GRADE 11 NOVEMBER 2015 PHYSICAL SCIENCES P1 · GRADE 11 NOVEMBER 2015 PHYSICAL SCIENCES P1 MARKS:...

NATIONAL SENIOR CERTIFICATE

GRADE 11

NOVEMBER 2015

PHYSICAL SCIENCES P1

MARKS: 150 TIME: 3 hours

This question paper consists of 18 pages, including data sheets.

*IPHSCE1*

2 PHYSICAL SCIENCES P1 (EC/NOVEMBER 2015)

Copyright reserved Please turn over

INSTRUCTIONS AND INFORMATION 1. Write your FULL NAME and SURNAME on your ANSWER SHEET or

ANSWER BOOK.

2. This question paper consists of THIRTEEN questions. Answer ALL the

questions.

3. Start EACH question on a new page in the ANSWER BOOK. 4. Number the answer correctly according to the numbering system used in this

question paper.

5. Leave ONE line between two sub-questions, for example between

QUESTION 2.1 and QUESTION 2.2.

6. You may use a non-programmable calculator. 7. You may use appropriate mathematical instruments. 8. YOU ARE ADVISED TO USE THE ATTACHED DATA SHEETS. 9. Show ALL formulae and substitution in ALL calculations. 10. Round off your FINAL numerical answers to a minimum of TWO decimal

places.

11. Give brief motivations, discussions, et cetera where required. 12. Write neatly and legible.

(EC/NOVEMBER 2015) PHYSICAL SCIENCES P1 3


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

1.1 Which of the following is true for the given sketch?

A ⃗⃗⃗⃗ ⃗ + ⃗⃗⃗⃗ ⃗ = ⃗⃗⃗⃗ ⃗ B ⃗⃗⃗⃗ ⃗ + ⃗⃗⃗⃗ ⃗ = ⃗⃗⃗⃗ ⃗ C ⃗⃗⃗⃗ ⃗ + ⃗⃗⃗⃗ ⃗ = ⃗⃗⃗⃗ ⃗ D ⃗⃗⃗⃗ ⃗ + ⃗⃗⃗⃗ ⃗ + ⃗⃗⃗⃗ ⃗ = 0 (2) 1.2 Two forces are acting on a block as shown in the diagram. The resultant force in

the horizontal direction is:

A zero B 10 N to the right C 10 N to the left D 20 N to the right (2) 1.3 Mary has a mass of 50 kg and is standing on a scale in a lift. The reading on the

scale is 555 N. The lift is … A accelerating upwards. B moving downwards at constant velocity. C moving upwards at a constant velocity. D accelerating downwards. (2)



1.4 Which one of the following is an example of a contact force? A Electrostatic force B Magnetic force C Gravitational force D Frictional force (2) 1.5 The diagram below shows water waves that spread out after passing through a

single slit.

The wave phenomenon observed after the water waves pass through the slit

is … A reflection. B diffraction. C dispersion. D total internal reflection. (2) 1.6 A ray of light moves from glass to air. The angle of incidence in glass is 30° and

refraction takes place. Which combination for speed and frequency of the refracted wave is correct?

Speed Frequency A Increases Remains constant B Decreases Remains constant C Increases Increases D Decrease Increases (2) 1.7 A central bright band is observed when light of wavelength passes through a

single slit of width a. Light of wavelength 3 is now used. Which ONE of the following slit widths would produce a central bright band of the same broadness?

A

a B

a C a D 3 a (2)



1.8 A bar magnet can be moved towards or away from a solenoid. Which ONE

of the following factors/actions will increase the deviation on the galvanometer?

A Use a solenoid with a smaller diameter. B Use a solenoid with fewer turns. C Use a weaker magnet. D Move the magnet faster towards and away from the solenoid. (2) 1.9 Which one of the following graphs correctly represents the relationship

between potential difference (V) and current (I) for an ohmic conductor? A B C D

(2) 1.10 Each of the four identical resistors is connected to the same emf. Which ONE

of the following circuits will have the highest resistance? A B

C D

(2) [20]



QUESTION 2 (Start on a new page.) During a mountain climbing exercise, Ferial, mass 50 kg, is suspended from an inelastic piece of nylon rope, fixed to a vertical cliff at X on the cliff. She pushes her legs against the cliff so that they make an angle of 45° with the cliff, as indicated in figure. The angle that the rope makes with the cliff is 20°.

Point Y is in equilibrium. 2.1 Explain what is meant by Point Y is in equilibrium. (1) 2.2 Draw a FORCE DIAGRAM showing all the forces acting on point Y. (3) 2.3 Determine by means of ACCURATE CONSTRUCTION and MEASUREMENT.

(use the scale 10 mm: 50 N and indicate at least TWO angles): 2.3.1 The magnitude of the force which the rope exerts on her. 2.3.2 The magnitude of the force exerted by her legs. (8) [12]



QUESTION 3 (Start on a new page.) A wooden cabinet of 60 kg rests on the back of a tip-up truck. The back tilts slowly, until it makes an angle of 30° with the horisontal. The cabinet does NOT move.

3.1 Calculate the magnitude of the frictional force. (3) 3.2 Calculate the coefficient of static friction. (3) 3.3 The angle of 30° is now increased. How will this change affect the following:

(Write down only INCREASES, DECREASES or REMAINS THE SAME)

3.3.1 The coefficient in QUESTION 3.2? (1) 3.3.2 The frictional force? Explain your answer. (3) [10]



QUESTION 4 (Start on a new page.) A 6 kg block on a horisontal rough surface is joined to a 2 kg block by a light, inelastic string running over a frictionless pulley. The frictional force between the 6 kg block and the table is 11,76 N. A downwards force of 2 N is applied to the 2 kg block as indicated in the diagram below.

4.1 State Newton’s Second Law of motion in words. (2) 4.2 Draw a free-body diagram showing ALL the forces acting on the 6 kg block. (4) 4.3 Calculate: 4.3.1 The magnitude of the acceleration of the 6 kg block. (5) 4.3.2 The magnitude of the tension (T) in the string connecting the two

blocks. (2) 4.4 The rough surface is replaced by a smooth frictionless surface. How will

this change affect the answer in QUESTION 4.3.1? Write only INCREASES, DECREASES or REMAINS THE SAME. (1)

[14]



QUESTION 5 (Start on a new page.) A truck is moving at a constant velocity and is suddenly hit from the rear (back) by a fast-moving bus. A loose-standing crate is placed on the back of the truck and a man, M, is sitting on the crate.

5.1 In which direction does the man, M, fall?

Write down only BACKWARDS or FORWARDS. (1) 5.2 State in words Newton’s laws of motion you used to answer QUESTION 5.1. (2) 5.3 Luxolo and Kay with masses 80 kg and 50 kg respectively are standing on

roller-skates facing each other. Initially they are at rest. They push against each other and both move away in opposite directions. (IGNORE FRICTION)

5.3.1 How does the force exerted on Kay compare to the force exerted on

Luxolo? Write only GREATER, SMALLER or EQUAL TO. (1)

5.3.2 Explain how their accelerations compare. (2) [6]



QUESTION 6 (Start on a new page.) NASA fires a space shuttle off the ground and into space. It accelerates from rest to 160 m.s-1 in the first 400s of its journey. The total mass of the space shuttle and its fuel tanks is 2 600 tons (2,6 x 106 kg). The space shuttle’s propulsion system consists of exhaust gases, which are pushed out of an outlet in its base, B.

B

6.1 Name an action-reaction pair of forces acting on the astronaut as the

space shuttle accelerates upwards. (2) 6.2 Calculate the resultant force on the space shuttle during the first 400 s of

its flight. (5) 6.3 Give TWO reasons why the space shuttle’s acceleration increases as it

moves away from the surface of the earth. (2) [9]



QUESTION 7 (Start on a new page.) The diagram below shows the earth (E) and the moon (M) a distance ‘r’ apart. The dotted line shows the path of the moon’s orbit around the earth. The mass of the moon (MM) is 7,35 x 1022 kg and the mass of the earth (ME) is 6 x 1024 kg. The distance r between their centres is 3,84 x 109 m.

7.1 Calculate the gravitational force, FEM, between the moon and the earth when

their centres are a distance r apart. (4) 7.2 Name Newton’s law that you applied in QUESTION 7.1. (1) 7.3 Give a reason why the gravitational force exerted by the earth on the moon

does not stay constant as the moon orbits the earth. (1) 7.4 An asteroid with a mass of ¼ MM moves at a distance of 2r from the centre

of the earth. Without using any of the values provided, determine the force between the earth and the asteroid in terms of FEM. (3)

[9]



QUESTION 8 (Start on a new page.) Learners passed light from a ray box through a rectangular glass block to verify Snell’s law. The results obtained are used to draw the graph below.

8.1 Write down the independent variable. (1) 8.2 Write down the variable that must be controlled. (1) 8.3 Write down the conclusion that can be obtained from the graph. (2) 8.4 USE THE GRAPH to determine the refraction index of the glass block. (4) [8]



QUESTION 9 (Start on a new page.) A diamond glitters in light because of its high refraction index of 2,42 and the small critical angle of the diamond-air boundary.

9.1 Calculate the critical angle of a diamond-air interface.

Take the refraction index of air as 1. (4) 9.2 The angle of incidence of light at the diamond-air interface is increased to 30°.

Redraw the diagram and complete the path of ray of light.

(2) 9.3 Name the phenomenon that will now be observed at the diamond-air interface. (1) 9.4 What are the TWO CONDITIONS that are necessary for the phenomenon in

QUESTION 9.3 to happen? (2) 9.5 Name the medical instrument used to examine internal parts of the body that

makes use of the phenomenon in QUESTION 9.3. (1) [10]



QUESTION 10 (Start on a new page.) Learners pass a green light with wavelength 534 nm through a single slit of width 1,8 x 10-4 m and shines it on a screen. They set up the apparatus as shown in the diagram below.

10.1 Define the term monochromatic light. (1) 10.2 Draw the observed pattern and indicate the positions of constructive and

destructive interferences. (4) 10.3 How will the broadness of the central band change if … (Write only INCREASES, DECREASES or REMAINS THE SAME) 10.3.1 the green light is replaced by a red light? (1) 10.3.2 the slit width is smaller?

Explain your answer. (2)

[8]



QUESTION 11 (Start on a new page.) Two charges P and Q with charges +4 nC and -2 nC respectively, are placed 8 cm apart. A point charge X is placed 2 cm from Q as shown in the sketch below.

11.1 Define the term electric field at a point. (2) 11.2 Draw the electric field pattern between two oppositely charged particles. (3) 11.3 Calculate the net electric field strength at point X due to P and Q. (7) 11.4 Calculate the magnitude and direction of the electrostatic force on an

electron if it is placed at point X. (3) [15] QUESTION 12 (Start on a new page.) A single circular loop of wire, 12 cm in diameter, is placed in a 0,6 T magnetic field. It is removed from the magnetic field in 0,04 s. 12.1 Calculate: 12.1.1 The flux which is linked to this coil. (4) 12.1.2 The average induced emf. (4) 12.2 How does the emf change if …

(Write only INCREASES, DECREASES or REMAINS THE SAME)

12.2.1 the magnetic field strength changes to 0,5 T? (1) 12.2.2 the coil is removed from the field in 0,02 s? (1) [10]



QUESTION 13 (Start on a new page.) In the circuit below, the resistance of the battery, ammeter and connecting wires can be ignored.

The power of the 8 Ω resistor is 0,5 W. 13.1 Calculate the reading on the: 13.1.1 voltmeter (V) (8) 13.1.2 ammeter (A) (4) 13.2 A television is labelled: 240 V; 750 W. 13.2.1 Calculate the resistance of the television’s resistor. (4) 13.2.2 Calculate the costs of using the television for 6 hours if electricity

costs R1,04 per kWh. (3) [19] TOTAL: 150



DATA FOR PHYSICAL SCIENCES GRADE 11

GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 11 PAPER 1 (PHYSICS) / 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 Swaartekragkonstante G 6,67 x 10-11 N.m².kg-2

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

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

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

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

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

Radius of earth Radius van aarde RE 6,38 x 106 m

Mass of earth Massa van aarde Mg 5,98 x 1024 kg

TABLE 2: FORMULAE/TABEL 2: FORMULES

MOTION/BEWEGING

vf = vi + a∆t ∆x = vi ∆t + ½a∆t² vf²= vi ²+ 2a∆x or/of vf²= vi ²+ 2a∆x

vf + vi ∆t ∆x = 2

FORCE/KRAG

Fnet = ma w = mg

fs(max) = μsN

fk = μkN WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG

n1 sin θ1 = n2 sin θ2



ELECTROSTATICS/ELEKTROSTATIKA

F = (k = 9,0 x 109 N.m².C-²) E =

E = (k = 9,0 x 109 N.m².C-²) E =

ELECTROMAGNETISM/ELEKTROMAGNETISME

Φ = BA cos θ ɛ = - N ∆Φ ∆t

CURRENT ELECTRICITY/STROOMELEKTRISITEIT R =

R = R1 + R2 + ......

I =

=

+ + + ......

W = Vq W = VI∆t W = I² R ∆t W = V² ∆t R

P =

P = VI P = I² R P = V² R

NATIONAL SENIOR CERTIFICATE

GRADE/GRAAD 11

NOVEMBER 2015

PHYSICAL SCIENCES P1/ FISIESE WETENSKAPPE V1

MEMORANDUM

MARKS/PUNTE: 150

This memorandum consists of 10 pages./ Hierdie memorandum bestaan uit 10 bladsye.

2 PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 (EC/NOVEMBER 2015)


QUESTION/VRAAG 1 1.1 A 99 1.2 A 99 1.3 A 99 1.4 D99 1.5 B 99 1.6 B 99 1.7 D 99 1.8 D 99 1.9 A 99 1.10 C 99 [20]

QUESTION/VRAAG 2 2.1 The resultant of the forces is zero. 9

Die resultant van die kragte is nul. 9 (1) 2.2

T 9 20° 45° FL/FB9

Fg 9

ACCEPTED LABELS/AANVAARDE BYSKRIFTE Fg

Fw/W/Weight/mg/gravitational force/force due to gravity Fw/W/gewig/mg/gravitasiekrag

T FT/Tension/Tension in rope FT/Spanning/Spanning in tou

FL FB

Flegs/Force of legs on wall Fbene/Krag van bene op muur (3)

(EC/NOVEMBER 2015) PHYSICAL SCIENCES P1/FISIESE WETENSKAPPE V1 3

Kopiereg voorbehou Blaai om asseblief

2.3 Fg = mg = (50)(9,8) = 490 N 9

SCALE/SKAAL 10 mm = 50 N Fg = 490 N = 98 mm (Accept/Aanvaar 97 – 99mm)

FL 9 45° T 9 Fg = 9,8mm 9

20°

T = 76,5 × 5 = 382,5 N 9 (Accept/Aanvaar 76,4 – 76,6 mm) FL = 37,0 × 5 = 185 N 9 (Accept/Aanvaar 184,5 – 186,5 mm) (8)

NOTES/AANTEKENINGE

x Mark awarded for correct label and direction./ x Punt toegeken vir korrekte benoeming en pyltjie.

x ANY TWO angles indicated./99 x ENIGE TWEE hoeke aangedui. 99

OR/OF CALCULATIONS instead of CONSTRUCTION/ max/maks

BEREKENINGE in plaas van KONSTRUKSIES Fg = FT FT = 4909 sin 45° 9 = 382,50 N 9

sin 115° sin 45° sin 115°

Fr = Fg FR = 490 sin 20° 9 = 184,91 N 9 sin 20° sin 115° sin 115° [12]



QUESTION/VRAAG 3

3.1 fs = Fg‖ = mg sin = (60)(9,8) sin 30° 9 = 294 N 9

Any one/Enige een 9 (3)

3.2 s = tan θ 9= tan 30° 9= 0,58 9

OR/OF fs = μs FN = μs Fg = μs mg cos 9 294 = μs (60)(9,8) cos 30° 9 μs = 0,58 9 (3)

3.3 3.3.1 REMAINS THE SAME/BLY DIESELFDE 9 (1) 3.3.2 DECREASES/NEEM AF 9

Θ Fg will decrease, 9 therefore FN will decrease/fk FN 9 Fg sal afneem, 9 dus sal FN afneem 9/fk FN 9 (3)

[10] QUESTION/VRAAG 4 4.1 When a resultant/net force acts on an object, the object accelerates in the

direction of the force. This acceleration is directly proportional to the force 9 and inversely proportional to the mass of the object. 9

Indien ŉ resulterende/netto krag op ŉ voorwerp inwerk, sal die voorwerp in die rigting van die krag versnel. Hierdie versnelling is direk eweredig aan die (resulterende/netto) krag 9 en omgekeerd eweredig aan die massa van die voorwerp. 9

OR/OF The net force acting on an object is equal to the rate of change of momentum 99 of the object (in the direction of the force). (2 or 0)

Die netto krag wat op ŉ voorwerp inwerk is gelyk aan die tempo van verandering in momentum 99 van die voorwerp (in die rigting van die krag). (2 of 0) (2)

4.2 N/FN9

f/Ff 9 T 9 w/Fg/mg 9 (4)

NOTES/AANTEKENINGE x Mark awarded for label and arrow./Punt toegeken vir benoeming en

pyltjie. x Any other additional force(s)/Enige ander addisionele krag(te).

Max/Maks ¾ x If force(s) do not make contact with body/Indien krag(te) nie met die

voorwerp kontak maak nie. Max/Maks ¾



4.3 4.3.1 OPTION/OPSIE 1

For the 6kg block./Vir 6kg blokkie: Fnet = ma9 T + (-ff ) = ma T – 11,769 = 6a ...........(1) For the 2kg block./Vir 2kg blokkie: 9 For either/vir of/6a or/of 2a Fnet = ma 2 + w + (-T) = ma 2 + (2 × 9,8) – T9 = 2a T = 21,6 – 2a .......(2) Substitute (2) into (1)/Stel (2) in (1): 21,6 – 2a – 11,76 = 6a a = 1,23 m.s-2 9

OPTION/OPSIE 2:

Fnet = ma9 2 + w +(-Ff ) 9 = ma 2 + (2 × 9,8) -Ff 9= 8a9 2 + (2 × 9,8) – 11,76 = 8a a = 1,23 m.s-2 9 (5)

4.3.2 POSITIVE MARKING from QUESTION 4.3.1

POSITIEWE NASIEN van VRAAG 4.3.1 T = 21,6 – 2a = 21,6 – 2(1,23) 9 = 19,14 N9

OR/OF T – 11,76 = 6a T = 11,76 + 6(1,23) 9 = 19,14 N9 (2)

4.4 INCREASES/TOENEEM 9 (1) [14] QUESTION/VRAAG 5 5.1 BACKWARDS/AGTERTOE 9 (1) 5.2 An object continues in a state of rest or uniform velocity, 9 unless it is acted

upon by an external unbalanced force (resultant force). 9 ŉ Voorwerp sal in ŉ toestand van rus bly of teen ŉ konstante snelheid bly voortbeweeg, 9 tensy ŉ eksterne ongebalanseerde (resulterende krag) daarop inwerk. 9

OR/OF The tendency of an object to maintain its state of rest 9 or constant motion in a straight line. 9 Die weerstand van ŉ voorwerp wat enige verandering in sy toestand rus 9 of beweging teenstaan. 9 (2)



5.3 5.3.1 EQUALS/GELYK 9 (1) 5.3.2 Kay will have a greater acceleration, 9 because her mass is

smaller. 9 Kay sal ŉ groter versnelling het, 9 want haar massa is kleiner. 9 (2)

[6] QUESTION/VRAAG 6

6.1 Action: Force of seat on astronaut (upwards) 9

Reaction: Force of astronaut on seat (downwards) 9 Aksie: Krag van sitplek op ruimtevaarder (opwaarts) 9 Reaksie: Krag van ruimtevaarder op sitplek (afwaarts) 9 (2)

6.2 FRES = m ∆v 9= (2,6 × 106)(160 – 0) 9 = 1,04 × 106 N 9 upwards/opwaarts 9

∆t 400 9

Mark direction independently/Merk rigting onafhanklik

(5) QUESTION/VRAAG 7 7.1 F = G m1.m2 9

r2

= (6,67 × 10-11)(7,35 × 1022)(6 × 1024) 9 (3 × 109)29 = 1,99 × 1018 N 9 (4)

7.2 NEWTON’s LAW OF GRAVITATION/NEWTON se GRAVITASIEWET. 9 (1) 7.3 The distance between the earth and the moon changes due to the elliptical

orbit. 9 Die afstand tussen die aarde en die maan verander as gevolg van elliptiese omwenteling. 9

OR/OF F α 1/r2 9 (1)

7.4 FEM(new) = GME1/4 MM 9 = FEM (1/4)(1/4) 9 = 1/16 FEM 9

(2r)2

OR/OF FEM(new) = 1/16 (1,99 × 1018) 9

OR/OF FEM(new) = 1,24 × 1017N 9 (1,24375 × 1017 N) (3)

[9]



QUESTION/VRAAG 8 8.1 Angle of incidence/ i/i/sin i 9

Invalshoek/ i/i/sin i 9 (1) 8.2 Type of block used/Temperature of the surroundings/Source of light/Surface

on which block is placed. 9. Tipe blok gebruik/Temperatuur van die omgewing/Bron van die lig/Oppervlak waarop blok geplaas word.9 (1)

8.3 Snell’s law has been verified:

* The graph is a straight line through the origin, because sin i sin r. 9 * Gradient = sin i = constant 9

sin r

Snell se wet is bewys: x Die grafiek vorm ŉ reguitlyn deur die oorsprong, omdat sin i sin r. 9

x Gradiënt = sin i = konstante 9 sin r (2)

8.4 Gradient/Gradiënt = n = Δ sin i9 = 0,77 – 0 9 = 1,51 9

Δ sin r 0,51 – 0 9 (4) [8] QUESTION/VRAAG 9 9.1 n sin i = n sin r 9

2,42 sin i 9 = 1 sin 90° 9 i = sin-1 2,42 = 24,2° 9 (4)

9.2

99

No arrows: max Geen pyltjies: maks

(2)

9.3 Total internal reflection 9/Totale interne weerkaatsing 9 (1) 9.4 Light moves from medium of higher optical density to medium of lower

optical density. 9 Angle of incidence must be larger than critical angle. 9 Lig beweeg van medium van hoër optiese digtheid na medium met laer optiese digtheid. 9 Invalshoek moet groter wees as kritiese hoek. 9 (2)

9.5 Endoscope/Endoskoop 9 (1) [10]



QUESTION/VRAAG 10 10.1 Light consisting of only one wavelength. 9

Lig wat uit net een golflengte bestaan. 9 (1) 10.2 B

A Broader Central band Breër sentraleband 9

Alternative light (bright) and dark bands.

Afwisselende helder (lig) en donker bande 9 A: central band and bright (light) bands = CONSTUCTIVE INTERFERENCES 9

sentraleband en helder (lig)bande = KONSTRUKTIEWE INTERFERENSIE/ STEURINGE 9

B: dark bands = DESTRUCTIVE INTERFERENCES 9

donkerbande = DESTRUKTIEWE INTERFERENSIE/STEURINGE 9 (4) 10.3 10.3.1 INCREASES/NEEM TOE 9 (1) 10.3.2 INCREASES/NEEM TOE 9

Θ The narrower the opening, the greater the degree of diffraction. 9 Hoe smaller die opening, hoe groter die mate van diffraksie. 9

OR/OF

Degree of diffraction 1 9 Width of opening Mate van diffraksie 1 9 Wydte van opening (2)

[8] QUESTION/VRAAG 11 11.1 Force experienced per unit charge at a point in the field. 99

Krag ondervind per eenheidslading by ŉ punt in die veld. 99

OR/OF It is the force experienced by a unit positive charge placed at that point. 99 Dit is die krag wat deur ŉ eenheids positiewe lading geplaas by daardie punt ondervind word. 99 (2)



11.2

CRITERIA FOR SKETCH/KRITERIA VIR SKETS MARKS/PUNTE Correct shape / Korrekte vorm 9 Direction from positive to negative

Rigting van positief na negatief 9

Fieldlines start on sphere and do not cross

Veldlyne begin op elke sfeer en kruis nie 9

(3) 11.3 Ep = kQ = (9 × 109)(4 × 10-9) 9 = 3 600 N.C-1 to the right/na regs

9 r2 (0,10)2 9 Any ONE Enige EEN EQ = kQ = (9 × 109)(2 × 10-9) 9 = 45 000 N.C-1 to the left/na links r2 (0,02)2 9 Enet = 3 600 – 45 000 = 41 400 N.C-1 9 to the left/na links 9 (7)

11.4 POSITIVE MARKING FROM QUESTION 11.3

POSITIEWE NASIEN VAN VRAAG 11.3 E = F 9 Q 41 400 = F 9 1,6 × 10-19 F = 6,62 × 10-15 N 9 (3)

[15] QUESTION/VRAAG 12 12.1 12.1.1 Φ = BA cos θ 9

= (0,6) r2 cos θ = (0,6)( × 0,062 ) 9cos 0° 9 = (0,6)( 0,01 ) = 6,79 × 10-3 Wb 9 (4)

12.1.2 POSITIVE MARKING FROM QUESTION 12.1.1

POSITIEWE NASIEN VAN VRAAG 12.1.1 = -N Δ Φ 9 Δt = (-1)(0 - 6,79 × 10-3) 9 0,04 9 = 0,17 V 9 (4)



1.2.2 12.2.1 DECREASES/NEEM AF 9 (1) 12.2.2 INCREASES/NEEM TOE 9 (1) [10] QUESTION/VRAAG 13 13.1 13.1.1 P = I2R 9

0,5 = I2 8 9 I = 0,25 A 9 1 = 1 + 1 9 = 1 + 1 9 = 3 Rp = 4 Ω 9 R p r1 r2 6 (6 + 6) 12 RT = 8 + 4 9 = 12 Ω VT = I RT = (0,25) (12) = 3V 9 (8)

13.1.2 Vp = I R 9 = (0,25)(4) 9 = 1V

I = Vp = 1 9 = 0,17 A 9 R 6 (4)

13.2 13.2.1 P = V2 9

R 7509 = 2402 9 R R = 76,80 Ω 9

OR/OF P = VI 750 9 = 240 I 9 9 I = 3,13 A R = V I = 240 3,13 = 76,80 Ω 9 (4)

13.2.2 E = Px t = 750 × 6 9 = 4 500 Wh = 4,5 kWh

Cost/Koste = 4,5 × 1,14 9 = R 5,13 9

OR/OF Cost/Koste = (0,75)(6) 9 (1,14) 9= R5,13 9 (3)

[19] TOTAL/TOTAAL: 150

GRADE 11 NOVEMBER 2015 PHYSICAL SCIENCES P1 · GRADE 11 NOVEMBER 2015 PHYSICAL SCIENCES P1 MARKS:...

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