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alekssr [168]
3 years ago
9

A 0.2 kg baseball moving with a velocity of 20 m/s is struck by a bat. The baseball reverses its direction and moves with a velo

city of 40 m/s . What is the average force exerted on the ball if the bat is in contact with the ball for 0.006 ?
Physics
1 answer:
krek1111 [17]3 years ago
7 0

Average force is 666.67 N

<u>Explanation:</u>

Given that :

m = Mass of the baseball = 0.2 kg

u = Initial velocity = 20 m/s  

v = Final velocity = 40 m/s

t = Time taken for change in velocity = 0.006 s

We know:

Force exerted = mass × acceleration = m ×a

Acceleration can be found by means of dividing the change in velocity measured in m/s by the time taken in seconds.

a = $\frac{v-u}{t} = $a = \frac{40-20}{0.006} = 3,333.3 m/s²

Now we have to find the force using the formula, F = mass × acceleration as,

F = 0.2 kg ×3333.3 ms⁻² = 666.67 N

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Answer:

The length is L = 6.206m and the angle is \theta = 37.752^o.

Explanation:

The period T of the pendulum is related to its length L by

T = 2\pi \sqrt{\dfrac{L}{g} },

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Solving for L we get

L = \dfrac{T^2g}{4\pi^2}

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L = \dfrac{(5.0s)^2*9.8m/s^2}{4\pi^2}

\boxed{L = 6.206m.}

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\theta = sin^{-1}( \dfrac{30N}{5.0kg*9.8m/s^2})

\boxed{\theta = 37.752^o.}

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