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A 5.5 kg bowling ball initially at rest is dropped from the top of a 12 m building. It hits the ground 1.75 s later. Find the net external force on the falling ball.

Respuesta :

meerkat18

We are given that:

m = 5.5 kg

d = 12 m

t = 1.75 s

Using the distance formula, we can get the acceleration of the ball:

d = (1/2) a*t^2

Rearranging:

a = 2d/t^2 = 2*(12 m)/(1.75 s)^2

a = 7.837 m/s^2

 

Using Newton’s 2nd law, force is defines as:

F = ma

F = 5.5 kg (7.837 m/s^2)

F = 43.10 N

osoriovanessa472

Answer:

52 N

That's your answer!!

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