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lesya692 [45]
3 years ago
13

In a little league baseball game, the 145 g ball enters the strike zone with a speed of 14.0m/s . the batter hits the ball, and

it leaves his bat with a speed of 23.0m/s in exactly the opposite direction.what is the magnitude of the impulse delivered by the bat to the ball?if the bat is in contact with the ball for 1.0ms , what is the magnitude of the average force exerted by the bat on the ball?
Physics
1 answer:
Novosadov [1.4K]3 years ago
8 0

Answer:

5365 N

Explanation:

v = Final velocity = 23 m/s

u = Initial velocity = -14 m/s (opposite direction)

m = Mass of ball = 145 g

t = Time taken = 1 ms

Impulse is given by

J=m(v-u)

Impulse is also given by

J=Ft

Ft=m(v-u)\\\Rightarrow F=\dfrac{m(v-u)}{t}\\\Rightarrow F=\dfrac{0.145\times (23-(-14))}{1\times 10^{-3}}\\\Rightarrow F=5365\ N

The magnitude of the average force exerted by the bat on the ball is 5365 N

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Answer:

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Explanation:

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\Delta E = \frac{1}{2}\cdot I\cdot (\omega_{1}^{2}-\omega_{2}^{2}) (2)

Where:

I - Moment of inertia of the flywheel, measured in kilograms per square meter.

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If we know that K_{1} = 30\,J, K_{2} = 15\,J and I = 18\,kg\cdot m^{2}, then the initial angular speed is:

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\omega_{1} = \sqrt{\frac{2\cdot (30\,J)}{18\,kg\cdot m^{2}} }

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K_{2} = \frac{1}{2}\cdot I \cdot \omega_{2}^{2} (4)

\omega_{2}=\sqrt{\frac{2\cdot K_{2}}{I} }

\omega_{2} = \sqrt{\frac{2\cdot (15\,J)}{18\,kg\cdot m^{2}} }

\omega_{2} \approx 1.291\,\frac{rad}{s}

\omega_{2} \approx 0.205\,\frac{rev}{s}

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t = \frac{\omega_{2}-\omega_{1}}{\alpha} (5)

If we know that \omega_{1}\approx 0.291\,\frac{rev}{s}, \omega_{2} \approx 0.205\,\frac{rev}{s} and \alpha = -0.200\,\frac{rev}{s^{2}}, then the time needed is:

t = \frac{0.205\,\frac{rev}{s}-0.291\,\frac{rev}{s}}{-0.200\,\frac{rev}{s^{2}} }

t = 0.43\,s

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Answer:

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Explanation:

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\text{Density of a substance}=\frac{\text{Mass of a substance}}{\text{Volume of a substance}}

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