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Licemer1 [7]
3 years ago
15

A tennis ball of mass 57 g travels with velocity < 70, 0, 0 > m/s toward a wall. After bouncing off the wall, the tennis b

all is observed to be traveling with velocity < -65, 0, 0 > m/s. -What is the magnitude of the change of momentum of the tennis ball? -What is the change in the magnitude of the tennis ball's momentum? Please explain
Physics
1 answer:
Lena [83]3 years ago
5 0

Answer:

1) The magnitude of change in momentum is 7.695 kgm/s.

Explanation:

We know that momentum of a particle with mass 'm' travelling with a speed of 'u' is mathematically given by

\overrightarrow{p}=m\overrightarrow{u}

Thus according to the given values initial momentum of the ball is

p_{1}=0.057kg\times 70m/s=3.990kgm/s

The final momentum of the ball after bouncing off becomes

p_{2}=0.057kg\times -65m/s=-3.705kgm/s

Thus the change in momentum is given by:

\Delta p=p_{2}-p_{1}\\\\=-3.705-3.990=-7.695kgm/s

Thus the magnitude of change in momentum is 7.695 kgm/s.

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Answer:

Explanation:

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E = e σ A  ( T⁴ - T₀⁴ )

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E = .35 x 5.67 x 10⁻⁸ ( 298⁴ - 268⁴ ) x 4π x .25²

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2 years ago
What is the acceleration of a car that goes from 0 to 20m/s in 7 seconds?
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Answer:

2m/s^2

Explanation:

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V = u +at

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Answer:

B. The escape speed of the Moon is less than that of the Earth; therefore, less energy is required to leave the Moon.

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Since the speed required to escape from the gravitational attraction of the Moon is less than the speed required to escape from the gravitational attraction of the Earth, less energy is required to travel from the Moon to the Earth, than is required to travel from the Earth to the Moon. This is because the kinetic energy is directly proportional to the square of the velocity.

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Hope this helps!
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