Tiêu đề 19. SKEMA JAWAPAN UJIAN SELARAS 2_DP024_2023_2024__FINAL.pdf ✅

PHYSICS 2 DP024 UNIT FIZIK KOLEJ MATRIKULASI NEGERI SEMBILAN Skema Jawapan dan Pemarkahan Ujian Selaras 2 Semester 2 Sesi 2023/2024 NO SKEMA JAWAPAN MARKAH 1(a) (i)Amplitude , A = 0.4 m JU1 (ii)Frequency, f = 1 T T = 4.0 s f = 1 4.0 = 0.25 Hz G1 JU1 (iii)Wave number, k = 2π λ λ = 0.4 m k = 2π 0.4 = 5π m-1 G1 JU1 (iv)Progressive wave equation y(x,t) = 0.4 sin(0.5πt − 5πx) where x, y are in metre and t is in second G1 JU1 (v)Wave propagation velocity, v = fλ v = 0.25(0.4) v = 0.1 m s −1 G1 JU1 (vi)Particle vibrational velocity, v = dy dt v = d(0.4 sin (0.5πt − 5πx)) dt v = 0.4(0.5π) cos (0.5π(2) − 5π(1)) v = 0.628 m s −1 K1 G1 JU1 1(b) (i) k = 1.5m−1 λ = 2π k λ = 2π 1.5 λ = 4.19 m G1 JU1 (ii) ω = 250 rad s −1 f = ω 2π f = 250 2π f = 39.8 Hz GJU1 (iii) v = fλ v = 39.8(4.19) v = 166.8 m s −1 GJU1 TOTAL 16
PHYSICS 2 DP024 NO SKEMA JAWAPAN MARKAH 2(a) (i) V = IR R = V I R = 240 12 R = 20.0 Ω GJU1 (ii) Q = It Q = (12)(5 × 60) Q = 3600C K1 GJU1 2(b) (i) R = ρl A , A = πd2 4 R = (2.8 × 10−8 )(10.0 × 10−2 ) π ( 2.0 × 10−2 2 ) 2 R = 8.9 × 10−6 Ω K1 GJU1 (ii) V = IR I = V R I = 24 8.9 × 10−6 I = 2.70 × 106 A GJU1 2 (c) (i) Consider R1=8.0 Ω, R2=9.0 Ω, R3=3.0 Ω, R4=20 Ω 1 R12 = 1 R1 + 1 R2 = 1 8 + 1 9 ⸫ R12 = 4.24 Ω R123 = R12 + R3 = 4.24 + 3.0 = 7.24 Ω 1 Reff = 1 R123 + 1 R4 = 1 7.24 + 1 20 Reff = 5.32 Ω G1 G1 GJU1 c (ii) VAB = V4 = V123 = 10 V I123 = I3 = I12 ⸫V123 = I123R123 10 = I123(7.24) I123 = I3 = 1.38 A K1 G1 JU1 TOTAL 12

PHYSICS 2 DP024 NO SKEMA JAWAPAN MARKAH 4 (a) B = μoI 2πr = (4π × 10−7 )(6.2) 2π(0.034) = 3.65 × 10−5 T G1 JU1 (b) B = μoI 2r 1.0 × 10−6 = (4π × 10−7 )(25 × 10−3 ) 2r r = 0.016 m G1 JU1 (c) (i) ε = Blv sin θ = (4.2)(1.5)(3.5) sin 90o = 22.05 V G1 JU1 (ii) - From Lenz’s law F oppose v - Fleming’s Left HR Direction →B to A J1 (d) (i) φ = BA cos θ = (30 × 10−3 )(0.15) cos 0 o = 4.5 × 10−3 Wb Φ = Nφ = 100(4.5 × 10−3 ) = 0.45 Wb K1 G1 G1 JU1 (ii) ε = −NB dA dt K1 G1 JU1 TOTAL 14 = −(100)(30 × 10−3 ) (0.08 − 0.15) 2 = 0.105 V