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frozen [14]
3 years ago
14

slader A girl of mass 55 kg 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 bounces 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:
Ivan3 years ago
6 0

Answer: 800N

Explanation:

Given :

Mass of ball =0.8kg

Contact time = 0.05 sec

Final velocity = initial velocity = 25m/s

Magnitude of the average force exerted on the wall by the ball is can be calculated using the relation;

Force(F) = mass(m) * average acceleration(a)

a= (initial velocity(u) + final velocity(v))/t

m = 0.8kg

u = v = 25m/s

t = contact time of the ball = 0.05s

Therefore,

a = (25 + 25) ÷ 0.05 = 1000m/s^2

Therefore,

Magnitude of average force (F)

F=ma

m = mass of ball = 0.8

a = 1000m/s^2

F = 0.8 * 1000

F = 800N

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3 years ago
14.A 90 kg quarterback gets tackled by being hit by a 120 kg lineman backwards
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The acceleration of the quarterback and the lineman is 5.55m/s² and 4.16m/s² respectively in the same direction.

As, we know, the 120 Kg lineman is moving with a force of 500N.

His net acceleration will be in the same direction as his motion.

It is already known that, If M is the mass of the body and a is the acceleration of the body, then the force F on the body can be calculated by using the formula,

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The weight of the quarterback is 90 Kg. He is being hit by a force of 500N.

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