Laurenn157idn
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A pitcher throws a ball with a mass of 0.15 kg at a batter with a velocity of 45 m/s towards the batter. The batter exerts a Force of 4500 N opposite the direction of the ball for a time of 0.003 s. What is the velocity of ball coming off the bat?

please if you can do the work i’ll mark you brainliest!!

Sagot :

Devarpitaidn

Answer:

-135m/s

Explanation:

Impulse P=change in momentum

Let the final velocity be -Vf (negative sign due to opposite direction)

Average force =Impulse/Time

=m(-Vf-Vi)/t

=>0.15m.g (-Vf-45)/0.003=4500N

=>0.15 (-Vf-Vi)=13.5

=>(-Vf-Vi)=90

Given,Vi=45

=>(-Vf) =90+45

=>(-Vf)=135

=>Vf = -135m/s

Hope it helps!!!

The velocity of ball coming off the bat will be -135m/s. Negative shows the direction.

What is velocity?

The change of distance with respect to time is defined as velocity. It is a time-based component. Its unit is m/sec.

The given data in the problem is:

m ( mass) =0.15 Kg

u is the initial velocity of fall =45 m/sec

v is the final velocity= ?

From the Newton's second law of motion;

Impulse = change in momentum

[tex]\rm I= \triangle P \\\\ Ft = m(v-u) \\\\ 4500=\frac{0.15(V-45)}{0.003} \\\\ V = -135 \ m/sec[/tex]

Hence,the velocity of ball coming off the bat will be -135m/s.

To learn more about the velocity refer to the link;

https://brainly.com/question/862972

#SPJ2

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