Skip to content

Technical Sciences Grade 12 P1 Eastern Cape June 2024 Errata

ErrataTechnical SciencesGrade 1220244 pages

Grade 12 Technical Sciences Paper 1 errata from the Eastern Cape June Exam 2024, 4 pages. Read online or get the free PDF download.

Free to read online, no sign-up needed

Loading document…

Loading document…

Document textSearch extracted text and jump to a page.
Downloaded from hlayiso.com 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) Page 1 of 2 Provincial Examination Errata
Downloaded from hlayiso.com (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✓. Page 2 of 5 Provincial Examination Errata
Downloaded from hlayiso.com 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✓ Page 2 of 5 Provincial Examination Errata
Downloaded from hlayiso.com (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 Provincial Examination Errata

Study this subject

More from Grade 12 Technical Sciences

View all