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NATIONAL
SENIOR CERTIFICATE/
NASIONALE
SENIOR SERTIFIKAAT
GRADE/GRAAD 11
PHYSICAL SCIENCES: PHYSICS (P1)
FISIESE WETENSKAPPE: FISIKA (V1)
NOVEMBER 2018
MARKING GUIDELINES/NASIENRIGLYNE
MARKS/PUNTE: 150
These marking guidelines consist of 16 pages.
Hierdie nasienriglyne bestaan uit 16 bladsye.
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Physical Sciences P1 Grade 11 Nov 2018 Memo Afr & Eng_hlayiso.com_.pdf
Physical Sciences · Grade 11 · North West June Exam · 2018. Memorandum, 16 pages. Read online or download the PDF.
- Subject
- Physical Sciences
- Grade
- Grade 11
- Document type
- Memorandum
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- 2018
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Physical Sciences P1/Fisiese Wetenskappe V1 2 DBE/November 2018
CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne
QUESTION 1/VRAAG 1
1.1 A (2)
1.2 C (2)
1.3 C (2)
1.4 D (2)
1.5 B (2)
1.6 D (2)
1.7 B (2)
1.8 A (2)
1.9 B (2)
1.10 C (2)
[20]
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Physical Sciences P1/Fisiese Wetenskappe V1 3 DBE/November 2018
CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne
QUESTION 2/VRAAG 2
2.1 Resultant (net) vector/Resultante (netto) vektor (1)
2.2.1 F y = Fsin θ
= 50sin 30° OR/OF 50cos 60º
= 25 N (2)
2.2.2 POSITIVE MARKING FROM QUESTION 2.2.1
POSITIEWE NASIEN VANAF VRAAG 2.2.1
F x = 50cos 30°
= 43,3 N
R x = 80 – 43,3
= 36,7 N Substitution marks awarded within the question
even if calculations for Fx and Rx are wrong
F net 2 = R x 2 + F y 2 Substitusiepunte toegeken in die vraag selfs indien
= 36,72 + 252 berekeninge vir Fx en Rx verkeerd bereken word.
= 44,41 N (5)
2.2.3 POSITIVE MARKING FROM QUESTION 2.2.1 AND 2.2.2
POSITIEWE NASIEN VANAF VRAAG 2.2.1 EN 2.2.2
OPTION 1/OPSIE 1
36,7
tan θ = Fnet = 44,41 N
Ry = 25 N
25
θ = 55,74° θ
ϕ
Rx = 36,7 N
OPTION 2/OPSIE 2
25
cos θ =
44,41
θ = 55,74°
OPTION 3/OPSIE 3
36,7
sin θ = Accept direction as /Aanvaar rigting as
44,41
θ = 55,74° ϕ = 90º - θ
= 34,26º
OPTION 4/OPSIE 4
25
cos θ =
44,41
θ = 55,74°
(2)
[10]
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Physical Sciences P1/Fisiese Wetenskappe V1 4 DBE/November 2018
CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne
QUESTION 3/VRAAG 3
3.1 The force that opposes the motion of a moving object relative to a surface.
Die krag wat die beweging van 'n bewegende voorwerp relatief tot 'n
oppervlak teenwerk.
[2 or/of 0] (2)
3.2 A body will remain in its state of rest or motion at constant velocity unless
a non-zero resultant/net force acts on it.
'n Liggaam sal in sy toestand van rus of beweging teen konstante snelheid
bly/volhard tensy 'n nie-nul resulterende/netto krag daarop inwerk.
[Penalise -1 if key words/phrase is omitted/
Penaliseer -1 indien sleutelwoorde/frase is uitgelaat] (2)
3.3 F x = 90cos 50° OR/OF 90sin 40º
= 57,85 N (2)
3.4 N = Fg - Fy NOTE/NOTA:
N = 45(9,8) - 90sin 50° Weight and the vertical component can be
N = 372,06 N calculated separately, award one mark each even (4)
if the formula for N is incorrect
Gewig en vertikale komponent kan apart bereken
word, een punt elk selfs indien die formule vir N
verkeerd is.
3.5 POSITIVE MARKING FROM QUESTION 3.3 and 3.4
POSITIEWE NASIEN VANAF VRAAG 3.3 en 3.4
fk = µkN
57,85 = µ k (372,06)
µ k = 0,16 (4)
3.6 No The coefficient is dependent on the (nature of) the surfaces / type of
material in contact.
Nee. Die koëffisiënt is afhanklik van die (tipe) oppervlakke / soort materiaal
in kontak. (2)
[16]
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Physical Sciences P1/Fisiese Wetenskappe V1 5 DBE/November 2018
CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne
QUESTION 4/VRAAG 4
4.1 When a resultant/net force acts on an object, the object will accelerate in the
direction of the force. The acceleration is directly proportional to the net force
and inversely proportional to the mass of the object.
Wanneer 'n resulterende/netto krag op 'n voorwerp inwerk, sal die voorwerp in
die rigting van die krag versnel. Die versnelling is direk eweredig aan die netto
krag en omgekeerd eweredig aan die massa van die voorwerp.
[Penalise -1 if key words/phrase is omitted/
Penaliseer -1 indien sleutelwoorde/frase is uitgelaat] (2)
4.2 Accept any set of coordinates from the graph, e.g.:
Aanvaar enige kombinasie van koördinate vanaf die grafiek, bv.:
2,5 – 0
Gradient/Helling = =2
1,25 – 0
OR/OF
2,1 – 1,7
Gradient/Helling = =2
1,05 – 0,85
(3)
4.3 OPTION 1/OPSIE 1
1 1
Gradient/Helling = = =2
ma Fnet
1
F net = = 0,5 N Accept/Aanvaar F net = 0,5 N
2
OPTION 2/OPSIE 2
F net = ma
Accept any coordinates from graph
= (1)(1/ 2 )
Aanvaar enige koördinate vanaf grafiek
= 0,5 N (2)
4.4 Acceleration is inversely proportional to the mass of an object (if the net force
is kept constant)
Accept: The inverse of acceleration is directly proportional to the mass of the
object (if the net force is kept constant)
1
OR a α m
Versnelling is omgekeerd eweredig aan die massa van die voorwerp (indien
die netto krag konstant bly)
Aanvaar: Die omgekeerde van die versnelling is direk eweredig aan die
massa van die voorwerp (indien die netto krag konstant bly)
1
OF a α m (2)
[9]
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Physical Sciences P1/Fisiese Wetenskappe V1 6 DBE/November 2018
CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne
QUESTION 5/VRAAG 5
5.1
N OR/OF N
f f
Fgll
Fg⊥
Fg
(3)
Accepted Labels/Aanvaarbare Byskrifte Mark/Punt
w weight/F G /F g OR Both components for one mark
gewig/gravitasiekrag/swaartekrag OF
Beide komponente vir een punt
N Normal force/F N
Normaalkrag/F N
f Friction/F f
Wrywingskrag/F f
Any additional force: deduct 1 mark (maximum ⅔)
Enige addisionele krag: trek 1 punt af (maksimum ⅔)
Omission of arrow heads: deduct 1 mark (maximum ⅔)
Pylpunte uitgelaat: trek 1 punt af (maksimum ⅔)
Lines must touch object otherwise (maximum ⅔ )
Lyne moet voorwerp raak anders (maksimum ⅔)
Do not penalise if vectors are not to scale
Moenie penaliseer indien vektore nie op skaal is nie
5.2 F net = ma Any one/Enige een
mgsin θ = ma
25(9,8)sin 15° = 25a NOTE/NOTA:
a = 2,54 m·s-2 Award one mark for the parallel
OR/OF component if calculated separately
25(9,8)cos 75º = 25a Ken een punt toe indien die parallel
-2
a = 2,54 m·s komponent apart bereken is (4)
5.3 Up the slope/Teen die helling op (1)
5.4 F net = ma Any one/Enige een
F g// + (-f) = ma
25(9,8)sin 15° - f = 25(-1,2)
f = 93,41 N
OR/OF
25(9,8)cos 75° - f = 25(-1,2)
f = 93,41 N
Note/Let wel:
Accept if calculation is done with direction up the slope as positive
Aanvaar indien berekening gedoen is met rigting teen die helling op as
positief (4)
[12]
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Physical Sciences P1/Fisiese Wetenskappe V1 7 DBE/November 2018
CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne
QUESTION 6/VRAAG 6
6.1 Each particle in the universe attracts every other particle with a
gravitational force that is directly proportional to the product of their masses
and inversely proportional to the square of the distance between their
centres.
Elke deeltjie in die heelal trek elke ander deeltjie aan met 'n krag wat direk
eweredig is aan die produk van hulle massas en omgekeerd eweredig is aan
die kwadraat van die afstand tussen hulle middelpunte.
[Penalise -1 if key words/phrase is omitted/
Penaliseer -1 indien sleutelwoorde/frase is uitgelaat] (2)
6.2 Gm1m 2
F=
r2
�6,67 x 10-11 ��6,39 x 1023 �(m)
3 338 =
3 2
(3 390 x 10 )
m = 900 kg
OR/OF
Gm
g=
r2
�6,67 x 10-11 ��6,39 x 1023 �
g=
3 2
(3 390 x 10 )
g = 3,71 m·s-2
F g = mg
3 338 = m(3,71)
m = 900 kg (899,73 kg) (4)
6.3 POSITIVE MARKING FROM QUESTION 6.2
POSITIEWE NASIEN VANAF VRAAG 6.2
w = mg
= 900(9,8)
= 8 820 N (2)
[8]
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Physical Sciences P1/Fisiese Wetenskappe V1 8 DBE/November 2018
CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne
QUESTION 7/VRAAG 7
7.1 Refraction/Refraksie (1)
7.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2
n i sin θ i = n r sin θ r sin θi
n=
1sin θ i = 1,33sin 40º sin θr
θ i = 58,75° sin θi
1,33 =
sin 40°
sin θ i = 1,33sin 40º
θ i = 58,75°
Therefore the angle between ray and surface/Daarom is die hoek tussen
invallende straal en oppervlak
θ = 90°- 58,75°
= 31,25° (4)
7.3 n i sinθ i = n r sinθ r
1,33sin 40° = n r sin 35º
n r = 1,49 (3)
7.4
58,75º air/lug
40º water
40º
glass/glas
35º
Allocation of marks/Toekenning van punte: (5)
Light ray bends towards normal in water
Ligstraal breek na die normaal in water
Light ray bends further towards normal in glass
Ligstraal breek nog meer na die normaal in glas
Angle of incidence 58,75º shown (OR 31,25º)
Invalshoek 58,75º aangedui (OF 31,25º)
Angles in water (40º)
Hoeke in water (40º)
Angle in glass (35º)
Hoeke in glas (35º)
If normal lines are not indicated, penalise with one mark
Indien normaal lyne nie aangedui is nie, penaliseer met
een punt
If arrows are omitted, penalise -1 (maximum 4/ 5 )
Indien pylpunte weggelaat word, penaliseer -1 (maks 4/ 5 )
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Physical Sciences P1/Fisiese Wetenskappe V1 9 DBE/November 2018
CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne
7.5 c
n=
v
3 × 10 8
1,5 =
v
v = 2 x 108 m·s-1 (3)
7.6 No/Nee (1)
[17]
QUESTION 8/VRAAG 8
8.1 Diffraction is the ability of a wave to spread out in wave fronts as the
wave passes through a small aperture or around a sharp edge.
Diffraksie is die vermoë van 'n golf om uit te sprei in golffronte soos wat die
golf deur 'n klein opening of om 'n skerp rand/kant beweeg. (2)
8.2 Criteria for investigative question/Riglyne vir ondersoekende vraag
The dependent and independent variables are stated correctly.
Die afhanklike en onafhanklike veranderlikes korrek genoem.
Ask the relationship between the dependent and independent variables
in a question, not as a statement. The question may not be written in a
way that the answer is yes or no.
Vra die verband tussen die afhanklike en onafhanklike veranderlike as 'n
vraag, nie 'n stelling nie. Die vraag mag nie op so 'n manier geformuleer
word dat die antwoord ja of nee is nie.
Dependent variable/Afhanklike veranderlike: degree of diffraction/mate
van diffraksie
Independent variable/Onafhanklike veranderlike: wavelength/golflengte
Examples/Voorbeelde:
What is the relationship between the wavelength of a light ray and the degree
of diffraction?
Wat is die verband tussen die golflengte van 'n ligstraal en die mate van
diffraksie?
OR/OF
How does a change in wavelength affect the degree of diffraction?
Hoe beïnvloed 'n verandering in golflengte die mate van diffraksie? (2)
8.3 Degree of diffraction is directly proportional to the wavelength.
Mate van diffraksie is direk eweredig aan die golflengte.
OR/OF
Degree of diffraction α λ.
Mate van diffraksie α λ. (2)
8.4 Red/Rooi (1)
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Physical Sciences P1/Fisiese Wetenskappe V1 10 DBE/November 2018
CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne
8.5 Line should indicate inverse proportionality
Lyn moet omgekeerde eweredigheid aandui (2)
Degree of
diffraction/
Mate van
diffraksie
Slit width (mm)/
Spleetwydte (mm)
[9]
QUESTION 9/VRAAG 9
9.1 The magnitude of the electrostatic force exerted by two point charges on each
other is directly proportional to the product of the (magnitudes of the) charges
and inversely proportional to the square of the distance between them.
Die grootte van die elektrostatiese krag wat deur twee puntladings op mekaar
uitgeoefen word, is direk eweredig aan die produk van die (groottes van die)
ladings en omgekeerd eweredig aan die kwadraat van die afstand tussen
hulle.
[Penalise -1 if key words/phrase is omitted/
Penaliseer -1 indien sleutelwoorde/frase is uitgelaat]
NOTE: If learners refers to masses, no marks awarded
NOTA: Indien leerder na massa verwys, geen punte (2)
9.2
OR/OF
20º T
20º
Fg T
FE
Fg
FE
Accepted Labels/Aanvaarbare Byskrifte Mark/Punt
w weight/F G /F g
gewig/gravitasiekrag/swaartekrag
T Tension/F T
Spanning/F T
FE Electrostatic force
Elektrostatiese krag
One angle indicated
Een hoek aangedui (4)
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Physical Sciences P1/Fisiese Wetenskappe V1 11 DBE/November 2018
CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne
9.3 OPTION 1/OPSIE 1
If F A and F B were used/Indien F A en F B gebruik word
kQ1Q 2
F= NOTE/NOTA:
r2 Due to information given in the question,
(9 x 109 )(8 x 10-9 )(7 x 10-9 ) accept all possible options
=
0,03 2
As gevolg van die inligting in die vraag
= 5,60 x 10-4 N gegee, aanvaar alle moontlike opsies
OPTION 2/OPSIE 2
If F g and F E were used/Indien F g en F E gebruik word
F g = mg
= (0,2 x 10-3)(9,8)
= 1,96 x 10-3 N
F E = (1,96 x 10-3)tan 70º
= 7,13 x 10-4 N
OPTION 3/OPSIE 3
If F g and F E were used/Indien F g en F E gebruik word
F g = mg
= (0,2 x 10-3)(9,8)
= 1,96 x 10-3 N
FE Fg
=
sin 20° sin 70°
FE (1,96 x 10-3 )
=
sin 20° sin 70°
F E = 7,13 x 10-4 N (4)
9.4 POSITIVE MARKING FROM QUESTION 9.3
POSITIEWE NASIEN VANAF VRAAG 9.3
OPTION 1/OPSIE 1 OPTION 2/OPSIE 2
Using F g and F E /Gebruik F g en F E Using F g and angle/Gebruik F g en
F g = mg hoek
= (0,2 x 10-3)(9,8) Fg
-3 T=
= 1,96 x 10 N sin 70°
T2 = (1,96 x 10-3)2 + (5,6 x 10-4)2 1,96 x 10-3
T =
sin 70°
T = 2,04 x 10-3 N -3
T = 2,09 x10 N
OPTION 3/OPSIE 3
NOTE/NOTA:
Using F E and angle/Gebruik F E en
Due to information given in the
hoek
FE question, accept all possible
T= options
cos 70°
As gevolg van die inligting in die
-4 vraag gegee, aanvaar alle
5,6 x 10
T= moontlike opsies
cos 70°
T = 1,64 x10-3 N
(3)
[13]
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Physical Sciences P1/Fisiese Wetenskappe V1 12 DBE/November 2018
CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne
QUESTION 10/VRAAG 10
10.1
+
Criteria for marking/Nasienkriteria
Shape of the field (minimum of 4 field lines)
Vorm van veld (minimum van 4 veldlyne)
Direction of the field
Rigting van veld
Lines don't touch charge/lines cross etc. (maximum ½)
Lyne raak nie lading/lyne kruis ens. (maksimum ½) (2)
10.2.1 16 : 1 (1)
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Physical Sciences P1/Fisiese Wetenskappe V1 13 DBE/November 2018
CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne
10.2.2 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2
EP : ET kQ
16 : 1 Ep = 2
r
9 × 10 9 Q
Because/Omdat 4 × 10 6 =
1 r2
Eα 2 9 x 109Q = (4 x 106)r2 …(1)
r
rP : rT kQ
1:4 ET =
r2
r : r + 3 mm 5 9 × 10 9 Q
2,5 x 10 =
r = 1 mm (0,001 m) (r + 0,003 ) 2
9 x 109Q = (2,5 x 105)(r + 0.003)2 …(2)
Equation/Vergelyking (1) = (2)
(4 x 106)r2 = (2,5 x 105)(r + 0,003)2
16r2 = r2 + 0,006r + 9 x10-6
r = 1 mm (0,001 m)
(4)
10.2.3 POSITIVE MARKING FROM QUESTION 10.2.2
POSITIEWE NASIEN VANAF VRAAG 10.2.2
kQ kQ
Ep = OR/OF ET =
r2 r2
9 × 10 9 Q 9 × 10 9 Q
4 × 10 6 = 2,5 × 10 5 =
(0,001) 2 (0,004 ) 2
Q = 4,44 x 10-10 C Q = 4,44 x 10-10 C (2)
[9]
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Physical Sciences P1/Fisiese Wetenskappe V1 14 DBE/November 2018
CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne
QUESTION 11/VRAAG 11
11.1 The magnitude of the induced emf across the ends of a conductor is directly
proportional to the rate of change in the magnetic flux linkage with the
conductor.
Die grootte van die geïnduseerde emk oor die punte van 'n geleier is direk
eweredig aan die tempo van verandering van die magnetiese vloedkoppeling
met die geleier.
[2 or/of 0] (2)
11.2 ε=
− N Δϕ
Δt
− 400 Δϕ
7=
0,08
-3
∆Φ = - 1,4 x 10 Wb (-0,0014) (3)
11.3 POSITIVE MARKING FROM QUESTION 11.2
POSITIEWE NASIEN VANAF VRAAG 11.2
∆Φ = AB(cos θ f – cos θ i )
-0,0014 = (0,03)2B(cos 45º - cos 0º)
B = 5,31 T (4)
11.4 Increase/Toeneem (1)
11.5 - 1
εα
Δt
OR/OF
Emf is inversely proportional to time.
Emk is omgekeerd eweredig aan tyd.
If the time decreases, the emf will increase.
Indien die tyd verminder, sal die emk toeneem. (1)
11.6 North/Noord (1)
11.7 From A to B/Van A na B (1)
[13]
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Physical Sciences P1/Fisiese Wetenskappe V1 15 DBE/November 2018
CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne
QUESTION 12/VRAAG 12
12.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2
1 1 1 R1R 2
= + Rp =
R p R1 R 2 R1 + R 2
1 1 1 4R × 6R
= + 4,8 =
4,8 4R 6R 4R + 6R
R=2Ω R=2Ω (3)
12.2 POSITIVE MARKING FROM QUESTION 12.1
POSITIEWE NASIEN VANAF VRAAG 12.1
OPTION 1/OPSIE 1 OPTION 2/OPSIE 2
V 4R = IR 4R V 4R = IR 4R
= 1,8(4)(2) = 1,8(4)(2)
= 14,4 V = 14,4 V
V V
I6 R = IT =
R 6R RT
14,4 14,4
I 6R = IT =
12 4,8
= 1,2 A =3A
V 2R = IR I 2R = 3 – 1,8
= 1,2(4) = 1,2 A
= 4,8 V
V 2R = IR
= 1,2(4)
= 4,8 V
OPTION 3/OPSIE 3 OPTION 4/OPSIE 4
R1 : R2 V 4R = IR 4R
4:6 = 1,8(4)(2)
I1 : I2 = 14,4 V
6:4
6 R : 2R : 3R
x I = 1,8 A 1:2:3
10
IT = 3 A V R : V 2R : V 3R
1:2:3
I 2R = 3 – 1,8
= 1,2 A 2
V 2R = x 14,4
6
V 2R = IR = 4,8 V
= 1,2(4)
= 4,8 V (5)
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Physical Sciences P1/Fisiese Wetenskappe V1 16 DBE/November 2018
CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne
12.3 POSITIVE MARKING FROM QUESTION 12.1 AND 12.2
POSITIEWE NASIEN VANAF VRAAG 12.1 EN 12.2
OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 OPTION 3/OPSIE 3
W = I2R ∆ t W = VIΔt V 2 ∆t
2
= 1,8 (8)(120) = (14,4)(1,8)(120) W=
R
= 1036,8 J = 3110,4 J
(14,4) 2 (120 )
W=
8
W = 3110,4 J (3)
12.4 Decrease/Neem af (1)
-
12.5 The ammeter has such a low resistance
It short circuits the parallel part and all current flows through the ammeter.
OR
The ammeter short circuits the resistors
No current flows through resistor 2R
Die ammeter het so 'n lae weerstand
Dit kortsluit die parallelgedeelte en al die stroom vloei deur die ammeter.
OF
Die ammeter kortsluit die resistors
Daar vloei geen stroom deur resistor 2R nie (2)
[14]
TOTAL/TOTAAL: 150
Copyright reserved/Kopiereg voorbehou
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