Here are the solutions to the physics problems. 9.A A body initially moving at 72 km/h accelerates uniformly to a velocity of 180 km/h in 3 seconds. Calculate its acceleration. 10.A A horizontal force

Here are the solutions to the physics problems. 9.A A body initially moving at 72 km/h accelerates uniformly to a velocity of 180 km/h in 3 seconds. Calculate its acceleration. Step 1: Convert initial and final velocities from km/h to m/s. u = 72 km/h = 72 × 1000 m3600 s = 20 m/s v = 180 km/h = 180 × 1000 m3600 s = 50 m/s Step 2: Use the formula for acceleration. a = (v - u)/(t) Substitute the values: a = 50 m/s - 20 m/s3 s a = 30 m/s3 s a = 10 m/s^2 The acceleration is 10 m/s^2. 10.A A horizontal force of 12 N is applied on a wooden block of mass 2 kg placed on a horizontal surface. It causes the block to accelerate at 5 m/s^2. Determine the frictional force between the block and the surface. Step 1: Apply Newton's Second Law of Motion. The net force (F_net) acting on the block is the applied force (F_applied) minus the frictional force (F_f). F_net = F_applied - F_f Also, F_net = ma. So, F_applied - F_f = ma Step 2: Substitute the given values and solve for F_f. Given: F_applied = 12 N, m = 2 kg, a = 5 m/s^2. 12 N - F_f = (2 kg)(5 m/s^2) 12 N - F_f = 10 N F_f = 12 N - 10 N F_f = 2 N The frictional force is 2 N. 11.A A car can be brought to rest from a speed of 200 m/s in a time of 2 s. i) Calculate the average deceleration. Step 1: Identify the given values. Initial velocity (u) = 200 m/s Final velocity (v) = 0 m/s (brought to rest) Time (t) = 2 s Step 2: Use the formula for acceleration. a = (v - u)/(t) Substitute the values: a = 0 m/s - 200 m/s2 s a = -200 m/s2 s a = -100 m/s^2 The average deceleration is 100 m/s^2. (Deceleration is the magnitude of negative acceleration). ii) If the driver reaction time is 0.2 s, Determine the shortest stopping Distance. Step 1: Calculate the distance covered during the reaction time. During reaction time, the car travels at a constant initial velocity. Distance during reaction time (s_1) = u × t_reaction s_1 = 200 m/s × 0.2 s s_1 = 40 m Step 2: Calculate the distance covered during braking. Using the deceleration calculated in part (i), a = -100 m/s^2. Initial velocity (u) = 200 m/s Final velocity (v) = 0 m/s v^2 = u^2 + 2as_2 0^2 = (200 m/s)^2 + 2(-100 m/s^2)s_2 0 = 40000 m^2/s^2 - 200 m/s^2 · s_2 200 m/s^2 · s_2 = 40000 m^2/s^2 s_2 = 40000 m^2/s^2200 m/s^2 s_2 = 200 m Step 3: Calculate the total shortest stopping distance. Total stopping distance = s_1 + s_2 Total stopping distance = 40 m + 200 m Total stopping distance = 240 m The shortest stopping distance is 240 m. 12.A A small iron ball is dropped from the top of a vertical cliff and takes 2.5 seconds to reach the sandy beach. Find the velocity with which it strikes the sand. Step 1: Identify the given values. Initial velocity (u) = 0 m/s (since it is dropped) Time (t) = 2.5 s Acceleration due to gravity (g) = 9.8 m/s^2 (assuming standard gravity) Step 2: Use the formula for final velocity under constant acceleration. v = u + gt Substitute the values: v = 0 m/s + (9.8 m/s^2)(2.5 s) v = 24.5 m/s The velocity with which it strikes the sand is 24.5 m/s. Last free one today — make it count tomorrow, or type /upgrade for unlimited.