EFSS 5105 4NA EOY solutions 2022

1 Edgefield Secondary School Sec 4NA Science Physics 2022 EOY Exam Solutions MCQ [20 marks] SECTION A [14 marks] Q No. Solutions Marks Remark s 1a Resultant force = 10 N Direction = to the right 1 b Water resistance, water drag, water friction (or drag fore or resistive force) 1 c As the forward thrust/force is equal to the water resistance, net force on swimmer is 0 N, So he will swim at constant speed 1 2a M = (1.5 x 40) + (3 x 10) = 90 Ncm 1 b Ma = Mc 90 = W x 30 W = 3.0 N 1 c Arrow drawn vertically down from CG at 50 cm mark The line of action of weight passes through the pivot, no perpendicular distance, no moment by the weight. 1 3a Q: Constant speed (zero acce leration) R: decreasing deceleration 1 Do not accept zero a for Q b Distance = ½(17.5 + 5)5 = 56.25 = 56.3 m or 56 m 1 c With the skis, the area in contact with the snow is bigger. Since P = F/A, since F = weight is constant, the pressure decreases, so wont sink into the snow. 1 B 1 2 3 4 5 6 7 8 9 10 Answer B D D D C D C A C A Question 11 12 13 14 15 16 17 18 19 20 Answer B C B B B B C B A A

2 4a b 1.6 min (accept 1.4 to 1.8 min) 1 c Shiny/white and smooth Shiny and smooth surface is a poor emitter of radiation. Rate of emission of radiation will be reduced, hence less heat loss through radiation. 1

3 SECTION B [16 marks] 5a gravitational potential energy 1 b GPE = mgh = 40 x 10 x 1.5 = 600 J 2 c P = E/t = 600/60 = 10 W 2 d Power from main = 100/70 x 10 = 14.28 = 14.3 W or 14 W 1 e E = Pt = 0.0143 x 12 = 0.1714 kWh Cost = 0.1714 x 0.28 = $0.048 (4.9 cents) 1 1 6a (i) Microwaves (ii) Gamma rays, x-ray or ultraviolet rays 1 1 b V = fλ 3.0 x 108 = 1800 x 106 x λ λ = 0.1667 = 0.167 m or 0.17 m 1 1 c (i) 20 Hz to 20kHz 1 (ii) (1) (2) B has lower amplitude than A 1 1 Do not accept B is softer as stated in question d Electromagnetic waves are transverse wave while sound is longitudinal. OR Electromagnetic waves can travel through vacuum while sound cannot pass through vacuum. 1

4 7a VR2 = 12 V 1 b R2 = V/I = 12/0.8 = 15 Ω 1 c PR2 = IV = 12 x 0.8 = 9.6 W 2 d IR2 = 2.4 – 0.8 = 1.6 A R1 = V/I = 12/1.6 = 7.5 Ω 1 1 e When R3 is connected parallel to R1, the total effective resistance of the circuit decreases. Hence the main current A1 reading will increase. 1 1