Answer:
Change in momentum of the stone is 3.673 kg.m/s.
Explanation:
Given:
Mass of the ball on the horizontal the surface, m = 0.10 kg
Velocity of the ball with which it hits the stone, v = 20 m/s
According to the question it rebounds with 70% of the initial kinetic energy.
We have to find the change in momentum i.e Δp
Before that:
We have to calculate the rebound velocity with which the object rebounds.
Lets say that the rebound velocity be "v1" and KE remaining after the object rebounds be "KE1".
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Joules (J).
Rebound velocity "v1".
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m/s ...as it rebounds.
Change in momentum Δp.
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Kg.m/s
The magnitude of the change in momentum of the stone is 3.673 kg.m/s.
The answer is 52.79 m.
I used this formula to get the formula for Vy:
v^2=vi^2+2(a)(x)
And got
Vy=square root (19.6 h)
Then I used that and put it in this formula:
tan(65) =Vy/Vx
tan(65) = square root (19.6 h)/15.0
Then I rearranged it to:
h=[(15.0)(tan65)]/19.6
h=52.79 m
PART A)
Here by force balance along Y direction

Force balance along X direction

now by torque balance



PART B)
now from above equations



now net reaction force of wall is given as


Answer:
an energy source (AC or DC), a conductor (wire), an electrical load (device), and at least one controller (switch).
Explanation:
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