NATIONAL
SENIOR CERTIFICATE/
NASIONALE SENIOR SERTIFIKAAT
GRADE/GRAAD 12
JUNE/JUNIE 2021
TECHNICAL SCIENCES P1/
TEGNIESE WETENSKAPPE V1
MARKING GUIDELINE/NASIENRIGLYN
(EXEMPLAR/EKSEMPLAAR)
MARKS/PUNTE: 150
This marking guideline consists of 9 pages./
Hierdie nasienriglyn bestaan uit 9 bladsye.
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Technical Sciences Grade 12 P1 EC June Exemplar 2021 Memo
Grade 12 Technical Sciences Paper 1 memorandum from the EC June Exemplar 2021 (Afrikaans), 9 pages. Read online or download the free PDF.
Technical Sciences · Grade 12 · EC June Exemplar · 2021 · Afrikaans. Memorandum, 9 pages. Read online or download the PDF.
- Subject
- Technical Sciences
- Grade
- Grade 12
- Language
- Afrikaans
- Document type
- Memorandum
- Year
- 2021
- Exam period
- EC June Exemplar
- Paper
- 1
- Curriculum
- CAPS
- Pages
- 9
- File size
- 1.0 MB
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EC June Exemplar 2021
Questions in this paper
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2 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/JUNE/JUNIE 2021)
QUESTION/VRAAG 1
1.1 A (2)
1.2 D (2)
1.3 C (2)
1.4 A (2)
1.5 B (2)
1.6 C (2)
1.7 D (2)
1.8 A (2)
1.9 D (2)
1.10 D (2)
[20]
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(EC/JUNIE/JUNIE 2021) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 3
QUESTION/VRAAG 2
2.1 2.1.1 F1 = Normal force/Normaalkrag (1)
2.1.2 F2 = Weight/Gewig (1)
2.2 2.2.1 Force of the block on the EARTH/Krag van die blok op die
AARDE (2)
2.2.2 F4 = mg
= 5 x 9,8
= 49 N (3)
2.3.1 Newton’s First Law/Newton se Eerste Wet
-
A body will remain at rest or continue moving with constant velocity unless
a net force acts on it.
ʼn Voorwerp sal in rus bly of beweeg teen ʼn konstante snelheid tensy ʼn
netto krag daarop inwerk. (3)
2.3.2 Fnet = 0 N (1)
[11]
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4 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/JUNE/JUNIE 2021)
QUESTION/VRAAG 3
3.1 3.1.1 To ensure that tension is uniform.
Om te verseker dat die spanning uniform is. (2)
3.1.2 A net force produces acceleration of an object in the direction of the net
force. The acceleration is directly proportional to the net force and
inversely proportional to the mass of the object.
ʼn Netto krag veroorsaak dat ʼn voorwerp versnel in die rigting van die
netto krag. Hierdie versnelling is direk eweredig aan die netto krag en
omgekeerd eweredig aan die massa van die
voorwerp. (2)
OR/OF
The net force acting on a body is equal to the rate of change of
momentum in the same direction as the net force.
Die netto krag wat op ʼn voorwerp inwerk is gelyk aan die tempo van
verandering van momentum in dieselfde rigting as die netto krag.
3.1.3 FH = F cosθ
= 40 cos 35°
= 32,77 N
Fnet = ma
80 kg mass/massa: 50 kg mass/massa:
Fnet = ma Fnet = ma
T – 6 = 80 a 32,77 – T = 50 a
a = 0,21 m.s-2 (7)
3.1.4 DECREASES Horisontal component (of 40 N force)
decreases
NEEM AF Horisontale komponent (van 40 krag) verminder. (3)
3.2 3.2.1 Mass (of the trolley)/Massa (van die trollie) (1)
3.2.2 a ∝ F (if m constant and net force decreases, the acceleration
decreases) OR (as net force increases acceleration increases)
a ∝ F (indien m konstant bly en netto krag afneem, neem die
versnelling af) OF (as netto krag toeneem sal versnelling toeneem) (2)
3.3 Experiment/Eksperiment B
Gradient of A > gradient of B / Gradiënt van A > gradiënt van B
1/mA > 1/mB
Therefore mA < mB / Dus mA < mB (4)
[21]
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(EC/JUNIE/JUNIE 2021) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 5
QUESTION/VRAAG 4
4.1 In an isolated system the total linear momentum remains constant.
(2 or 0)
In ʼn geïsoleerde sisteem bly die totale lineêre momentum konstant.
(2 of 0) (2)
4.2 Σ pi = Σ p f
m1v1i + m2v2i = m1v1f + m2v2f
16 000 + (2 000)(10,61) = (800)(12) + 2 000 v2f
v2f = 13,81 m.s-1 (4)
4.3 Ektotal before/voor = ½ mvA2 +½mvB2
= ½ x 800 x 202 + 112 500
= 272 500 J
Ektotal after/na = ½(800)122 + ½ (2 000)(13,81)2
= 248 316,1 J
Inelastic: Ek before > Ek after OR Ek before ≠ Ek after
Onelasties: Ek voor> Ek na OF Ek voor≠ Ek na (5)
4.4 Fnet Δt = Δp
Fnet = (9 600 – 16 000 )/0,15
= - 42 666,67 N
= 42 666,67 N, left/links (4)
[15]
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6 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/JUNE/JUNIE 2021)
QUESTION/VRAAG 5
5.1 Energy a body has because of its position above the ground/
Energie wat ʼn liggaam het as gevolg van sy posisie bo die grond. (2)
5.2 Ek = ½ mv2
= ½ x 0,5 x 19,82
= 98,01 J (3)
5.3 Ep lost/verloor = - Ek gained/bygekry
= - 98,01
98,01 = - mg(yf –yi)
98,01 = - 0,5 x 9,8(yf – 0)
yf = - 20 m
Emech bottom = Emech top (Emeg onder = Emeg bo)
(mgh + 1/2mv2)bottom/onder = (mgh + 1/2mv2)top/bo
0 + ½ (0,5 x 252) = ½ x 9,8 x hP
hP = 31,89 m
h = 31,89 + 20 = 51,89 m (8)
5.4 P =W/Δt
= 37 350/20
= 1 867,5 W
P = 1867,5 / 746 = 2,50 hp (4)
[17]
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(EC/JUNIE/JUNIE 2021) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 7
QUESTION/VRAAG 6
6.1 Stress/Druk
(1)
6.2 6.2.1 Strain/Rekking = Δl/L
= (405 – 400) /400
= 0,0125 (3)
6.2.2 Strain/Rekking = F/A
A = π r2
= π x (5/1000)2
= 7,85 x 10 -5 m2
Strain/Rekking = 30 000/7,85 x 10 -5
= 3,82 X 10 8 Pa (5)
6.3 6.3.1 Modulus of elasticity/Elastisiteitsmodulus (1)
6.3.2 Hooke’s law/Hooke se wet (1)
6.3.3 Within the limits of elasticity, stress is directly proportional
to strain
Binne die beperkings van elastisiteit is rekking direk eweredig aan
druk. (2)
6.4 6.4.1 B (1)
6.4.2 A (1)
6.5 6.5.1 Elasticity/Elastisiteit (1)
6.5.2 Elastic band, bow, trampoline, spring diving board (Any TWO)
Elastiese band, boog, trampolien, spring-duikplank (Enige TWEE) (2)
6.5.3 Restoring force/Herstelkrag (1)
[19]
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8 TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 (EC/JUNE/JUNIE 2021)
QUESTION/VRAAG 7
7.1 7.1.1 Viscosity is the property of a fluid to oppose relative motion between
two adjacent layers.
Viskositeit is die eienskap van ʼn vloeistof om die relatiewe beweging
tussen twee aangrensende lae teen te werk. (2)
7.1.2 R
Time of flow in R is highest / Vloeityd is die hoogste in R. (3)
7.1.3 smaller than/kleiner as
Viscosity increases with increase in temperature.
Viskositeit neem toe met ʼn toename in temperatuur. (3)
7.1.4 p = ρgh
= 1000 x 9.8 x 10
= 98 00 Pa (3)
7.2 7.2.1 In a continuous liquid at equilibrium, the pressure applied at a point
is transmitted equally to the other parts of the liquid.
In ʼn kontinue vloeistof in ewewig, word die druk wat by enige punt
toegepas word eweredig na die ander dele van die vloeistof
versprei. (2)
7.2.2 F1/A1= F2/A2
F1/(0,045) = 20 000/0,28
F1 = 3 214,29 N (4)
[17]
QUESTION 8/VRAAG 8
8.1 8.1.1 Communication/Kommunikasie (1)
8.1.2 Images of bones /Security machines at airports etcetera./Beelde
van bene /Sekuriteitmasjiene by lughawe ensovoorts (1)
8.2 Gamma rays/Gamma strale (1)
8.3 c = fλ
3 x 108 = 5 x 1017 λ
λ = 6 x 10-9 m (3)
8.4 E = hf
= h c/ λ
= 6.63 x10-34 x 3 x108/ 502x10-9
= 3.96 x 10-19 J (3)
[9]
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(EC/JUNIE/JUNIE 2021) TECHNICAL SCIENCES P1/TEGNIESE WETENSKAPPE V1 9
QUESTION/VRAAG9
9.1 9.1.1 Reflection/Weerkatsing (1)
Angle of incidence equal to angle of reflection/Invalshoek gelyk aan
9.1.2
weerkatsinghoek
Angle of incidence, angle of reflection and normal are in the same
plane/ Invalshoek, weerkaatsinghoek en normal is in dieselfde
beraming. (2)
b is the angle of reflection b = 60o /
9.1.3
b is die hoek van weerkatsing (2)
9.2 9.2.1 Breaking up of white light into its constituent colours /
Opbreek van wit lig in sy samestellende kleure (2)
9.2.2 Red Highest wavelength The higher the wavelength the
smaller the refraction/
Rooi Hoogste golflengte Hoe hoër die golflengte, hoe kleiner
die ligbreking (3)
9.3 9.3.1 Refraction /Ligbreking (1)
9.3.2 Critical angle /Kritieke hoek (1)
9.3.3 Total internal reflection /Totale interne weerkatsing (1)
9.3.4 Communications and medicine /Kommunikasie en medisyne (2)
9.4 9.4.1
(5)
9.4.2 Upright / larger than the object/Virtual (Any one) /
Opreg /groter as die voorwerp/Virtueel (Enige een) (1)
[21]
TOTAL/TOTAAL: 150
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