A 2 kg object released from rest at the top of a tall cliff reaches a terminal speed of 37.5 m/s after it has fallen a height of
1 answer:
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1) The grapefruit is in free fall, so it moves by uniformly accelerated motion, with constant acceleration 
. Calling h its height at t=0, the height at time t is given by
We are told thatn when
the grapefruit hits the ground, so h(0.75 s)=0. If we substitute these data into the equation, we can find the initial height h of the grapefruit:
2) The speed of the grapefruit at time t is given by
where
is the initial speed of the grapefruit. Substituting t=0.75 s, we find the speed when the grapefruit hits the ground:
We are told thatn when
2) The speed of the grapefruit at time t is given by
where
Answer:
W = 0 J
Explanation:
It is given that,
Weight of the crate, W = 100 N
Distance moved by the crate, d = 1.5 m
Let W is the work done to hold the crate 1.5 m above the ground in this way. It is given by the product of force and the displacement. Its formula is given by :
Here, as it is lifted vertically
W = 0
So, the work done to hold the crate is 0. Hence, this is the required solution.