θ = Fnet ΔxCos0° = (32,3) (20)(1) = 646 J (4) 8.1.3 Wnet = ∆Ek Wnet = Ekf – Eki Any one/Enige een 1 1 Wnet = mv 2f - mv i2 2 2 1 1 646 = (8)v 2f - (8)(0) 2 2 (2)(646) v 2f = 8 = 161,5 vf = 12,71 m.s-1 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 (EC/JUNE/JUNIE 2021) PHYSICAL SCIENCES P1 /FISIESE WETENSKAPPE V1 13 8.2 1 1 ∆Ek = mv 2f - mv i2 2 2 1 1 = (1 500)(26)2 - (1 500)(0)2 2 2 = 507 000 J ΔEk P= Δt 507000 = 4 = 126 750 W = 126,75 kW (5) [15] QUESTION/VRAAG 9 9.1 9.1.1 It is the apparent change in frequency (or pitch/wavelength) of the sound detected by a listener because the sound source and the listener have different velocities relative to the medium of sound propagation. Dit is die verandering in frekwensie (toonhoogte/golflengte) 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/ wavelength) as a result of relative motion between a source and an observer. Dit is oënskynlike verandering in frekwensie (toonhoogte/golflengte) wat waargeneem word as gevolg van die relatiewe beweging tussen die bron en waarnemer. (2) 9.1.2 v ± vL fL = fS v ± vS Any one/Enige een v vL fL = fS v 340 vL 0,85x = (x) 340 289x = (340 – vL)(x) 289 = 340 – vL vL = 51 m.s-1 (4) 9.1.3 When the listener moves away from the source the sound waves that reach him have longer wavelength and a sound with a lower pitch is registered because of a lower frequency. As die luisteraar van die bron af wegbeweeg, het die klankgolwe wat hom bereik langer golflengte en ’n klank met ’n laer toonhoogte geregistreer weens 'n laer frekwensie. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 14 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/JUNE/JUNIE 2021) 9.1.4 It would be the same Dit sal dieselfde bly (2) 9.1.5 FL = 0,85fS = 0,85 x 190 = 161,5 Hz (2) 9.2 To measure the velocity of blood flowing through blood vessels. To scan a fetus/foetus. Om die snelheid van bloed wat deur are vloei, te meet. Om ’n fetus te skandeer. (2) [14] QUESTION/VRAAG 10 10.1 The magnitude of the electrostatic force exerted by one point charge (Q) on another point charge (Q) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (r) between them. Die grootte van die elektrostatiese krag wat een puntlading (Q 1) op ’n ander puntlading (Q2) uitoefen, is direk eweredig aan die produk van ladings en omgekeerd eweredig aan die kwadraat van die afstand (r) tussen hulle. (2) 10.2 East is positive/Oos is positief kQ A QN FAN = r2 (9 x109 )(4 x10 9 )(2 x10 9 ) = (0,014)2 = 0,0005 N, East/Oos kQBQN FBN = r2 (9 x109 )(8 x10 9 )(2 x10 9 ) = (0,049)2 = 0,00006 N, West/Wes, Fnet = FAN + FBN = 0,0005 – 0,00006 = 0,00044 N, East/Right / Oos/regs (5) [7] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ## Related documents - [PHYSICAL SCIENCES P1 GR12 QP JUNE2022 Afrikaans](https://www.hlayiso.com/library/physical-sciences-p1-gr12-qp-june2022-afrikaans/7da2a467-33ba-4d42-9de8-a501165ff1e0) — Physical Sciences · QP · 2022 · 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 - [NW NSC GR 12 PS P1 memo 21](https://www.hlayiso.com/library/nw-nsc-gr-12-ps-p1-memo-21/edaa5a71-c3e8-43a6-b32a-09bfea08a19f) — Physical Sciences · Memo · 2021 - [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) — Physical Sciences · QP · 2025 · Afr - [Physical Sciences P1 May June 2021 Eng](https://www.hlayiso.com/library/physical-sciences-p1-may-june-2021-eng/f4b97e24-79cc-40f8-86c0-9185416a66a1) — Physical Sciences · QP · 2021 · Eng - [Gr 12 Physical Sciences P2 (English and Afrikaans) June 2024 Possible Answers](https://www.hlayiso.com/library/gr-12-physical-sciences-p2-english-and-afrikaans-june-2024-possible-answers/7e563c7f-5d30-46c2-a50f-2942505df8e3) — Physical Sciences · Memo · 2024 · Afr ## More from Grade 12 Physical Sciences - [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 - [Physical Sciences Physics Grade 12 NSC QP P1 May June 2025 Mpumalanga](https://www.hlayiso.com/library/physical-sciences-physics-grade-12-nsc-qp-p1-may-june-2025-mpumalanga/d9c5f1c3-bd6c-468e-a50a-f6b40782c695) — 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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