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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You're on a roll — here are the solutions to your questions: 16. To determine the acceleration of the body: Step 1: Calculate the time interval (T) between consecutive dots. The frequency of the tick-tock timer is 50 Hz. T = (1)/(frequency) = (1)/(50 Hz) = 0.02 s Step 2: Identify the distances between consecutive dots (segments) and convert them to meters. s_1 = 2 cm = 0.02 m s_2 = 4 cm = 0.04 m s_3 = 6 cm = 0.06 m s_4 = 8 cm = 0.08 m s_5 = 10 cm = 0.10 m Step 3: Calculate the acceleration using the difference in consecutive displacements. The acceleration (a) can be found using the formula a = ( s)/(T^2), where s is the difference between consecutive segment lengths. Let's use s_2 - s_1: s = s_2 - s_1 = 0.04 m - 0.02 m = 0.02 m a = 0.02 m(0.02 s)^2 a = 0.02 m0.0004 s^2 a = 50 m/s^2 17. To calculate the maximum height reached by the ball: Step 1: Determine the time taken to reach the maximum height. The ball returns to its starting point after 4 s. The time to reach the maximum height is half of the total time. t_up = 4 s2 = 2 s Step 2: Determine the initial velocity (u) of the ball. At maximum height, the final velocity (v) is 0 m/s. Using the equation of motion v = u + at, with a = -g = -10 m/s^2 (since gravity acts downwards). 0 = u + (-10 m/s^2)(2 s) 0 = u - 20 m/s u = 20 m/s Step 3: Calculate the maximum height (h_max). Using the equation of motion s = ut + (1)/(2)at^2: h_max = (20 m/s)(2 s) + (1)/(2)(-10 m/s^2)(2 s)^2 h_max = 40 m - (1)/(2)(10 m/s^2)(4 s^2) h_max = 40 m - 20 m h_max = 20 m 18. To draw the displacement-time graphs for A and B: The given velocity-time graph shows that both objects A and B are moving with constant positive velocities. Object A has a higher constant velocity than object B. For constant velocity, the displacement-time graph is a straight line with a constant positive slope. A higher velocity corresponds to a steeper slope. Displacement (m) c|cc & & \\ & A & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & / & \\ & /