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avanturin [10]
1 year ago
11

-A coconut falls out of a tree 12.0 m above the ground and hits a bystander 3.00 m tall on the top of the head. It bounces back

up 1.50 m before falling to the ground. If the mass of the coconut is 2.00 kg, calculate the Eg of the coconut relative to the ground at each of the following sites: Assume no loss due to friction or air resistance.
a.Eg while it is still in the tree
b.Eg when it hits the bystander on the head
c.Eg when it bounces up to its maximum height
d. At what point will it have the most Ek?
Physics
1 answer:
Llana [10]1 year ago
3 0

Explanation:

Eg = mgh

a. Eg = (2.00 kg) (9.8 m/s²) (12.0 m)

Eg = 235 J

b. Eg = (2.00 kg) (9.8 m/s²) (3.00 m)

Eg = 58.8 J

c. Eg = (2.00 kg) (9.8 m/s²) (4.50 m)

Eg = 88.2 J

d. When the coconut hits the bystander:

Ek = 235 J − 58.8 J = 176 J

When the coconut hits the ground:

Ek = 88.2 J − 0 J = 88.2 J

Ek is the greatest when the coconut hits the bystander.

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From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

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