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3 years ago

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A 0.145-kg baseball pitched horizontally at 27.0 m/s strikes a bat and pops straight up to a height of 31.5 m. If the contact ti

me between bat and ball is 2.5 ms, calculate the average force between the ball and bat during contact.

1 answer:

Digiron [165]3 years ago

7 0

Answer:

The average force between the ball and bat during contact is 3006.72 N.

Explanation:

Given that,

Mass of the baseball, m = 0.145 kg

Initial speed of the ball, u = 27 m/s

Time of contact between bat and ball is 2.5 ms, $t=2.5\times 10^{-3}\ s$

After striking the bat, it pops straight up to a height of 31.5 m. The final velocity of the ball is given by using third equation of motion as :

$v^2-u^2=2as$

a = -g

And initially, u = 0

$v=\sqrt{2gs}$

$v=\sqrt{2\times 9.8\times 31.5}$

v = -24.84 m/s (as it pops straight up)

Let F is the average force between the ball and bat during contact. It is given by :

$F=\dfrac{m(v-u)}{t}$

$F=\dfrac{0.145\times (24.84 -(-27))}{2.5\times 10^{-3}}$

F = 3006.72 N

So, the average force between the ball and bat during contact is 3006.72 N. Hence, this is the required solution.

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Coherent light with wavelength 540 nm passes through narrow slits with a separation of 0.370 mm .Part AAt a distance from the sl

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Answer:

The phase difference between waves 31.7 rad

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