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CHIEF DIRECTORATE: EXAMINATIONS AND ASSESSMENT
Steve Vukile Tshwete Complex, Zone 6 Zwelitsha, 5608, Private Bag X0032, Bhisho, 5605 REPUBLIC OF SOUTH AFRICA:
Enquiries: Mrs P. Japhta. Tel: 040 602 7031. Fax: 040 602 7295. E-mail: Penelope.Japhta@ecdoe.gov.za
Website: www.ecdoe.gov.za
Ref. no. 13/P Tel.: (040) 608 7031/060 523 8658
Enquire: Mrs P. Japhta Fax: 040 608 7295
ERRATUM
TO: CHIEF EDUCATION SPECIALISTS
DISTRICT CURRICULUM COORDINATORS
DISTRICT ASSESSMENT OFFICIALS (DAOs
DISTRICT SUBJECT ADVISORS (DSAs)
PROVINCIAL SUBJECT COORDINATORS
CIRCUIT MANAGERS
DEPUTY CHIEF EDUCATION SPECIALISTS
SENIOR EDUCATION SPECIALISTS
PRINCIPALS OF SCHOOLS IN THE FET BAND
SUBJECT: ERRATUM – TECHNICAL SCIENCES P1 GRADE 12 JUNE COMMON 2024
DATE: 11 JUNE 2024
ERRATA TECHNICAL SCIENCES P1 GRADE 12 JUNE 2024
QUESTION 2
2.2.1 When a net (resultant) force is exerted on an object, the object will accelerate in the
direction of the force. This acceleration is directly proportional to the net (resultant) force✓, and
inversely proportional to the mass of the object. ✓
2.2.6
For 40 kg block/Vir 40 kg blok
𝐹𝑛𝑒𝑡 = 𝑚𝑎
𝑚𝑎 = 𝐹1 cos 𝜃 − 𝐹10 𝑘𝑔 − 𝑓𝑘 ✓
40𝑎 = 40 cos 35° − 𝐹10 𝑘𝑔 − 16,60 ✓ ………(1)
For 10 kg block/Vir 10 kg blok
F𝑛𝑒𝑡 = 𝑚𝑎
𝑚𝑎 = 𝐹40 𝑘𝑔 + 𝐹2 − 𝑓𝑘
10𝑎 = 𝐹40 𝑘𝑔 + 20 − 2,5 ✓………….(2)
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Technical Sciences Grade 12 P1 Eastern Cape June 2024 Errata
Grade 12 Technical Sciences Paper 1 errata from the Eastern Cape June Exam 2024, 4 pages. Read online or get the free PDF download.
Technical Sciences · Grade 12 · Eastern Cape June Exam · 2024. Errata, 4 pages. Read online or download the PDF.
- Subject
- Technical Sciences
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- Grade 12
- Document type
- Errata
- Year
- 2024
- Exam period
- Eastern Cape June Exam
- Paper
- 1
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- CAPS
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(1) + (2)
40𝑎 + 10𝑎 = 40 cos 35° − 𝐹10 𝑘𝑔 − 16,60 + 𝐹40 𝑘𝑔 + 20 − 2,5 ✓
𝑎 = 0,67 𝑚. 𝑠-2✓
QUESTION 3
3.1.2
OPTION 1 Away from batsman / OPSIE 1 Weg van die kolwer
∆𝑝 = 𝑚 𝑣𝑓 − 𝑣𝑖 ✓
= (0,175) ✓ (−30 − 12) ✓
= −7,35 N ∙ s
= 7,35 kg.m.s-1✓ away from the batsman/weg van die kolwer ✓
OPTION 2 Towards the batsman / OPSIE 1 Na die kolwer
∆𝑝 = 𝑚 𝑣𝑓 − 𝑣𝑖 ✓
= (0,175) ✓ (30 − (−12))✓
= 7,35 N ∙ s
= 7,35 kg.m.s-1✓ away from the batsman/weg van die kolwer ✓
3.1.3
POSITIVE MARKING FROM 3.1.2/POSITIEWE MERK VAN 3.1.2
FnetΔt = Δp ✓
Fnet(0,05) = 7,35 ✓
Fnet = 147 N ✓
OR /OF
FnetΔt = Δp ✓
Fnet(0,05) = − 7,35 ✓
Fnet = −147 N
= 147 N ✓
3.1.4
1
Fnet α ∆𝑡✓ if Δp is constant ✓ crumple zones increase the contact time ∆t✓ Fnet decreases✓.
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1
Fnet α ∆𝑡✓ as Δp konstant is✓ Frommelsones veroorsaak toename in kontak tyd✓Fnet neem
af✓.
3.2.2
OPTION 1 Right + / OPSIE 1 Regs +
∑ 𝑝𝑏𝑒𝑓𝑜𝑟𝑒 = ∑ 𝑝𝑎𝑓𝑡𝑒𝑟 ✓
𝑚1𝑣1𝑖 + 𝑚2𝑣2𝑖 = 𝑚1𝑣1𝑓 + 𝑚2𝑣2𝑓
(2) (3) + (3,5) (0) ✓ = (2) (−1) + 3,5𝑣 ✓
𝑣 = 2,29 𝑚 ∙ 𝑠-1
𝑣 = 2,29 𝑚 ∙ 𝑠-1✓ ; right/regs ✓
OPTION 2 Left + / OPSIE 2 Links +
∑ 𝑝 𝑏𝑒𝑓𝑜𝑟𝑒 = ∑ 𝑝 𝑎𝑓𝑡𝑒𝑟 ✓
𝑚1𝑣1𝑖 + 𝑚2𝑣2𝑖 = 𝑚1𝑣1𝑓 + 𝑚2𝑣2𝑓
(2) (−3) + (3,5) (0) ✓ = (2) (1) + 3,5𝑣 ✓
𝑣 = −2,29 𝑚 ∙ 𝑠-1
𝑣 = 2,29 𝑚 ∙ 𝑠-1 ✓ ; right/regs ✓
QUESTION 4
Suppose a learner states in
4.2.3 𝑊𝐹 = 𝐹∆𝑥 cos 𝜃 ✓ = mgh
𝑊𝐹 = (1 500)(9,8) cos 0° ✓
𝑊𝐹 = 7350 𝐽✓
4.2.4 Ep = 7350 J✓✓✓ full marks be awarded.
QUESTION 6
6.1.1 It is the thrust or (force) acting on the unit area around that point. ✓✓/ Dit is die stukrag
of (krag) op die eenheidsoppervlakte rondom daardie punt ✓✓
NB: (Foce be accepted for thrust)
6.1.2
𝐹
P = 𝐴✓
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(250)(9,8)✓
P=
(3,15𝑥 104 )(10−6 )✓
P = 7,78 × 104Pa ✓
6.2.2 Correction of P= h to P = ρgh
NB: EXAMINER & MODERATOR to decide about 6.1.2 and 6.2.1 as the Formulae were not
included in the Data-Sheet for the calculations.
11 June 2024
MRS P.E. JAPHTA DATE
(A) CES: AIDIBM SUBDIRECTORATE
Page 2 of 5
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