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stellarik [79]
3 years ago
10

The acceleration of a baseball pitcher's hand as he delivers a pitch is extreme. For a professional player, this acceleration ph

ase lasts only 50 ms, during which the ball's speed increases from 0 to about 90 mph, or 40 m/s.
What is the force of the pitcher's hand on the 0.145 kg ball during this acceleration phase? Express your answer with the appropriate units.
Physics
1 answer:
SashulF [63]3 years ago
4 0

Answer:

116 N.

Explanation:

Given,

mass of the ball, m = 0.145 Kg

initial speed = 0 m/s

Final speed = 40 m/s

time = 50 ms = 0.05 s

Force is equal to change in momentum per unit time

F = \dfrac{m\Delta v}{\Delta t}

F = \dfrac{0.145(40-0)}{0.05}

F = 116 N

Force of the pitcher hand on the ball is equal to 116 N.

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