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galben [10]
3 years ago
7

A 50.0-g Super Ball traveling at 29.5 m/s bounces off a brick wall and rebounds at 19.0 m/s. A high-speed camera records this ev

ent. If the ball is in contact with the wall for 4.75 ms, what is the magnitude of the average acceleration of the ball during this time interval
Physics
1 answer:
Dima020 [189]3 years ago
8 0

Answer:

a=-10210.52\ m/s^2

Explanation:

Given that,

Mass of a ball, m = 50 g

It is traveling at 29.5 m/s bounces off a brick wall and rebounds at 19.0 m/s.

Initial velocity, u = 29.5 m/s

Finl velocity, v =-10 m/s (as it rebounds)

We need to find the magnitude of the average acceleration of the ball during this time interval.

Acceleration = rate of change of velocity

a=\dfrac{v-u}{t}\\\\a=\dfrac{(-19)-29.5}{4.75\times 10^{-3}}\\\\=-10210.52\ m/s^2

So, the required acceleration is 10210.52\ m/s^2.

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a. The spring is compressed  by  0.174 m.

b.  The vertical distance the dart travels from its position when the spring is compressed to its highest position is 9.6 m.

c. The horizontal velocity of the dart at that time is  13.74 m/s.

d. The horizontal distance from the equilibrium position at which the dart hits the ground is 19.236 m.

<h3>What is potential energy?</h3>

The energy by virtue of its position is called the potential energy.

a. Given is the potential energy stored in the compressed spring of a dart gun, with a spring constant of 62.00 N/m, is 0.940 J.

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Horizontal distance = 13.74 m/s x 1.4s

Horizontal distance = 19.236 m

Thus, the horizontal distance is 19.236 m.

Learn more about potential energy.

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