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Technical Sciences Grade 12 P1 Gauteng Prelim 2024 Memo

MemorandumTechnical SciencesGrade 1220249 pages

Grade 12 Technical Sciences Paper 1 memorandum from the Gauteng Prelim Exam 2024 (English), 9 pages. Read online or download the free PDF.

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Downloaded from hlayiso.com PREPARATORY EXAMINATION 2024 MARKING GUIDELINES TECHNICAL SCIENCE (PAPER 1) (11101) 9 pages
Downloaded from hlayiso.com 11101/24 QUESTION 1: MULTIPLE-CHOICE QUESTIONS 1.1 A  (2) 1.2 REMOVED (2) 1.3 B  (2) 1.4 D  (2) 1.5 A  (2) 1.6 C  (2) 1.7 A  (2) 1.8 B  (2) 1.9 D  (2) 1.10 D (2) [18] QUESTION 2: MATCHING ITEMS 2.1 D (1) 2.2 H (1) 2.3 A (1) 2.4 F (1) 2.5 B (1) 2.6 G (1) 2.7 C (1) 2.8 E (1) [8] 2
Downloaded from hlayiso.com 11101/24 QUESTION 3 3.1 3.1.1 The body will remain at rest or continue to move in a straight line at a constant velocity unless it is acted by a net force.  (2) 3.1.2 OR (3) CRITERIA MARKS Tension van A (T)  Tension of B (Fb)  Gravitational force (Fg)  3.1.3 Option 1 W = mg  9000 = m x 9,8 9000 𝑚𝑚 = 9,8 = 918,37 kg  (3) 3.1.4 Option 1 Option 2 Fnet = 0 Fnet = 0 FAx + TBx = 0  Fay + Fby + Fg = 0 FAcosƟ + (- TBcosα) = 0 FAsinƟ + (- Fbsinα) = 0 6 500 cos30o – Tcos45o  = 0 6500sin300 + Fbsin450 – 9000 = 0 T = 6 500 cos30o Fbsin450 = 5 750 cos45o = 7 960,84 N  = 8 131,73 N (4) 3
Downloaded from hlayiso.com 11101/24 3.2 3.2.1 When the net force is exerted on an object, the object will accelerate in the direction of the net force with the acceleration that is directly proportional to the net force and inversely proportional to the mass of the object.  (2) 3.2.2 Fnet = ma FA + Fbx + Ff = ma  FA + FBcosƟ + Ff = ma FA + 70cos30o  – 25 = 200 x 2  FA = 364,38 N to the right  (4) 3.2.3 Increases  (1) 3.2.4 Increasing the angle will increase the y component of FB that is acting downward which will add to the gravitational force. Adding the component of FB to gravitational force will increase the normal force.  (2) [21] QUESTION 4 4.1 4.1.1 The total linear momentum of an isolated system remains constant in magnitude and direction.  (2) 4.1.2 Σp before = Σp after or (m1 + m2)vi = m1v1f + m2v2f  (2 100 + 1 200)7  = 2 100vf + 1 200 x 10  23 100 = 2 100vf + 12 000 Vf = 5,29 m.s-1  (4) 4.2 4.2.1 The product of the net force acting on an object and the time the net force acts on the object. (2) 4.2.2 Fnet Δt = ΔP Increasing time decreases the net force./The relationship is inversely proportional.  (2) 4.2.3 Equal  (1) 4.2.4 During the collision the car is still moving at the same speed as it was. The airbags will be deployed and the body will hit the airbag with the same speed.  The airbags will slow down the body by increasing the time of impact and decreasing the force of the impact.  Time is inversely proportional to the force. (3) 4
Downloaded from hlayiso.com 11101/24 4.2.5 OPTION 1 OPTION 2 Fnet Δt = Δp  Fnet Δt = Δp Fnet Δt = m(vf – vi) Fnet Δt = m(vf – vi) -55000 x Δt = 1000(0 – 12)  55000 x Δt = 1000(0 – (-12)) Δt = 0,22 s  Δt = 0,22 s (4) [18] QUESTION 5 5.1 The total mechanical energy in an isolated system remains constant.  (2) 5.2 Ep = mgh  Ep = 250 x 9,8 x 30  = 73 500 J  (3) 5.3 Ek = ½mv2  Ek = ½ x 250 x 52  = 3 125 J  (3) 5.4 The sum of kinetic energy and potential energy.  (2) 5.5 EM = mgh + ½mv2  = 2 x 9,8 x 2 + 0  (If 0 is omitted then 2/3) = 39,2 J  (3) 5.6 Positive marking from 5.5 MEtop = MEbottom EP + EK = EP + EK  mgh + ½mv2 = mgh + ½mv2 2 x 9,8 x 2 + ½ x 2 x 02  = 2 x 9,8 x 0 + ½ x 2 x v2  39,2 + 0 = 0 + v2 v = 6,26 m.s-1  (4) [17] QUESTION 6 6.1 Within the elastic limit, the stress is directly proportional to the strain.  (2) 6.2 • Stress ball. • Spring. (any two)  • Kicking a rugby or soccer ball. (2) 5
Downloaded from hlayiso.com 11101/24 6.3 • Power steering. • Shock absorbers. • Windshields. (any two)  • Brake. • Dentists' chair. (2) 6.4 6.4.1 σ=F A = F  πr2 = 6 000  π x 0,025  2 = 3 055 744,91 Nm-2  = 3,06 x 106 Nm-2 (4) 6.4.2 ε = Δl  L = 0,003  5 = 0,0006  = 6 x 10-4 (3) 6.4.3 POSITIVE MARKING FROM 6.4.1 and 6.4.2 OPTION 1 OPTION 2 K=σ K=σ ε ε = 3,06 x 106 = 3 055 744,91 6 x 10-4 6 x 10-4 = 5,1 x 109 N.m-2  = 5,09 x 109 N.m-2 (2) [15] QUESTION 7 7.1 Distance between the plates, area of the plates and dielectric of material used. (any two).  (2) 7.2 7.2.1 C = εo A  d = 8,85 x 10-12 x 0,04  0,002 = 1,77 x 10-10 CV-1  (Or the unit F can be used) (3) 6
Downloaded from hlayiso.com 11101/24 7.2.2 POSITIVE MARKING FROM 7.2.1 C=Q V 1,77 x 10-10 = Q  12 Q = 2,12 x 10-9 C  (3) [8] QUESTION 8 8.1 The electric current is proportional to the voltage and inversely proportional to the resistance when the temperature is constant. OR the potential difference across the conductor is directly proportional to the current in the conductor at constant temperature  (2) 8.2 8.2.1 Rp = R5 x R3  OR R5 + R3 1 1 1 = +  = 5x3 RP R5 R3 5+3 1 1 Req = Rp + R2 = +  5 3 = 1,88 + 2  = 3,88 Ω  Rp = 1,88 Ω Req = Rp + R2 = 1,88 + 2  = 3,88 Ω  (4) 8.2.2 R = V  I 3,88 = 3  I V = IR I = 0,77 A = 0,77 x 2  = 1,54 V  (4) 8.3 8.3.1 The rate at which energy is transferred. /The rate at which work is done.  (2) 7
Downloaded from hlayiso.com 11101/24 8.3.2 P = V2  R = 2202  18 = 2 688,89 W  = 2,67 kW (3) 8.3.3 • Stove • Heater (any two)  • Kettle (Any other relevant appliances are correct) (2) [17] QUESTION 9 9.1 CRITERIA MARKS Object (at 2F)  Inverted Image  Image position (2F)  Light ray passing through the centre  Light ray passing through F2.  Correct Lens (Convex lens)  (6) 9.2 9.2.1 Red  (1) 9.2.2 Violet  (1) 9.2.3 Dispersion  (1) 9.2.4 The phenomenon whereby light breaks up into its component colours.  (2) [11] 8
Downloaded from hlayiso.com 11101/24 QUESTION 10 10.1 • They originate from accelerated electric charges. • Propagate as changing electric field and magnetic fields that are perpendicular to each other. • They are transverse waves. • They transfer energy. • They can travel through the vacuum and a medium. • They have a speed of 3 x 108 m·s-1 (any two) (2) 10.2 10.2.1 Infrared  (1) 10.2.2 Infrared  (1) 10.3 10.3.1 E = hf  = 6,63 × 10-34 × 5,56 × 1014  = 3,69 × 10-19 J  (3) 10.3.2 E = hc  ʎ = 6,63 × 10-34 × 3 × 108  390 x 10-9  = 5,1 × 10-19 J  (4) [11] QUESTION 11 11.1 Commutator (split ring)  (1) 11.2 Anticlockwise  (1) 11.3 Electrical energy to mechanical energy  (2) 11.4 DC generator has commutator (split rings) and an AC generator has slip rings.  (2) [6] TOTAL: 150 9

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