Physical Sciences Grade 12 P1 Eastern Cape Prelim 2024 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GREYIDI 12 SEPTEMBA 2024 IFIZIKHALI SAYENSIZ P1 ISIKHOKHELO SOKUMAKISHA AMANQAKU: 150 Esisikhokhelo sokumakisha sinamaphepha ayi18. Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 2 IFIZIKHALI SAYENSIZ EC/SEPTEMBA 2022) 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 verkeerd 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 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 formules ingestel word en nie vir waardes wat voor ʼn berekening gelys is nie. Ushicilelo lugciniwe tyhila Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com EC/SEPTEMBA 2024) IFIZIKHALI SAYENSIZ 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. Ushicilelo lugciniwe tyhila Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 4 IFIZIKHALI SAYENSIZ EC/SEPTEMBA 2022) 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, not 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 QUESTION 3.1 is incorrect, no marks can be awarded for QUESTION 3.2. However, if the answer for e.g. QUESTION 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 VRAAG 3.1 is verkeerd, kan geen punte vir VRAAG 3.2 toegeken word nie. Indien die antwoord op bv. VRAAG 3.1 egter op ʼn berekening gebaseer is, kan die motivering vir die verkeerde antwoord in VRAAG 3.2 oorweeg word. Ushicilelo lugciniwe tyhila Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com EC/SEPTEMBA 2024) IFIZIKHALI SAYENSIZ 5 UMBUZO 1: IMIBUZO YEMALTIPLI TSHOYISI 1.1 B  (2) 1.2 C  (2) 1.3 C  (2) 1.4 A  (2) 1.5 B  (2) 1.6 A  (2) 1.7 D  (2) 1.8 D  (2) 1.9 B  (2) 1.10 A  (2) [20] Ushicilelo lugciniwe tyhila Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 6 IFIZIKHALI SAYENSIZ EC/SEPTEMBA 2022) UMBUZO 2 2.1 Xa irizalthenti/inethi fosi isebenza kwiobhjekhthi, iobhjekhthi izakuakhselereyithela ngakwidayirekhshini yefosi. Iakhselereyishini idayirekhthli propowushinali kwifosi ze ibe-invesili propowushinali kwimesi yeobjekhthi.  OKANYE Irizalthenti/inethi fosi esebenza kwiobhjekhthi ilingana nereyithi yetsheyinji yemomentam yeobhjekhthi kwidayirekhshini yerizalthenti/nethi fosi, kwidayirekhshini yerizalthenti/nethi fosi  (2) 2.2 Ezinye iileyibheli w Fg/Fw/weyithi/gravitheyishinali fosi  T FT/Thenshini/ ifosi ekwisitringi  N FN/N/Nomali fosi  Fk fk/Frikhshinali fosi  (4) 2.3.1 fk = μkN  fk = μkFg fk = (0,25)(4)(9,8)  fk = 9,8 N  (3) Ushicilelo lugciniwe tyhila Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com EC/SEPTEMBA 2024) IFIZIKHALI SAYENSIZ 7 2.3.2 INDLELA 1 Fnet = ma Fnet = T – f Nayiphi na ✓ Fnet = Fg – T Ibhlokhi u4 kg T – 9,8 = 4a T = 4a + 9,8 Nayiphi na (4a okanye 2a) ✓ ibhlokhi u2 kg (2)(9,8) – T✓ =2a T = 19,6 – 2a 19,6 – 2a = 4a + 9,8 a = 1,63 m∙s-2 vf2 = vi2 + 2aΔy  vf2 = 02 + 2(1,63)(1)  vf = 1,81 m∙s-1  (7) 2.4 Igravitheyishinali fosi (weyithi) akuyiyo yodwa ifosi esebenza kwi bhlokhi u2 kg  OKANYE Iakhselereyishini ayingo 9,8 m.s-2.  (2) [18] Ushicilelo lugciniwe tyhila Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 8 IFIZIKHALI SAYENSIZ EC/SEPTEMBA 2022) UMBUZO 3 3.1 Ukuya ezantsi  (1) 3.2 Ukuya phezulu phozithivu Ukuya ezantsi phozithivu v f = vi + a t  v f = vi + a t  = 30  + (-9,8)(2,135) = - 30 + (9,8)(2,135) = 9,08 m·s-1, ukuya phezulu  = - 9,078 m·s-1 = 9,08 m·s-1, ukuya phezulu  (3) 3.3 INDLELA 1 Ukuya phezulu phozithivu Ukuya ezantsi phozithivu Ibhola uA: Δy = v i Δt + 21 at 2  Δy = v i Δt + 21 at 2  ∆y = 12(2,5) + ½(9,8)(2,5)2  ∆y = -12(2,5) + ½(-9,8)(2,5)2  ∆y = 60,625 m ∆y = - 60,625 m (ihayithi = 29,375 m) (hayithi = 29,375 m) Ibhola uB: Ibhola uB: Δy = v i Δt + 21 at 2 Δy = v i Δt + 21 at 2 = -30(2,5) + ½(9,8)(2,5)2  .. = 30(2,5) + ½(-9,8)(2,5)2  = - 44,375 m = 44,375 m Idistensi = 44,375 – 29,375  Idistensi = 44,375 - 29,375  = 15 m  = 15 m  INDLELA 2 Ibhola uA v f = vi + a t vf = -12 + (-9,8)(2,135) vf = 32,923 m·s-1, ukuya ezantsi Δy = v i Δt + 21 at 2  ∆y = -32(2,5-2,135) + ½(-9,8)(2,5-2,135)2  ∆y = - 12,6696 m Ibhola uB Δy = v i Δt + 21 at 2 ∆y = 9,08(2,5 – 2,135) + ½(-9,8)(2,5 – 2,135)2 ∆y = 2,66 m idistensi = 2,66 + 12,6696  15,33 m  Ushicilelo lugciniwe tyhila Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com EC/SEPTEMBA 2024) IFIZIKHALI SAYENSIZ 9 3.4 INDLELA 1 Ukuya phezulu phozithivu INDLELA 2 Ukuya phezulu negethivu Indlela yegrafu Amanqaku Isheyiphu ngokwebhola uA de iyokufika  kuziro pozishini Isheyiphu ngokwebhola uB de iyokufika kwi-intasekhshini yelayini yamaxesha.  2,135 s Ukubonisa ixesha u2,135 s  Igrawundi engengo ziro pozishin (xa yonke into ichanekile): 2/3 (3) [12] Ushicilelo lugciniwe tyhila Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 10 IFIZIKHALI SAYENSIZ EC/SEPTEMBA 2022) UMBUZO 4 4.1 Itotali liniya momentam ye-ayisoleyithedi sistim ihlala injalo/igciniwe . (2) 4.2 Ngasekunene nguphozithivu Σpi = Σpf Any one / Enige een ✓ (mvi)1 + (mvi)2 = (mvf)1 + (mvf)2 (5 000)(15) + (2 000)(-20 ✓ = (5 000)vf + (2 000)(5) ✓ vi = 5 m·s-1✓ Ngasekunxele nguphozithivu Σpi = Σpf Any one / Enige een ✓ (mvi)1 + (mvi)2 = (mvf)1 + (mvf)2 (5 000)(-15) + (2 000)(20) ✓ = (5000)vf + ( 2000)(-5) ✓ vi = -5 m·s-1 Imagnityhudi yevelosithi = 5 m∙s-1 ✓ (4) 4.3 INDLELA 1 Fnethi∆t = ∆p Nayiphi na ✓ Fnethi Δt = mvf – mvi Fnethi (0,4) ✓ = (5 000)(5) – (5 000)(15) ✓ Fnethi = -125 000 N Fnethi = 125 000 N Ukuya ngasekunene ✓ INDLELA 2 Fnet∆t = ∆p Ukuya ngasekunene ✓ Fnet Δt = mvf – mvi Fnet (0,4) ✓ = (2000)(-5) – (2000)(20) ✓ Fnet = -125 000 N Fnet = 125 000 N Ukuya ngasekunene ✓ (4) [10] Ushicilelo lugciniwe tyhila Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com EC/SEPTEMBA 2024) IFIZIKHALI SAYENSIZ 11 UMBUZO 5 5.1 Ukuya ngasemva/ngasemva kwakhe  (1) 5.2 Nyuthonz thed lo of mowushini Xa ibhodi yokuqala isebenzisa ifosi kwibhodi yesibini, ibhodi yesibini isebenzisa ifosi enemagnithyudi elinganayo  kwidayirekhshini eophozithi kwibhodi yokuqala . (3) 5.3 INDLELA 1 Wnethi = ∆K Nayiphi na ✓ Wg + Wf = ∆K Fg∆xcosӨ + f∆xcosӨ = ∆K (57)(9,8)(4)cos180°  + 40∆xcos180o = 0 – ½(57)(62)  ∆x = - 30,21 m 4 Sin Ө = 30,21 Ө = 7,61°  INDLELA 2 Wnc = ΔU + ΔK / Wnc = ΔEp + ΔEk  40∆xcos180o  = (57)(9,8)(4) – (57)(9,8)(0)  + ½ (57)(0)2 – ½ (57)(6)2  ∆x = -30,21 m 4 Sin Ө = 30,21 Ө = 7,61°  (5) 5.4 INDLELA 1 Wnet = ∆K Nayiphi na ✓ WT + Wg + Wf = ∆K (80)(5)(4)cos0° + (4)(9,8)sin30° •(5)cos180° + (15)(5)cos180°  = ½(4)vf2 – ½(4)(32)  vf = 11,07 m·s-1  INDLELA 2 Wnc = ΔU + ΔK Nayiphi na ✓ WT + Wf = ΔU + ΔK (80)(5)(4)cos0° + (15)(5)cos180°  = (4)(9,8)(sin30°)(5) – (4)(9,8)(0)  + ½ (4)vf2 – ½ (4)(3)2  vf = 11,07 m·s-1  (6) [15] Ushicilelo lugciniwe tyhila Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 12 IFIZIKHALI SAYENSIZ EC/SEPTEMBA 2022) UMBUZO 6 6.1 U (apharenti) kutshintsha kwefrikhwensi/pitshi  ye (sawundi) weyivu eviwa yilisina xa kukho irelethivu mowushini phakathi kwelisina nesosi eyenza iisawundi weyivu  (2) 6.2.1 IYANDA  (1) 6.2.2 IYANCIPHA  (1) 6.3 v  vL v+v fL = f s  or fL = v L 𝑓𝑆  v  vs 330 + 30 fL=  x 1 800  330 − 0 fL = 1 963,6 Hz  (5) 6.4 Indlela Yokumakisha Iileyibheli zee-ekhziz  Ifrikhwensi ephezulu ekuqaleni  Ifrikhwensi esezantsi ngokuya  linyuka/lisanda ixesha (3) 6.5 Bhladiflowu mitha  / Dopla flow mitha  (1) [13] Ushicilelo lugciniwe tyhila Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com EC/SEPTEMBA 2024) IFIZIKHALI SAYENSIZ 13 UMBUZO 7 7.1 Ifosi ye-athrakhshini okanye iriphalzhini phakathi kweetshaji ezimbini idayirekhthli propowushinail kwiphrodakhthi yemagnityhudi yeetshaji ze ibe invesli prpopowushinali kwiskwe sedistensi phakathi kwazo (2) 7.2 Kuba uFK on M = noFG on M  (Fnet)2 = (FK on M)2 + (FG on M)2 (2,864 x 10-6)2  = F2 + F2 F = 2,025 x 10-6 N  (4) 7.3 F = kQ1Q2  r2 2,025 x 10-6  = 9 x 109 x 6 x 10-9 x 6 x 10-9  (X)2 X = 0,4 m  (4) [10] Ushicilelo lugciniwe tyhila Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 14 IFIZIKHALI SAYENSIZ EC/SEPTEMBA 2022) UMBUZO 8 8.1 (3) Indlela yokumakisha Isheyiphu echanekileyo  Idayirekhshini echanekileyo  Iilayini ezifika/ezibamba kwitshaji kwaye  zinganqumlani/khrosani 8.2 Yi-elekhthrostatikhi fosi esetyenziswayo kwiphozothivu tshaji nganye ebekwe kuloopoyinti . (2) 8.3 EP = kQ1  r2 EP = 9 x 109 x 200 x 10-9  (0,2)2 EP = 45 000 N.C-1 Ukuya ngasekunene EQ = kQ2 r2 EQ = 9 x 109 x 200 x 10-9  (0,4)2 EQ = 11 250 N.C-1 ukuya ngasenxele Enethi = 45 000 + (-11 250)  Enethi = 33 750 N.C-1 Ukuya ngasekunene  (5) [10] Ushicilelo lugciniwe tyhila Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com EC/SEPTEMBA 2024) IFIZIKHALI SAYENSIZ 15 UMBUZO 9 9.1 Ukuqinisekisa ukuqhubeleka kokujikeleza kwekhoyili  (1) 9.2 (i) Tshintsha isosi yephothenshiyali diferensi ngelowudi (ii) Tshintsha ispliti ringi (khomyutheyitha) ngee (ezimbini) slip ringi (2) 9.3 I I rms = max  2 0,54 Irms =  √2 Irms = 0,38 A Pave = VrmsIrms  60 = Vrms x 0,38  Vrms = 157,89 V  (5) [8] Ushicilelo lugciniwe tyhila Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 16 IFIZIKHALI SAYENSIZ EC/SEPTEMBA 2022) UMBUZO 10 10.1 Iphothenshiyali diferensi idayirekhthli propowushinali kwikharenti xa itempiritsha ihlala injalo/ingaguqukiyo. OKANYE Ireyshiyo yephothenshiyali diferensi kwikharenti  Xa itempiritsha ihlala injalo ayiguquki/ihlala injalo  (2) 10.2 INDLELA 1 INDLELA 2 V2 P = VI  P =  IR = P/V = 13,5/18  = 0,75 A R (18)2 V = IR 13,5 =  R = V/IR = 18/0,75 = 24 Ω  R R = 24 Ω  (3) 10.3 INDLELA 1 INDLELA 2 V 1 1 1 R =  = +  I Rp 𝑅1 R2 1 1 1  = +  24 = (18) RP 12 24 RP = 8 Ω  I V = IRP 18 = I (8)  I = 0,75 A I = 2,25 A  Vp R= I12 18 12 = I12 I12 = 1,5 A  ITotali = 1,5 + 0,75 = 2,25 A  (5) 10.4 I-inthenali reszistensi yi-ophozishini kwiflowu yetshaji ngaphakathi kwibhethri / (2) 10.5 V10 = IR10  V10 = 2,25 x 10  V10 = 22,5 V  (3) Ushicilelo lugciniwe tyhila Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com EC/SEPTEMBA 2024) IFIZIKHALI SAYENSIZ 17 10.6 INDLELA 1 V1 = Vext V1 = VP + V10 V1 = 18 + 22,5  V1 = 40,5 V Xa iswitshi ivuliwe iemfu = 45,9 V I‘losti volthi’ ingu: Vlosti = emf – Vext = 45,9 – 40,5  = 5,4 V V r= L  I 5,4 r=  2,25 r = 2,4 Ω  INDLELA 2 1 1 1 Emfu= I (R + r)  = + Rp r1 r2 1 1 1 45,9 = (2,25) (8 + 10 + r)  = +  RP 12 24 r = 2,4 Ω  RP = 8 Ω (5) 10.7 Iyanda (1) [21] Ushicilelo lugciniwe tyhila Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 18 IFIZIKHALI SAYENSIZ EC/SEPTEMBA 2022) UMBUZO 11 11.1 c = f x λ Yamkela v = f λ 3 x 108 = f x 229 x 10-9 f = 1,31 x 1015 Hz  (3) 11.2 Ithresholdi frikhwensi  (1) 11.3 Iwekhi fankhshini ngowona mlinganiselo umncinci we-eneji efunekayo ukijekhtha ielekhthroni ukusuka kwisafeeyisi yesolidi enikiweyo, kudla ngokuba yimethali.  (2) 11.4 E = Wo + Ek ( makhz )  hf = Wo + ½ mv2 6,63 x 10-34 x f  = (6,63 x 10-34 x 1,31 x 1015) + ½ (9,11 x 10-31) (1,57 x 106)2  f = 3 x 1015 Hz  (4) 11.5 IYANDA Ukuncipha kweweyivulenthi kwenza ukuba ifrikhwensi yande.  Iwekhi fankhshini iyanda.  Okanye 1 Wo ∝ λ  (3) o [13] AMANQAKU XA EWONKE: 150 Ushicilelo lugciniwe tyhila Hosted by www.ecexams.co.za.