Physics: Apitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. the bat does wnc = 75.0 j of work on the ball in hitting it. ignoring air resistance, determine the speed of the ball after the ball leaves the bat and is 25.0 m above the point of impact.(answer in m/s)

YExplanation:the net force on y is 45 +35 = 75 N [right]...

Apitcher throws a 0.140 kg baseball, and it approaches

Ответ:

aeshaalhemri

07.01.2022

Physics

The speed of the ball is 42.5 m/s

Explanation:

The initial kinetic energy of the ball is:

$K_1=\frac{1}{2} m v_0^2=\frac{1}{2}*0.140 kg*(35.0 m/s)^2$ = 85.75 J

The speed of the ball after leaving the bat is:

$K_2=K_1+W_{nc}\\ \frac{1}{2}mV^2= 85.75 J + 75 J\\ (\frac{1}{2}mV^2)2=( 160.75 J)2\\ mV^2= 321.5 J\\ V^2= \frac{321.5 J}{0.140kg} \\ V=\sqrt{\frac{321.5 J}{0.140kg}}$

V=47.92 m/s

Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:

V_f^2-V^2=-2gh

V_f^2-(47.92 m/s)^2=-2*9.81m/s^2*25m

V_f^2=-2*9.81m/s^2*25m+(47.92 m/s)^2

$V_f=\sqrt{-2*9.81m/s^2*25m+(47.92 m/s)^2}$

V_f=42.5m/s

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Ответ:

christyr2002

28.11.2022

Physics

Explanation:

the net force on y is 45 +35 = 75 N [right]

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Apitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. the bat does wnc = 75.0 j of work on the ball in hitting it. ignoring air resistance, determine the speed of the ball after the ball leaves the bat and is 25.0 m above the point of impact.(answer in m/s)

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