Physics, asked by debadritoraj1436, 10 months ago

During a tennis match, a player hits the ball reaching a speed of 58 m/s. What is the average force exerted on the 0.057-kg tennis ball by the player, assuming that the ball’s speed just after impact is 58 m/s and that the initial horizontal component of the velocity before impact is negligible and that the ball remained in contact with the racquet for 5.0 ms (milliseconds)? (With explaination please)

Answers

Answered by galilearh11
1

Answer:

idk im sorry

Explanation:

im sorry

Answered by amitnrw
3

661.2 N  is the average force exerted on the 0.057-kg tennis  ball by the player

Explanation:

F  = m dv/dt

m = 0.057 kg

dv =  58 m/s

dt =  5 ms = 5 * 10⁻³ s

F = 0.057 *  (58/ 5 * 10⁻³)

=> F = 57 * 58 / 5

=> F = 661.2 N

or v = u + at

=> 58 = 0  + a (5 *  10⁻³)

=> a = 58/ (5 *  10⁻³)

F = ma  =   0.057 *  (58/ 5 * 10⁻³) = 661.2 N

average force exerted on the 0.057-kg tennis  ball by the player,  = 661.2 N

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