This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Hey, good to see you again. 8. a) i. The principle of moments states that for a body to be in rotational equilibrium, the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments about the same point. 8. a) ii. Step 1: Identify the forces and distances. Let R_M be the reaction force from the mother and R_F be the reaction force from the father. Pole length L = 1.7 m. Weight of pole W_P = 150 N, acting at the center of the pole, 0.85 m from each end. Weight of boy W_B = 650 N, sitting 0.75 m from one end (let's assume the mother's end). Weight of girl W_G = 490 N, sitting 0.5 m from the other end (father's end). This means the girl is at 1.7 - 0.5 = 1.2 m from the mother's end. Step 2: Apply the principle of moments about the mother's support (R_M). Clockwise moments: Moment due to boy = W_B × 0.75 = 650 N × 0.75 m = 487.5 Nm Moment due to pole = W_P × 0.85 = 150 N × 0.85 m = 127.5 Nm Moment due to girl = W_G × 1.2 = 490 N × 1.2 m = 588 Nm Anticlockwise moment: Moment due to father's support = R_F × 1.7 Sum of clockwise moments = Sum of anticlockwise moments 487.5 + 127.5 + 588 = R_F × 1.7 1203 = 1.7 R_F R_F = (1203)/(1.7) R_F = 707.647 N Step 3: Apply the condition for translational equilibrium (sum of upward forces = sum of downward forces). Upward forces: R_M + R_F Downward forces: W_B + W_P + W_G R_M + R_F = 650 + 150 + 490 R_M + R_F = 1290 N Substitute R_F: R_M + 707.647 = 1290 R_M = 1290 - 707.647 R_M = 582.353 N Step 4: State the reactions. The reaction of the mother is R_M and the reaction of the father is R_F. R_M ≈ 582.4 N R_F ≈ 707.6 N 8. b) i. Relative motion describes the motion of an object with respect to another object or a specific frame of reference. It means that the observed velocity or position of an object depends on the motion of the observer. 8. b) ii. Given: Velocity of car A, V_A = 100 ms^-1. Velocity of car B, V_B = 60 ms^-1. α. If the two cars are travelling in the same direction: The relative velocity of car A with respect to car B is the difference between their velocities. V_AB = V_A - V_B V_AB = 100 ms^-1 - 60 ms^-1 V_AB = 40 ms^-1 β. If the two cars are travelling in opposite directions: The relative velocity of car A with respect to car B is the sum of their magnitudes of velocities. V_AB = V_A + V_B V_AB = 100 ms^-1 + 60 ms^-1 V_AB = 160 ms^-1 8. c) Given: Mass m = 4.0 kg. Initial velocity u = 20 ms^-1. Final velocity v = 30 ms^-1. Time t = 5 s. i. Acceleration of the body: Using the equation of motion v = u + at: 30 ms^-1 = 20 ms^-1 + a × 5 s 10 ms^-1 = a × 5 s a = 10 ms^-15 s a = 2 ms^-2 ii. Work done by the body: Work done is equal to the change in kinetic energy. W = KE = (1)/(2)mv^2 - (1)/(2)mu^2 W = (1)/(2)m(v^2 - u^2) W = (1)/(2) × 4.0 kg × ((30 ms^-1)^2 - (20 ms^-1)^2) W = 2.0 kg × (900 m^2s^-2 - 400 m^2s^-2) W = 2.0 kg × 500 m^2s^-2 W = 1000 J iii. Change in momentum of the body: Change in momentum p = mv - mu. p = m(v - u) p = 4.0 kg × (30 ms^-1 - 20 ms^-1) p = 4.0 kg × 10 ms^-1 p = 40 kg ms^-1 9. a) Two factors that determine the quantity of heat in a body are: • The mass of the body. • The specific heat capacity of the material of the body. • The change in temperature of the body. (Any two of these are sufficient) 9. b) Given: Mass of lead m = 100 kg. Initial temperature T_1 = 40 °C. Melting point T_m = 320 °C. Power of furnace P = 10 kW = 10 × 10^3 W. Specific heat capacity of lead c = 120 Jkg^-1K^-1. Specific latent heat of fusion of lead L_f = 2.5 × 10^4 Jkg^-1. i. Quantity of heat required to heat the lead to its melting point: The change in temperature T = T_m - T_1 = 320 °C - 40 °C = 280 °C. Since 1 °C change is equal to 1 K change, T = 280 K. Q_1 = mc T Q_1 = 100 kg × 120 Jkg^-1K^-1 × 280 K Q_1 = 12000 × 280 J Q_1 = 3.36 × 10^6 J ii. Additional heat required to melt the lead: This is the latent heat of fusion. Q_2 = mL_f Q_2 = 100 kg × 2.5 × 10^4 Jkg^-1 Q_2 = 250 × 10^4 J Q_2 = 2.5 × 10^6 J iii. Time taken to supply this additional energy: The power of the furnace is P = (Q_2)/(t). We need to find t. t = (Q_2)/(P) t = 2.5 × 10^6 J10 × 10^3 W t = (2.5 × 10^6)/(10^4) s t = 2.5 × 10^2 s t = 250 s Send me the next one 📸