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Natali5045456 [20]
3 years ago
11

A girl throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s, and it bounc

es back with this same speed. The ball is in contact with the wall 0.050 s. What is the magnitude of the average force exerted on the wall by the ball
Physics
1 answer:
erma4kov [3.2K]3 years ago
4 0

Answer:

F = 800N

the magnitude of the average force exerted on the wall by the ball is 800N

Explanation:

Applying the impulse-momentum equation;

Impulse = change in momentum

Ft = m∆v

F = (m∆v)/t

Where;

F = force

t = time

m = mass

∆v = v2 - v1 = change in velocity

Given;

m = 0.80 kg

t = 0.050 s

The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed.

v2 = 25 m/s

v1 = -25 m/s

∆v = v2 - v1 = 25 - (-25) m/s = 25 +25 = 50 m/s

Substituting the values;

F = (m∆v)/t

F = (0.80×50)/0.05

F = 800N

the magnitude of the average force exerted on the wall by the ball is 800N

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1)   λ = 24.7 cm,  2) f = 13.88 Hz, 3)  L = 117.3 cm

Explanation:

1) This is a resonance process, that is, the wave going downwards will interfere with the wave going upwards.

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3) the next resonance occurs for n = 9, so the prime number is

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         L = 19 λ / 4  

We must suppose a value for the wavelength, if the wavelength is present in the tube and the length of the column increases, the resonance number increases

         L = 19 24.7/4

         L = 117.3 cm

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