A 6.1 kg flowerpot drops off the top of a tall wall. 1) 2) 3) 4) What is the mag

Question

A 6.1 kg flowerpot drops off the top of a tall wall.

1)

2)

3)

4)

What is the magnitude of the force acting on the pot while it is in the air 1.6 s after it begins to fall? After the pot has fallen 35 m, what is the magnitude of its velocity? After the pot has fallen h= 35 m, it enters a pool of viscous liquid, which brings it to rest over a distance of 1.5 m. Assuming constant deceleration over this distance, what is the magnitude of this deceleration? What is the force exerted on the liquid by the pot?

Explanation / Answer

A)magnitude of force=weight of the pot=59.78 N

b)sqrt(2*9.8*35)=26.2 m/s

c)initial velocity=26.2 m/s

final velocity=0

distance=1.5 m

so final velocity^2=initial velocity^2+2*deceleration*distance

deceleration=228.66 m/s^2

4)force=mass*deceleration=1394.8 N

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