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 nkemdivo — ready when you are. Q47. Step 1: Identify the given values and relevant formulas. Mass of car, m = 1200 kg Power, P = 54 kW = 54000 W Angle of inclination, = ^-1 ((1)/(10)), so = (1)/(10) Non-gravitational resistance, R = 300 N Acceleration due to gravity, g = 10 m s^-2 At maximum speed, the net force is zero, and power is given by P = Fv, where F is the total force opposing motion. Step 2: Calculate the component of gravitational force acting down the incline. F_g = mg F_g = (1200 kg)(10 m s^-2)((1)/(10)) F_g = 1200 N Step 3: Calculate the total opposing force. The total opposing force is the sum of the gravitational component and the non-gravitational resistance. F_total = F_g + R F_total = 1200 N + 300 N F_total = 1500 N Step 4: Calculate the maximum speed v. Using the power formula P = F_total v: v = (P)/(F_total) v = 54000 W1500 N v = 36 m s^-1 The options provided are A. 20 m s^-1, B. 25 m s^-1, C. 100 m s^-1, D. 900 m s^-1. There seems to be a discrepancy between the calculated answer and the given options. Let me recheck the calculation. P = 54 kW = 54000 W. m = 1200 kg. = 1/10. R = 300 N. g = 10 m s^-2. Force due to gravity down the incline: F_g = mg = 1200 × 10 × (1/10) = 1200 N. Total resistive force: F_total = F_g + R = 1200 N + 300 N = 1500 N. Maximum speed: v = P / F_total = 54000 W / 1500 N = 36 m s^-1. Given the options, none of them match 36 m s^-1. It's possible there's a typo in the question's values or the options. However, based on the provided information, the calculated maximum speed is 36 m s^-1. If I had to choose the closest option, it would be B. 25 m s^-1 or A. 20 m s^-1, but 36 m s^-1 is the direct calculation. I will state the calculated answer. The maximum speed of the car is 36 m s^-1. Q48. Step 1: Identify the given probabilities and number of trials. Probability of landing head-up, P(H) = (1)/(4). Number of tosses, n = 3. Step 2: Calculate the probability of landing tail-up. P(T) = 1 - P(H) = 1 - (1)/(4) = (3)/(4) Step 3: Use the binomial probability formula to find the probability of getting exactly one tail in 3 tosses. The probability of exactly k successes in n trials is given by P(X=k) = nk p^k (1-p)^n-k. Here, "success" is getting a tail, so p = P(T) = (3)/(4). We want k=1 tail in n=3 tosses. P(exactly one tail) = 31 ((3)/(4))^1 (1 - (3)/(4))^3-1 P(exactly one tail) = 3 × ((3)/(4)) × ((1)/(4))^2 P(exactly one tail) = 3 × (3)/(4) × (1)/(16) P(exactly one tail) = (9)/(64) The correct option is A. The probability of getting exactly one tail is A. (9)/(64). Q49. Step 1: Identify the masses and positions of the particles. Particle 1: m_1 = 2 kg, (x_1, y_1) = (3, 1) Particle 2: m_2 = 3 kg, (x_2, y_2) = (8, 6) Step 2: Calculate the x-coordinate of the center of mass (X_CM). X_CM = (m_1 x_1 + m_2 x_2)/(m_1 + m_2) X_CM = (2 kg)(3) + (3 kg)(8)2 kg + 3 kg X_CM = (6 + 24)/(5) X_CM = (30)/(5) X_CM = 6 Step 3: Calculate the y-coordinate of the center of mass (Y_CM). Y_CM = (m_1 y_1 + m_2 y_2)/(m_1 + m_2) Y_CM = (2 kg)(1) + (3 kg)(6)2 kg + 3 kg Y_CM = (2 + 18)/(5) Y_CM = (20)/(5) Y_CM = 4 The position of their center of mass is (6, 4). The correct option is A. The position of their center of mass is A. (6,4). Q50. Step 1: Identify the given values. Force, F = 14 N Mass, m = 70 kg Initial speed, u = 3 m s^-1 Final speed, v = 7 m s^-1 Step 2: Apply the Work-Energy Theorem. The work done by the force is equal to the change in kinetic energy. W = KE Fs = (1)/(2)mv^2 - (1)/(2)mu^2 Step 3: Substitute the values and solve for the distance s. (14 N)s = (1)/(2)(70 kg)(7 m s^-1)^2 - (1)/(2)(70 kg)(3 m s^-1)^2 14s = (1)/(2)(70)(49) - (1)/(2)(70)(9) 14s = 35(49) - 35(9) 14s = 1715 - 315 14s = 1400 s = (1400)/(14) s = 100 m The correct option is D. The distance travelled is D. 100 m. Send me the next one 📸