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Leokris [45]
3 years ago
14

"A 0.15 kg ball moving at 40 m/s is struck by a bat. The bat reverses the ball's direction and gives it a speed of 50 m/s. What

average force does the bat apply to the ball if they are in contact for 6.0 ×10 -3 s?"
Physics
1 answer:
ser-zykov [4K]3 years ago
6 0

Answer:

-2250 N

Explanation:

From the question,

Using.

F = m(v-u)/t .................. Equation 1

Where F = Average force the bat apply to the ball, m = mass of the ball, v = final velocity of the ball, u = initial velocity of the ball, t = time.

Given: m = 0.15 kg, v = -50 m/s(reversed direction), u = 40 m/s, t = 6.0×10⁻³ s = 0.006 s.

Substitute into equation 1

F = 0.15(-50-40)/0.006

F = 0.15(-90)/0.006

F = -2250 N.

The force is negative because it opposes the initial motion of the ball.

Hence the bat average force = -2250 N

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Answer:

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Explanation:

The index of expansion is the relationship of pressures:

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The variation of specific internal energy is:

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35 bar = 3500000 Pa

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q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg

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Shock waves travel faster than sound and their speed increases as the amplitude of the wave is increased but their intensity fades faster due to the fact that some of its energy gets expended in the form of heat due to the resistance of the medium.

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