NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12

Size: px

Start display at page:

Download "NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12"

Felix Pearson
6 years ago
Views:

1 NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 1 PHYSICAL SCIENCES: PHYSICS (P1) FISIESE WETENSKAPPE: FISIKA (V1) NOVEMBER 016 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 19 pages. Hierdie memorandum bestaan uit 19 bladsye.

2 Physical Sciences P1/Fisiese Wetenskappe V1 DBE/November 016 QUESTION 1/VRAAG A () 1. C () 1.3 C () 1.4 D () 1.5 B () 1.6 A () 1.7 C () 1.8 A () 1.9 B () 1.10 B () [0] QUESTION /VRAAG.1 When a resultant/net force acts on an object, the object will accelerate in the (direction of the net/resultant force). The acceleration is directly proportional to the net force and inversely proportional to the mass of the object. Wanneer 'n netto krag op 'n voorwerp inwerk, versnel die voorwerp in die rigting van die netto krag teen 'n versnelling direk eweredig aan die krag en omgekeerd eweredig aan die massa van die voorwerp. The resultant/net force acting on the object is equal (is directly proportional to) to the rate of change of momentum of an object (in the direction of the force). Die resulterende/netto krag wat op 'n voorwerp inwerk, is gelyk aan (is direk eweredig aan) die tempo van verandering van momentum van die voorwerp (in die rigting van die netto krag). (). f k = μ k N= μ k mg = (0,15)(3)(9,8) = 4,41 N (3).3 N 5 N N T f k f k T 5 N w w

3 Physical Sciences P1/Fisiese Wetenskappe V1 3 DBE/November 016 w N T f k 5 N Accepted Labels/Aanvaarde benoemings F g/ F w /force of Earth on block/weight/14,7 N/mg/gravitational force F g/ F w /krag van Aarde op blok/gewig/14,7 N/mg/gravitasiekrag F N /F normal /normal force F N /F normaal /normalekrag Tension/F T Spanning/F T f kinetic friction/kinetiesewrywing /f f/w /f//f f/w kinetic friction/kinetiesewrywing F applied /F A /F F toegepas /F A /F.4.1 OPTION 1/OPSIE 1 f k = μ k N = μ k (5sin 30 o + mg) = 0,15[(5sin30 o ) + (1,5)(9,8)] = 4,08 N OPTION /OPSIE f k = μ k N = μ k (5cos 60 o + mg) = 0,15[(5cos60 o ) + (1,5)(9,8)] = 4,08 N (3).4. POSITIVE MARKING FROM QUESTION. AND QUESTION.4.1 POSITIEWE NASIEN VANAF VRAAG. EN VRAAG.4.1 OPTION 1/OPSIE 1 For the 1,5 kg block/vir die 1,5 kg blok F net = ma F x + (-T) +(- f k ) = ma 5 cos 30 o T f k = 1,5a (5 cos 30 o T) 4,08 = 1,5a 17,571 T = 1,5a.(1) either one enigeen For the 3 kg block Vir die 3 kg blok T f k = 3a T 4,41 = 3a.() 13,161 = 4,5 a a =,95 m s - T = 13,19 N (13,17 N 13,19 N)

4 Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/November 016 OPTION /OPSIE For the 1,5 kg block/vir die 1,5 kg blok F net = ma F x + (-T) +(- f k ) = ma 5 cos 30 o T f k = 1,5a (5 cos 30 o T) 4,08 = 1,5a 17,571 T = 1,5a.(1) For the 3 kg block Vir die 3 kg blok T f k = 3a T 4,41 = 3a.() 35,14 T = T 4,41 T = 13,18 N either one enigeen OPTION 3/OPSIE 3 For the 1,5 kg block/vir die 1,5 kg blok F net = ma F x + (-T) +(- f k ) = ma 5 cos 30 o T f k = 1,5a (5 cos 30 o T) 4,08 = 1,5a 17,571 T = 1,5a.(1) 17,571 T a = 1,5 For the 3 kg block Vir die 3 kg blok T f k = 3a T 4,41 = 3a.() either one enigeen T 4,41 a = 3 17,571 T T 4,41 = 1,5 3 T = 13,18 N (5) [18] QUESTION 3/VRAAG The motion of an object under the influence of gravity/weight/gravitational force only / Motion in which the only force acting is the gravitational force. Die beweging van 'n voorwerp slegs onder die invloed van swaartekrag/gewig gravitasiekrag. Beweging waarin die enigste krag wat op die liggaam inwerk, die gravitasiekrag is. () 3..1 OPTION 1/OPSIE 1 Upwards positive/opwaarts positief: v f = v i + aδy = 0 + ()(-9,8)(-0) v f = 19,80 m s -1 Downwards positive Afwaarts positief v f = v i + aδy = 0 + ()(9,8)(0) v f = 19,80 m s -1 OPTION /OPSIE Upwards positive/opwaarts positief: Δy = v i Δt + ½ aδt -0 = 0 + ½ (-9,8) Δt Δt =,0 s v f = v i + aδt = 0 + (-9,8)(,0) = -19,80 m s -1 = 19,80 m s -1 Downwards positive Afwaarts positief Δy = v i Δt + ½ aδt 0 = 0 + ½ (9,8) Δt Δt =,0 s v f = v i + aδt = 0 + (9,8)(,0) = 19,80 m s -1 either one enigeen either one enigeen

5 Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/November 016 OPTION 3/OPSIE 3 (E mech ) Top/Bo = (E mech ) Ground/Grond (E P +E K ) Top = ( E P +E K ) Bottom/Onder 1 mark for any (mgh + ½ mv ) Top/Bo = (mgh + ½ mv ) 1 punt vir enige Bottom/Onder (9,8)(0) + 0 = (0 + ½v f ) v f = 19,80 m s -1 OPTION 4/OPSIE 4 W nc = ΔE p + ΔE k 0 = mgδh + ½ mδv 0 = m(9,8)(0 0) + ½ m(v f 0) v f = 19,80 m s -1 OPTION 5/OPSIE 5 W net = ΔE k mgδxcos0 o = ½ m(v f 0) m(9,8)(0)(1) = ½ mv f v f = 19,80 m s -1 (4) 3.. POSITIVE MARKING FROM QUESTION 3..1/POSITIEWE NASIEN VANAF VRAAG 3..1 OPTION 1/OPSIE 1 Downwards positive/afwaarts positief v f = v i + aδt 19,80 = 0 + (9,8)Δt Δt =,0 s Upwards positive/opwaarts positief v f = v i + aδt -19,80 = 0 + (-9,8)Δt Δt =,0 s OPTION /OPSIE Upwards positive/opwaarts positief: Δy = v i Δt + ½ aδt -0 = 0 + ½ (-9,8) Δt Δt =,0 s Downwards Positive/Afwaarts positief Δy = v i Δt + ½ aδt 0 = 0 + ½ (9,8) Δt Δt =,0 s OPTION 3/OPSIE 3 Downwards positive/afwaarts positief: v i + v f Δy = Δt ,80 0 = ( t) Δt =,0 s Upwards positive/opwaarts positief: v i + v f Δy = Δt 0 19,80-0 = ( t) Δt =,0 s (3)

6 Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/November Downward positive/afwaarts positief velocity/snelheid (m s -1 ) 0 Time/Tyd (s) Upward positive/opwaarts positief Time/Tyd (s) 0 velocity/snelheid (m s -1 ) Notes/Aantekeninge Straight line through the origin. Reguitlyn deur die oorsprong () [11]

7 Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/November 016 QUESTION 4/VRAAG A system on which the resultant/net external force is zero/'n Sisteem waarop die resulterende krag/netto eksternekrag nul is A system which excludes external forces /'n Sisteem wat eksterne kragte uitlaat OPTION 1/OPSIE 1 p = mv = (1 500)v v = 0 m s -1 OPTION /OPSIE Δp = mv f mv i 0 = (1 500)v f v = 0 m s POSITIVE MARKING FROM QUESTION 4..1/POSITIEWE NASIEN VANAF VRAAG 4..1 OPTION 1/OPSIE 1 p i = p f m 1 v 1i + m v i = m 1 v 1f + m v f 1 mark for any/1 punt vir enige (900)(-15) = v B v B =,78 m s -1 east/oos (Accept/Aanvaar: to the right/na regs) OPTION /OPSIE Δp A = -Δp B p f p i = -(mv f - mv i ) 1 mark for any/1 punt vir enige = 900v f 900(-15) v f =,78 m s -1 east/oos (Accept/Aanvaar: to the right/na regs) (1) (3) (5) 4..3 OPTION 1/OPSIE 1 Δ p Slope/Helling = = Fnet Δt = ( ) ( 0, - 0,1) = F net = N OPTION /OPSIE F net Δt = Δp F net (0,1) = F net = N F net = N POSITIVE MARKING FROM QUESTION 4../POSITIEWE NASIEN VANAF VRAAG 4.. OPTION 3/OPSIE 3 F net Δt = Δp F net (0,1) = 900[(,78) (-15)] F net = N F A = - F B F net = N

8 Physical Sciences P1/Fisiese Wetenskappe V1 8 DBE/November 016 OPTION 4/OPSIE 4 p = mv = 1 500v f v f = 9,33 m s -1 m(v f v i ) F net = = t = = N 1500(9,33 0) 0,1 v f = v i + aδt 9,33 = 0 + a(0,1) a = m s - F net = ma = 1 500(-106,7) F net = N F net = N (4) [13] QUESTION 5/VRAAG E k /K = ½ mv = ½ ()(4,95) = 4,50 J (3) 5.1. POSITIVE MARKING FROM QUESTION 5.1.1/POSITIEWE NASIEN VANAF OPTION 1/OPSIE 1 E mech before = E mech after [(E mech ) bob + (E mech)block ] before/voor = [(E mech ) Block + (E mech ) bob ] after/na (mgh + ½ mv ) before/voor = (mgh + ½ mv ) after/na (5)(9,8)h = 5(9,8)¼h ,50 h = 0,67 m OPTION /OPSIE W nc = ΔE p + ΔE k Any one/ 0 = ΔE p + ΔE k Enige een -ΔE p = ΔE k -[(5)(9,8)(¼h) (5)(9,8)h] = 4,50 h = 0,67 m OPTION 4 /OPSIE 4 Before/Voor (mgh + ½ mv ) top/bo = (mgh + ½ mv ) bottom/onder (5)(9,8)h + 0 = (5)(9,8)h o + ½ (5)v v i = 19,6h - 19,6h o After/Na (mgh + ½ mv ) bottom/onder = (mgh + ½ mv ) top/bo (5)(9,8)h o + ½(5)v f = (5)(9,8)(¼h) + 0 v f = 4,9h 19,6h o OPTION 3/OPSIE 3 Loss Ep bob = Gain in Ek of block mg(¾h) = 4,5 (5)(9,8)(¾h) = 4,5 h = 0,67 m E mech/meg before collision/voor botsing = E mech/meg after collision/na botsing ½ mv i (bob/skietlood) + 0 = ½ mv f (bob/skietlood)+ ½ mv (block/blok) ½ (5)(19,6h 19,6h o ) = ½ (5)(4,9h -19,6h o ) + 4,5 h = 0,67 m Any one/ Enige een (4)

9 Physical Sciences P1/Fisiese Wetenskappe V1 9 DBE/November The net/total work done on an object is equal to the change in the object's kinetic energy Die netto/totale arbeid op 'n voorwerp verrig is gelyk aan die verandering in die kinetiese energie van die voorwerp. The work done on an object by a resultant/net force is equal to the change in the object's kinetic energy. Die arbeid verrig op 'n voorwerp deur 'n resulterende/netto krag is gelyk aan die voorwerp se verandering in kinetiese energie. () 5.3 OPTION 1/OPSIE 1 W net = ΔE K W f + mgδycosθ = ½ m(v - v ) W f +()(9,8)(0,5)cos180 o = ½ ()( 4,95 ) W f = - 10,7 J OPTION /OPSIE W nc = ΔE K + ΔU W nc = ΔE K + ΔE P f i W f = ½ ()( 4,95 ) + ()(9,8)(0,5-0) = - 10,7 J (4) [13] QUESTION 6/VRAAG It is the (apparent) change in frequency (or pitch) 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 (of toonhoogte) van die klank (waargeneem deur 'n luisteraar) omdat die klankbron en die luisteraar verskillende snelhede relatief tot die medium van klankvoortplanting het. An (apparent) change in (observed/detected) frequency (pitch), (wavelength) as a result of the relative motion between a source and an observer (listener). 'n Skynbare verandering in (waargenome) frekwensie (toonhoogte),(golflengte) as gevolg van die relatiewe beweging tussen die bron en 'n waarnemer/luisteraar. () 6.1. v = fλ 340 = f(0,8) f s = 1 14,9 Hz (3)

10 Physical Sciences P1/Fisiese Wetenskappe V1 10 DBE/November POSITIVE MARKING FROM QUESTION 6.1./POSITIEWE NASIEN VANAF VRAAG 6.1. v ± v L v ± vl v v fs fl = fs fl = fl = fs fl = v ± v s v ± vs λs v v v s s 1 v ,9 f L = ( )114,9 f ( ) L = f L = , = 1 331,80 Hz (1 331,80 Hz 1 335,7 Hz) (5) Decreases/Verlaag (1) 6. The spectral lines of the star are/should be shifted towards the lower frequency end, which is the red end (red shift) of the spectrum. Die spektraallyne van die van die ster is verskuif na die laer frekwensie ent, wat die rooi ent van die spektrum is. () [13] QUESTION 7/VRAAG The (magnitude of the) electrostatic force exerted by one (point) charge on another is directly proportional to the product of the charges and inversely proportional to the square of the distance between their (centres) them. Die (grootte) van die elektrostatiese krag wat een (punt) lading op 'n ander uitoefen, is direk eweredig aan die produk van die ladings en omgekeerd eweredig aan die kwadraat van die afstand tussen hul middelpunte. () 7.1. F E /Electrostatic force/elektrostatiese krag (1) The electrostatic force is inversely proportional to the square of the distance between the charges Die elektrostatiese krag is omgekeerd eweredig aan die kwadraat van die afstand tussen die ladings The electrostatic force is directly proportional to the inverse of the square of the distance between the charged spheres (charges). Die elektrostatiese krag is direk eweredig aan omgekeerde van die kwadraat van die afstand tussen die gelaaide sfere (ladings). 1 F α r They are inversely proportional to each other /Hulle is omgekeerd eweredig aan mekaar (1)

11 Physical Sciences P1/Fisiese Wetenskappe V1 11 DBE/November OPTION 1/OPSIE 1 1 mark for using slope/ FE (0,07-0) Slope/Helling = = 1 punt vir die gebruik van helling 1 (5,6-0) r = 4,8 x 10-3 N m (4,76 x x 10-3 ) 7..1 Slope/Helling = F E r = kq 1 Q = kq 4,8 x 10-3 = 9 x 10 9 Q Q = 7,3 x 10-7 C OPTION /OPSIE Accept any pair of points on the line/aanvaar enige paar punte op die lyn kq1q F = r 9 (9 10 )Q ( ) = ( ) Q = 7,3 x 10-7 C (7,3 x ,45 x 10-7 C) Examples/Voorbeelde 9 (9 10 )Q (0,005) = ( 1 ) Q = 7,45 x 10-7 C 9 (9 10 )Q (0,07) = 1 ( ) 5,6 Q = 7,3 x 10-7 C (6) Criteria for drawing electric field: Kriteria vir teken van elektriese veld: Direction /Rigting Field lines radially inward/veldlyne radiaal inwaarts Marks/Punte

12 Physical Sciences P1/Fisiese Wetenskappe V1 1 DBE/November kq E = r Take right as positive/neem regs as positief 9-6 ( 9 10 ) E PA ( 0,75 10 ) = ( 0,09 ) = 8,33 x 10 5 N C -1 to the left/na links 9-6 ( 9 10 ) E PB ( 0,8 10 ) = 0,03 ( ) = 8 x 10 6 N C -1 to the left/na links E net = E PA + E PC = [-8,33 x (- 8 x 10 6 )] = -8,83 x 10 6 = 8,83 x 10 6 N C -1 Take left as positive/neem links as positief 9-6 ( 9 10 ) E PA ( 0,75 10 ) = ( 0,09 ) 1 mark for the addition of same signs/ 1 punt vir optelling van dieselfde tekens = 8,33 x 10 5 N C -1 to the left/na links 9-6 ( 9 10 ) E PB ( 0,8 10 ) = ( 0,03) = 8 x 10 6 N C -1 to the left/na links E net = E PA + E PC = (8,33 x x mark for the addition of same signs/ ) = 8,83 x 10 6 N C -1 1 punt vir optelling van dieselfde tekens (5) [17]

13 Physical Sciences P1/Fisiese Wetenskappe V1 13 DBE/November 016 QUESTION 8/VRAAG (Maximum) energy provided (work done) by a battery per coulomb/unit charge passing through it / Energie verskaf (arbeid verrig) deur 'n battery per coulomb/eenheid lading wat daardeur vloei. () (V) (1) (V) / Zero/nul (1) ε = I(R + r) ε = V ext + V int 1 = 11,7 +Ir 0,3 = I tot (0,) I tot = 1,5 A V = IR (Accept/Aanvaar: V lost = Ir) 0,3 = I tot (0,) I tot = 1,5 A (3) OPTION 1/OPSIE = + R// R1 R Any one = + Enigeen R R = 6 Ω OPTION /OPSIE R1R R = R1 + R Any one (10)(15) Enigeen = = 6 Ω () POSITIVE MARKING FROM QUESTIONS AND 8.1.5/POSITIEWE NASIEN VANAF VRAE EN OPTION 1/OPSIE 1 V = IR 11,7 = 1,5(6 + R) R = 1,8 Ω V = IR 11,7 = 1,5R R = 7,8 Ω R R = 7,8 6 = 1,8 Ω

14 Physical Sciences P1/Fisiese Wetenskappe V1 14 DBE/November 016 OPTION /OPSIE ε = I(R + r) 1 = 1,5(R + 0,) R = 7,8 Ω R R = 7,8 6 = 1,8 Ω OPTION 3/OPSIE 3 V = IR = (6)(1,5) = 9 V V R = IR (11,7-9) = (1,5)R R = 1,8 Ω (4) 8..1 P ave/gemid = Fv ave/gemid = mg(v ave/gemid ) = (0,35)(9,8)(0,4) = 1,37 W W P = nc t = Ek + Ep 0 + (0,35)(9,8)(0,4 = t 1 0) = 1,37 W W P = t = E p t = (0,35)(9,8)(0,4) = 1,37 W 1 (3) 8.. POSITIVE MARKING FROM QUESTION 8..1/POSITIEWE NASIEN VANAF VRAAG 8..1 OPTION 1/OPSIE 1 OPTION /OPSIE P = VI V 1,37 = (3)I P = R I = 0,46 A ε = V ext + V int = V T + V X + V int 1 = V T (0,)(0,46) V T = 8,91 V V T = IR T 8,91 = (0,46)R T R T = 19,37 Ω Any one Enigeen 1,37 = R 3 R = 6,57 Ω P = VI 1,37 = (3)I I = 0,46 A Any one Enigeen ε = I(R + r) 1 = 0,46(6,57 + R T + 0,) R T = 19,38 Ω

15 Physical Sciences P1/Fisiese Wetenskappe V1 15 DBE/November 016 OPTION 3/OPSIE 3 P = VI 1,37 = (3)I I = 0,46 A P tot = P r + P motor + P T (1)(0,46) = (0,46) (0,) + 1,37 + (0,46) R T R T = 19,41 Ω P = VI 1,37 = (3)I I = 0,46 A P tot = P r + P motor + P T (1)(0,46) = (0,46) (0,) + 1,37 + P T P T = 4,07 W P=I R 4,07 = (0,46) R T R T = 19,49 Ω OPTION 4/OPSIE 4 P = VI 1,37 = (3)I I = 0,46 A Any one Enigeen ε = I(R + r) 1 = (0,46)(R + 0,) R = 5,87 Ω V = IR 3 = (0,46)R R = 6,5 Ω R T = 5,87 6,5 = 19,35 Ω P = I R 1,37 = (0,46) R R = 6,47 Ω R T = 5,87 6,47 = 19,4 Ω V Pmotor = R 3 1,37 = R R = 6,56 Ω R T = 5,87 6,56 = 19,31 Ω (5) [1]

16 Physical Sciences P1/Fisiese Wetenskappe V1 16 DBE/November 016 QUESTION 9/VRAAG DC/GS-generator Uses split ring/commutator/gebruik spleetring/kommutator () 9.1. Induced emf (V) Geïnduseerde emk (V) time/tyd (s) Induced emf (V) Geïnduseerde emk (V) time/tyd (s) () 9..1 OPTION 1/OPSIE 1 Vmax/maks Vrms / wgk = Pave/gem = V rms/wgk Irms/wgk = ( I rms / wgk ) I rms/wgk = 3,33 A Vmax/ maks Vrms / wgk = = P = I ave/ gem V rms/wgk rms/wgk 800 = I rms/wgk (40,416) I rms/wgk = 3,33 A 340 = 40,416

17 Physical Sciences P1/Fisiese Wetenskappe V1 17 DBE/November 016 OPTION /OPSIE V rms / wgk P = ave gem R 800 = ( ) 340 ( ) ( R) R = 7,5 Ω ( ) V max/ maks / = ( )( R) P ave/gem = I rms/wgk R 800 = I rms/wgk (7,5) I rms/wgk = 3,33 A V rms/wgk = I rms/wgk R 40,416 I rms/wgk = 7, 5 = 3,33 A (3) 9.. POSITIVE MARKING FROM QUESTION 9..1 POSITIEWE NASIEN VANAF VRAAG 9..1 OPTION 1/OPSIE 1 Pave/ gemid = V rms/wgk Irms/wgk for the kettle/vir die ketel: = ( I rms / wgk ) I rms/wgk = 8,3 A I tot = (8,3 + 3,33) = 11,65 A OPTION /OPSIE V rms / wgk P = / gemid R 800 = ( ) 340 ( ) ( R) R = 7,5 Ω ( 340) 000 = ( ) ( R ) 000 R = 8,9 Ω = + R R ( ) V max/ maks ave = 1 R V rms/wgk = I rms/wgk R 40,4 = I rms/wgk (0,64) I rms/wgk = 11,65 A ( )( R) ( 8,9 )( 7,5 ) R = ( ) 8,9 + 7,5 = 0,64 Ω

18 Physical Sciences P1/Fisiese Wetenskappe V1 18 DBE/November 016 OPTION 3/OPSIE 3 P ave/gemid = V rms/wgk I rms/wgk = ( 340) Imax/ maks 800 = I max/maks = 16,47 A Imax/ maks 16,47 Irms = = I rms/wgk = 11,65 A OPTION 4/OPSIE 4 P ave/gemid = V rms/wgk I rms/wgk = Irms/wgk I rms/wgk = 11,65 A OPTION 5/OPSIE 5 P T : P K 800 : :,5 V I max/ maks max/ maks I T : I K 3,33 : 8,35 I rms = 3,33 + 8,35 = 11,66 A (4) [11] QUESTION 10/VRAAG The minimum frequency (of a photon/light) needed to emit electrons from (the surface of) a metal. (substance) Die minimum frekwensie (van 'n foton/lig) benodig om elektrone vanaf die (oppervlakte van)'n metaal (stof) vry te stel The frequency (of a photon/light) needed to emit electrons from (the surface of) a metal. (substance) with zero kinetic energy Die frekwensie (van 'n foton/lig) benodig om elektrone vanaf die (oppervlakte van)'n metaal (stof) met nul/geen kinetiese energie vry te stel ()

19 Physical Sciences P1/Fisiese Wetenskappe V1 19 DBE/November Silver/Silwer Threshold/cutoff frequency (of Ag) is higher/drumpel/afsnyfrekwensie (van Ag) is hoër W o α f o / W o = hf o To eject electrons with the same kinetic energy from each metal, light of a higher frequency/energy is required for silver. Since E = W o + E k(max) (and E k is constant), the higher the frequency/energy of the photon/light required, the greater is the work function/w o. Om elektrone met dieselfde kinetiese energie van elke metal vry te stel,is lig van hoër frekwensie benodig vir silwer. Aangesien E = W o + E k(maks) (en E k(maks) is konstant) word fotone/lig van hoër frekwensie/energie benodig, dus is arbeidsfunksie hoër (3) Planck s constant /Planck se konstante (1) Sodium/Natrium (1) Energy radiated per second by the blue light /Energie per sekonde uitgestraal 5 deur die bloulig = ( )(60 x 10-3 ) = 3 x 10-3 J s hc E photon/foton = λ 34 8 (6,63 10 )(3 10 ) = = 4,3 x J Total number of photons incident per second/totale aantal fotone wat per sekonde inval = -19 4,3 10 = 7,09 x (5) 10.. POSITIVE MARKING FROM QUESTION POSITIEWE NASIEN VANAF VRAAG ,09 x (electrons per second/elektron per sekonde) Same number as that calculated in Question above/dieselfde as die in Vraag hierbo bereken (1) [13] TOTAL/TOTAAL: 150

NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12

NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 NATIONAL SENIOR CERTIICATE NASIONALE SENIOR SERTIIKAAT GRADE/GRAAD 1 PHYSICAL SCIENCES: PHYSICS (P1) ISIESE WETENSKAPPE: ISIKA (V1) EXEMPLAR 014 MODEL 014 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists