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Physical Sciences P1 1 FS/September 2014 Grade 12 Prep. Exam.

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PREPARATORY EXAMINATION

GRADE 12

PHYSICAL SCIENCES P1 (PHYSICS)

SEPTEMBER 2015

MARKS: 150

TIME: 3 HOURS

This question paper consists of 13 pages and 3 data sheets.



INSTRUCTIONS AND INFORMATION 1. Write your name in the appropriate space on the ANSWER BOOK. 2. 3. 4. 5. 6. 7. 8.

This question paper consists of TEN questions. Answer ALL the questions in the ANSWER BOOK. Begin EACH question on a NEW page. Number the answers correctly according to the numbering system used in this question paper. Leave ONE line between two sub-questions, for example between QUESTION 3.1 and QUESTION 3.2. You may use a non-programmable calculator. You may use appropriate mathematical instruments. YOU ARE ADVISED TO USE THE ATTACHED DATA SHEETS.

9. Show the formulae and substitutions in ALL calculations. 10. Round off your final numerical answers to TWO decimal places. 11. Give brief motivations, discussions, et cetera where required. 12. Write neatly and legibly.



QUESTION 1: MULTIPLE-CHOICE QUESTIONS Four options are provided as possible answers to the following questions. Each question has only ONE correct answer. Write only the letter (A – D) next to the question number (1.1 – 1.10) in the ANSWER BOOK.

1.1 An object moving with a constant speed v has a kinetic energy E. Which one

of the following will be true for the kinetic energy if the object has a constant velocity of 2v?

A

B

C

D

½E

E

2E

4E

1.2 A sphere is attached to a string, which is suspended from a fixed horisontal

bar as shown in the sketch. The reaction force to the gravitational force exerted by the earth on the sphere is …

A

B

C

D

the force of the bar on the sphere.

the force of the string on the sphere.

the force of the sphere on the earth.

the force of the bar on the string.

(2)

1.3 A ball is dropped from height h above the ground and reaches the ground

with kinetic energy E. From which height must the ball be dropped to reach the ground with kinetic energy 2E? (Ignore all effects of friction.)

A

B

C

D

2h

3h

4h

8h

(2)



3t

1.4 The velocity versus time graph below represents the movement of an object under the influence of gravitational force. v v (m·s-1) t

-2v The displacement of the object in time 3t is …

A

B

C

D

vt. zero.

-vt.

23

− v t.

(2)

If the amount of work done by net force F on object P equals W, the amount of

work done on Q will be …

A

B

C

D

W.

½ W.

2 W.

4 W.

(2)

1.5 A net force F accelerates two isolated objects, P and Q, from rest on a straight line for time t as shown below. Object P experiences an acceleration of a and object Q an acceleration of 2a.

F F P Q



1.6 A block, being pulled by a force F, and moving to the left on a rough horizontal surface, is slowing down.

F

The directions of the resultant force and the acceleration are …

DIRECTION OF RESULTANT FORCE

DIRECTION OF ACCELERATION

A to the right to the left B to the right to the right C to the left to the left D to the left to the right

(2)

1.7 Three small identical spheres, A, B and C is charged as shown in the diagram. The distance between sphere B and C is x. +16 µC -4 µC +4 µC A B C x For sphere B to experience no resultant electrostatic force, the distance between A and B must be …

A

B

C

D

¼ x.

½ x.

2 x.

4 x.

(2)



Q

1.8 Two metal balls A and B, mass m and 2m respectively, are allowed to roll down two different frictionless slopes as indicated in the diagram below. A m B 2m h h 2Ө Ө Which ONE of the following is true for the work done and the force acting on balls A and B respectively?

MAGNITUDE OF FORCE WORK DONE

A FA = FB WA ˃ WB B FA ˂ FB WA ˂ WB C FA ˃ FB WA ˂ WB D FA ˂ FB WA ˂ WB

(2)

1.9 In the circuits shown below all resistors and cells are identical. P A

A

Which ONE of the following gives the correct comparison between the voltmeter

and ammeter readings in circuit P and Q. VOLTMETER READING AMMETER READING A VP ˃ VQ Ap ˃ A Q B VP ˃ VQ Ap ˂ A Q C VP ˂ VQ Ap = A Q D VP = VQ Ap ˂ A Q

(2)

V

V



300

1.10 In the circuit shown below the resistance of X is R and that of Y is 2R.

If the power dissipated by X equals P, then the power dissipated by Y will

be …

A

B

C

D

¼ P.

½ P.

2P.

4P.

(2)

TOTAL SECTION A: 20 QUESTION 2 (Begin on a new page.) Two blocks of masses 5 kg and 3 kg respectively are connected by a light inextensible string that runs over a light frictionless pulley as shown in the diagram below. The 5 kg block experience a frictional force of 8 N and the coefficient of kinetic friction between the 3 kg block and the surface of the inclined plane is 0,15.

T 3 kg 2.1 Define the term frictional force. (2) 2.2 Draw a labelled free-body diagram to indicate all the forces acting on

the 3 kg block.

(3) 2.3 Calculate the: 2.3.1 Magnitude of the frictional force acting between the

3 kg block and the surface of the inclined plane

(3) 2.3.2 Magnitude of the tension T in the string (6) [14]

Y X V

5 kg



QUESTION 3 (Begin on a new page.) Ball A is thrown vertically downwards from the top of a building, 80 m high, at a velocity of 12 m·s-1. At the same instant a second identical ball B is thrown upwards at a velocity of 30 m·s-1. Ball A and ball B pass each other after 2,135 s. Ignore all effects of air friction. A 12 m·s-1 80 m 30 m.s-1 B

3.1 Give the direction of the acceleration of ball B while moving upwards. (1) 3.2 Calculate the velocity of ball B the moment it passes ball A. (3) 3.3 Calculate the distance between ball A and B 2,5 s after it was

projected.

(6) 3.4 Sketch a position-time graph for the motion of ball A till it reaches the

ground as well as for the motion of ball B until it passes ball A. Use the ground as zero position. Clearly indicate the time at which the balls pass each other.

(3) [13] QUESTION 4 (Begin on a new page.) A trolley, mass 5 kg, moves at 4 m·s-1 east across a frictionless horizontal surface. A brick of mass 1,5 kg is dropped onto the trolley. 1,5 kg 5 kg

4.1 Define in words the Law of Conservation of Momentum. (2) 4.2 State the condition for an elastic collision. (1) 4.3 Calculate the change in momentum of the 5 kg trolley. (5) [8]

4 m·s-1



C

3 m·s-1

QUESTION 5 (Begin on a new page.) 5.1 A boy on roller-skates moves at a constant velocity in an easterly direction along a frictionless horizontal part AB of a track carrying a parcel. He decides to increase his velocity by throwing the parcel horizontally away from him. X 4 m 6 m·s-1 Ө A B

5.1.1 In which direction must the parcel be thrown to cause a maximum

increase in the velocity of the boy? (1)

5.1.2 Name and define in words the law in physics that you have applied

in QUESTION 5.1.1. (3)

On reaching point B at a velocity of 6 m·s-1, the boy on the roller-skates, with total mass 57 kg, continues to move up a rough section BC of the track and comes to rest at position X, height 4m. The magnitude of the frictional force acting on the roller-skates, is 40 N.

5.1.3 Calculate value Ө of the inclined plane. (6)

5.2 A remote controlled car is driven up an inclined plane at 300 to the

horizontal as shown below. The car of mass 4 kg, experiences an average forward force of 80 N. A frictional force of 15 N is acting on the car as it moves up the plane. The speed of the car at the bottom of the inclined plane is 3 m·s-1.

80 N

30o Use energy principles to calculate the speed of the car after it has

travelled 5 m up the inclined plane.

(7) [17]



QUESTION 6 (Begin on a new page.) Light emitted from distant stars demonstrates the phenomenon known as red shift. 6.1 Explain how the phenomenon known as red shift can be used to

explain an expanding universe.

(2) 6.2 A submarine can use the Doppler effect to detect the speed of ship. A

submarine at rest and just below the surface of the water, detects the frequency of a moving ship as 437 Hz, 0,985 times the actual frequency of the sound emitted by the ship. The speed of sound in water is 1470 m·s-1.

6.2.1 Is the ship moving away from or towards the submarine? Give

a reason for your answer.

(2) 6.2.2 Calculate the speed of the ship. (5) 6.3 Name two applications of the Doppler effect in Medical Science. (2) [11] QUESTION 7 (Begin on a new page.) The diagram below shows two identical insulated metal spheres. Spheres P an Q each carry a charge of 6 nC.

+6 nC +6 nC

P Q

20 cm

7.1 Define Coulomb’s Law in words. (2) 7.2 Draw the electric field pattern due to the two spheres P and Q. (3) 7.3 Calculate the magnitude of the electrostatic force between spheres P

and Q.

(4) A third sphere, R, of charge -2 nC is now placed at a position relative to the other spheres and a chosen point X as shown in the diagram below.

+6 nC +6 nC -2 nC

P Q R X● 20 cm 20 cm 10 cm

7.4 Calculate the net electric field at point X due to spheres P, Q and R. (6) [15]



S2

0,4

QUESTION 8 (Begin on a new page.) 8.1 The graph below is obtained from an experiment to calculate the internal

resistance of a battery.

Graph of current versus potential difference

0,6 0,2 0 0,5 1,0 1,5 2,0 2,5 3,0 V(V)

8.1.1 Calculate Vinternal if the current in the circuit is equal to 0,2 A . (2)

8.1.2 Calculate the internal resistance of the battery. (4) 8.2 A circuit is connected as shown below. When switch S1 is closed, Vexternal is

equal to 22,5 V. The internal resistance of the battery is 0,8 Ω.

● ● V1

8.2.1 Define Ohm’s Law in words. (2)

8.2.2 Calculate the power dissipated by the 16 Ω resistor. (7)

S1

I (A)

emf = 24 V

R

16 Ω 3 Ω



8.2.3 Calculate the resistance of R. (5)

8.2.4 Switch S2 is now closed. How will voltmeter reading V1 be influenced? (Write down only INCREASE, DECREASE or STAYS THE SAME.) Give an explanation to your answer.

(4) [24]

QUESTION 9 (Begin on a new page.)

9.1 The diagram below shows a coil that is rotated through a magnetic field. ► N ● S 9.1.1 Name the principle demonstrated in the above diagram? (1) 9.1.2 The maximum emf is generated at position A of the rotation

cycle. Give an explanation for this observation.

(2) 9.1.3 Name one structural difference between a DC and AC

generator.

(2) 9.1.4 Use the positions indicated in the diagram above and sketch a

graph of current versus position for one complete rotation of a DC generator. (Indicate the positions A, B, C and D on the graph.)

(4) 9.2 When an AC supply is connected to a lamp, it lights up with the same

brightness as it does when connected to a 18 V battery (DC source). The power dissipated by the lamp is equal to 60 W.

9.2.1 What is the rms voltage of the AC supply? (1) 9.2.2 Calculate the peak current delivered by the AC source. (5) [15]

A

B

C

D



QUESTION 10 (Begin on a new page.) The diagram below shows a circuit in which a photocell is irradiated alternately with red and blue light to demonstrate the photo-electric effect.

10.1 An ammeter reading is recorded when the photocell is irradiated with

red light. Give an explanation for this observation.

(2) 10.2 Blue light with the same intensity as the red light is now used to

irradiate the photocell. How will this influence the following:

10.2.1 The kinetic energy of the photo-electrons (Write down only

INCREASE, DECREASE or STAYS THE SAME.)

(1) 10.2.2 The ammeter reading. (Write down only INCREASE,

DECREASE or STAYS THE SAME.) Give an explanation for your answer.

(4) 10.3 The wavelength of the blue light used in the demonstration is 4,5 x 10-7 m.

Calculate the threshold frequency (cut-off frequency) of the metal used in the photo cell if the average speed of an emitted photo-electron is equal to 4,78 x 105 m·s-1.

(6) [13] TOTAL SECTION B:

GRAND TOTAL: 130 150!

A

Incident light

e-


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DATA FOR PHYSICAL SCIENCES GRADE 12 PAPER 1 (PHYSICS)

GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 12

VRAESTEL 1 (FISIKA) TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES

NAME/NAAM SYMBOL/SIMBOOL VALUE/WAARDE Acceleration due to gravity Swaartekragversnelling g 9,8 m·s-2

Universal gravitational constant Universele gravitasie konstante G 6,67 x 10-11 N·m2·kg-2

Speed of light in a vacuum Spoed van lig in 'n vakuum c 3,0 x 108 m·s-1

Planck's constant Planck se konstante h 6,63 x 10-34 J·s

Coulomb's constant Coulomb se konstante k 9,0 x 109 N·m2·C-2

Charge on electron Lading op elektron e -1,6 x 10-19 C

Electron mass Elektronmassa me 9,11 x 10-31 kg

Mass of Earth Massa van Aarde M 5,98 x 1024 kg

Radius of Earth Radius van Aarde RE 6,38 x 103 km


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TABLE 2: FORMULAE/TABEL 2: FORMULES MOTION/BEWEGING

tavv if Δ+= 2

21

i taΔtvΔx Δ+= or/of 2

21

i taΔtvΔy Δ+=

xa2vv 2i

2f Δ+= or/of ya2vv 2

i2f Δ+= Δt

2vvΔx fi !"

#$%

& += or/of Δt

2vvΔy fi !"

#$%

& +=

FORCE/KRAG

maFnet = mvp=

Nfk

k =µ Nf )maks(s

s =µ

if

net

mvmvpptF−=Δ

Δ=Δ mgw =

221

rmGmF=

2r

Gmg= WORK, ENERGY AND POWER/ARBEID, ENERGIE EN DRYWING

xFW Δ= cosθ mghU= or/of mghEP = 2mv

21K= or/of 2

k mv21E = KWnet Δ= or/of knet EW Δ=

if KKK −=Δ or/of kikfk EEE −=Δ

Wnc = ∆K + ∆U or/of Wnc = ∆Ek + ∆Ep avav FvP = /

gemidgemid FvP =

tWPΔ

=

WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG

λ= fv f1T =

ss

LL f

vvvvf

±

±= or/of b

b

LL f

vvvvf

±

±=

hfE=

λ=

chE

maxo K WE += or/of (max)ko E WE += where/waar

hf E = and/en 00 hfW = and/en or/of........max2

(max) 21mvEk =

max2

(max) 21 mvK =


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ELECTROSTATICS/ELEKTROSTATIKA

221

rQkQF= 2r

kQE =

qFE =

qWV =

n = eQ

or/of n =

eqQ

ELECTRIC CIRCUITS/ELEKTRIESE STROOMBANE

IVR= emf ( ε ) = I(R + r)

...RRR 21s ++=

...R1

R1

R1

21p

++= I=q Δ t

W = Vq W = VIΔ t W= I2RΔ t

W= RΔtV2

ΔtWP=

P = VI P = I2R

RVP2

=

ALTERNATING CURRENT/WISSELSTROOM

2max

rms

II = /

2maks

wgkII =

2

VV maxrms = /

2VV maks

wgk =

rmsrmsaverage VP I= / wgkwgkgemiddeld VP I=

RP 2rmsaverage I= / RP 2

wgkgemiddeld I=

RV

P2rms

average = / RV

P2wgk

gemiddeld =

Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

PREPARATORY EXAMINATION VOORBEREIDENDE EKSAMEN

GRADE 12/GRAAD 12

PHYSICAL SCIENCES P1 (PHYSICS) FISIESE WETENSKAPPE V1 (FISIKA)

SEPTEMBER 2015

MEMORANDUM

MARKS/PUNTE: 150

This memorandum consists of 13 pages. Hierdie memorandum bestaan uit 13 bladsye.

Physical Sciences P1 (Physics) 2 FS/September 2015 Grade 12 Prep. Exam. Memorandum


SECTION A/AFDELING A QUESTION 1/VRAAG 1 1.1 D !! (2) 1.2 C !! (2) 1.3 A !! (2) 1.4 D !! (2) 1.5 C!! (2) 1.6 B !! (2) 1.7 C!! (2) 1.8 B !! (2) 1.9 B !! (2) 1.10 B !! (2)

[20]

TOTAL SECTION A/TOTAAL AFDELING A: 20



SECTION B/AFDELING B QUESTION 2/VRAAG 2 2.1 The force that opposes the motion! of an object and which act parallel

to the surface! Die krag wat die beweging van 'n voorwerp teenstaan en parallel aan die oppervlak inwerk.

(2) 2.2

f/Ff/Ffriction/friction!/Fwrywing,wrywing ! T Fg/w/weight/force of gravity!/gewig,gravitasiekrag

(3) 2.3.1 fk(max) = µkFN!

= 0,15(3)(9,8)(cos 300)! = 3,82 N!

(3) 2.3.2 Positive marking from 2.3.1/ Positiewe merk van 2.3.1

Right/downwards as positive:/ Regs/afwaarts as positief 5 kg block: Fnet = ma ! T + f = ma T - (8) = 5a! 1

3 kg block : T + f + Fg// = ma - T - 3,82 + (3)(9,8)sin30o != 3a! 2

-T + 10,88 = 3a Substitute 2 into 1: a = 0,36 m·s-2

Substitute a into 1: T - 8 = (5)(0,36) ! T = 9,8 N!

(6) [14]

!



QUESTION 3/VRAAG 3 3.1 Downwards/Afwaarts! (1) 3.2 Upwards positive/Opwaarts

positief:

tavv if Δ+= ! = 30! + (-9,8)(2,135) = 9,08 m·s-1, upwards!

Downwards positive/Afwaarts positief:

tavv if Δ+= ! = - 30! + (9,8)(2,135) = - 9,078 m·s-1 = 9,08 m·s-1, upwards !

(3) 3.3 Upwards positive/Opwaarts

positief: Ball A:

221

i taΔtvΔy Δ+= ! = -12(2,5) + ½(-9,8)(2,5)2! = - 60,625 m (Height /Hoogte= 19,375 m) Ball B:

221

i taΔtvΔy Δ+= = 30(2,5) ! + ½(-9,8)(2,5)2! = 44,375 m Distance = 44,375 - 19,375! = 25 m !

Downwards positive/Afwaarts positief:

221

i taΔtvΔy Δ+= ! = 12(2,5) + ½(9,8)(2,5)2! = 60,625 m (Height /Hoogte= 19,375 m) Ball B:

221

i taΔtvΔy Δ+= = -30(2,5) ! + ½(9,8)(2,5)2! = - 44,375 m Distance = 44,375 - 19,375! = 25 m! (6)

3.4 OPTION 1/OPSIE 1

Upwards positive/Opwaarts positief:

t (2,135s)

Time/tyd(s)

∆y (m

)

0

B

A



OPTION 2/OPSIE 2 Upwards negative/ Opwaarts negatief:

(2,135s)

[13]

Criteria for graph/Kriteria vir grafiek: Marks/ Punte

Shape for ball A up till zero position. Vorm vir bal A tot zero posisie. !

Shape for ball B up till intersection of lines time. Vorm vir bal B tot grafieklyne kruis. 2,135 s. !

Indication of time 2,135 s. Aanduiding van tyd 2,135 s. !

Ground not zero position (provided everything else is correct): 2/3 Grond nie zero posisie nie (op voorwaarde die res is korrek) : 2/3

(3)

Criteria for graph/Kriteria vir grafiek: Marks/ Punte

Shape for ball A up till zero position. Vorm vir bal A tot zero posisie. !

Shape for ball B up till intersection of lines time. Vorm vir bal B tot grafieklyne kruis. 2,135 s. !

Indication of time 2,135 s. Aanduiding van tyd 2,135 s. !

Ground not zero position (provided everything else is correct): 2/3 Grond nie zero posisie nie (op voorwaarde die res is korrek) : 2/3

(3)

Time/tyd(s) ∆y (m

)

0

B

A



QUESTION 4/VRAAG 4 4.1 The total linear momentum of a closed system! remains constant !

Die totale linieëre momentum in 'n geslote sisteem bly konstant

(2) 4.2 The kinetic energy remains constant. ! OR

The kinetic energy before the collision equals kinetic energy after the collision. Die kinetiese energie bly konstant. OF Die kinetiese energie voor botsing is gelyk aan die kinetiese energie na botsing.

(1) 4.3 Σ pbefore = Σ pafter !

(5)(4 ) = (6,5)vf ! vf = 3,077 m·s-1 ∆p = m(vf - vi) = 5 ( 3,077 – 4) ! = - 4,62 kg·m·s-1 = 4,62 kg·m·s-1!, left/west/ opposite to direction of motion ! Links/wes/teenoorgesteld aan bewegingsrigting (5)

[8] QUESTION 5/VRAAG 5 5.1.1 Backwards/behind him!/Terugwaarts/agter hom. (1) 5.1.2 Newton’s third Law!of motion: 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. ! Newton se derde bewegingswet: Wanneer een liggaam 'n krag op 'n tweede liggaam uitoefen sal die tweede liggaam 'n krag van gelyke grootte in die teenoorgestelde rigting op die eerste liggaam uitoefen.

(3)



5.1.3 OPTION 1/ OPSIE 1 Wnet = ∆K ! Wg + Wf = ∆K Fg∆xcosӨ + f∆xcosӨ = ∆K (57)(9,8)(4)cos180o ! + 40∆xcos180o! = 0 - ½(57)(62) ! ∆x = - 30,21 m

Sin Ө = 30,214

!

Ө = 7,61o!

(6)

5.2 OPTION 1/ OPSIE 1

Wnet = ∆K ! WT + Wg + Wf = ∆K (80)(5)(4)cos0o !+ (4)(9,8)sin30o!(5)cos180o! + (15)(5)cos180o != ½(4)vf

2 -½(4)(32) ! vf = 11,07 m·s-1 ! !!!!!!!!!!!!!!!!!!!!!!!!!!!

(7) [16]

OPTION 2/ OPSIE 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

Sin Ө = 30,214

!

Ө = 7,61o!

OPTION 2/ OPSIE 2 Wnc = ΔU + ΔK ! WT + Wf = ΔU + ΔK

(80)(5)(4)cos0o!+ (15)(5)cos180o! = (4)(9,8)(sin30o)(5) ! - (4)(9,8)(0) ! + ½ (4)vf

2 - ½ (4)(3)2 ! vf = 11,07 m·s-1 !



QUESTION 6/VRAAG 6 6.1 Red shift implies that light emitted by stars shows a shift towards the

lower frequencies" of the spectrum. According to the Doppler effect this means that the source (star) is moving away from the observer. " Rooiverskuiwing impliseer dat lig vrygestel deur sterre 'n verskuiwing na die laer frekwensies van die spektrum toon. Volgens die Doppler effek dui dit daarop dat die bron (ster) weg van die waarnemer af beweeg.

(2) 6.2.1 Away ! (from submarine)

The detected/observed frequency is lower than the actual frequency. ! Weg ! (van duikboot) Die waargenome frekwensie is laer as die werklike frekwensie. !

(2)

6 2.2 OPTION 1/ OPSIE 1

ss

LL f

vvvvf

±

±= OR/OF s

sL f

vvvf+

= !

sS

L

vvv

ff

+=

sv+=14701470985,0

vs = 22,39 m·s-1 !

(5) 6.2.2 To measure the velocity of blood flowing through blood vessels.!

To scan a foetes.! Om die snelheid van bloedvloei deur bloedvate te bepaal. Om 'n fetus te skandeer.

(2) [11]

OPTION 2/ OPSIE 2 fL = (0,985) fs 437 = (0,985) fs fs = 443,655 Hz

ss

L fvvvf+

= !

655,44314701470437

sv+= !

vs = 22,39 m·s-1 !

!

!

! !

!



QUESTION 7/VRAAG 7 7.1 The magnitude of the electrostatic force exerted by one point charge on

another point charge is directly proportional to the product of the magnitude of the charges! and inversely proportional to the square of the distance between them. ! Die grootte van die elektrostatiese krag uitgeoefen deur een puntlading op 'n ander puntlading is direk eweredig aan die produk van die grootte van die ladings en omgekeerd eweredig aan die kwadraat van die afstand tussen die ladings.

(2) 7.2

Criteria for field pattern/Kriteria vir veldpatroon: Marks/ Punte

Correct direction away from the spheres. Korrekte vorm weg vanaf sphere. !

Correct shape of field pattern. Korrekte vorm vir veldpatroon. !

Field lines not crossing/ not drawn inside the sphere Veldlyne kruis nie/ nie binne-in spheer geteken. !

(3)

7.3

2QP

ZY rQkQ

F = !

2

99

0,21010 9 91066 −− ×××××

= !

= 8,1 x 10-6 N!

(4) 7.4

2P

net rkQE = !+ 2

Q

rkQ

+ 2R

rkQ

2

99

0,51010 9 −×××

=6

!+ 2

99

0,310610 9 −××× ! -

2

99

0,11010 9 −××× 2 !

= -9,84 x 102 N.C-1 = 9,84 x 102 N.C-1!, left!links

(6) [15]

!



QUESTION 8/VRAAG 8 8.1.1 Vi = (3,0 – 2,0) = 1,0 V !! (2) 8.1.2

VΔΔ

=IGradient !

0-10,6-0,4

=

= -0,2 Ω-1 ri = 5 Ω!

(4) 8.2.1 The potential difference across a conductor is directly proportional to

the current! in the conductor at constant temperature!. Die potensiaalverskil oor 'n geleier is direk eweredig aan die stroom deur die geleier by konstante temperatuur.

(2) 8.2.2 Vi = 24 – 22,5

= 1,5 V

Ii

iVr =

!

I1,5

=�� !

I = 1,875 A

V3Ω = IR

= (1,875)(3) !

= 5,625 V

V// = Vext - Vs

= 22,5 – 5,625!

= 16,875 V

RVP2

= !

1616,8752

=

!!!!!= 17,80 W!! (7)

!

!

!



8.2.3 Positive marking from 8.2.2/Positiewe nasien van 8.2.2

R//VI =

1616,875 = !

= 1,055 A IR = 1,875 – 1,055! = 0,82 A

IV

R =

��16,875

=

!!!!= 16,06 Ω!

(5) 8.2.4 Decrease!

Total resistance in circuit decrease and the total current increase. ! Vinternal will increase ! Therefore: Vexternal will decrease because emf stays constant. ! Afneem Totale weerstand in stroombaan neem af en die totale stroom neem toe. Vintern sal toeneem. Dus: Vekstern sal afneem omdat die emk konstant bly. (4)

[24] QUESTION 9/VRAAG 9 9.1.1 Electromagnetic Induction"/elektromagnetiese induksie (1) 9.1.2 The rate of change in the magnetic flux" is a maximum" at position A.

Die tempo van verandering in die magnetiese vloed is 'n maksimum by punt A.

(2) 9.1.3 DC generator : split ring commutator"

AC generator : slip rings" GS generator: splitring kommutator WS generator: sleepringe

(2)

!

!



9.1.4 A B C D

(4) 9.2.1 18 V " (1) 9.2.2 Pave = VrmsIrms!"!

60 " = 18(Irms)!"!(Irms) = 3,33 A

2max

rms

II =

2I

3,33 max= "

Imax = 4,71 A"

(5) [15] QUESTION 10/VRAAG 10 10.1

The energy of the photons of red light is greater! than the work function of the metal in the photocell.!OR The frequency of red light is higher than the threshold/cut-off frequency of the metal in the photocell. Die energie van die fotone van rooi lig is groter as die werksfunksie van die metaal in die fotosel.OF Die frekwensie van rooi lig is groter as die drumpel/afsnyfrekwensie van die metaal in die fotosel.

(2) 10.2.1 Increase!/ Neem toe (1) 10.2.2 Stays the same !

The change in colour/frequency only has an influence on the kinetic energy of the photo electrons. ! Only the intensity of the light has an influence on the number of photo electrons emitted per time unit. ! The intensity of the light stays the same and therefore the number of photo electrons emitted per unit time /current stays the same. !

(4)

!

t(s)!

! !!!!!!!!!!I!(A)

"!

! !

"!

"!

"!



Bly dieselfde Die verandering in kleur/frekwensie beïnvloed slegs die kinetiese energie van die foto-elektrone. Slegs intensitiet van lig het 'n invloed op die aantal foto-elektrone wat per tydeenheid vrygestel word. Die intensiteit van die lig het dieselfde gebly en daarom het die aantal foto-elektrone per tydeenheid/stroom konstant gebly.

10.3 OPTION 1/OPSIE 1

(max)ko E WE += !

=λch

=

×××

−7104,510310 x 6,63 8-34

fo = 5,10 x 1014 Hz ! !

(6) [13]

TOTAL SECTION B/TOTAAL AFDELING B:

GRAND TOTAL/GROOTTOTAAL: 130 150

hfo + ½mv2

6,63 x 10-34(fo) ! + ½(9,11 x 10-31)(4,78 x 105)2! !

!

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