Electric energy is generated at a nuclear power plant using which type of nuclear reaction
2 answers:
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Consult the attached free body diagram.
By Newton's second law, the net force on the crate acting parallel to the surface is
∑ F[para] = (370 N) cos(-20°) - f = 0
(this is the x-component of the resultant force)
where
• (370 N) cos(-20°) = magnitude of the horizontal component of the pushing force
• f = magnitude of kinetic friction
The crate is moving at a constant speed and thus not accelerating, so the crate is in equilibrium.
Solve for f :
f = (370 N) cos(-20°) ≈ 347.686 N
The net force acting perpendicular to the surface is
∑ F[perp] = n - 1480 N - (370 N) sin(-20°) = 0
(this is the y-component of the resultant force)
where
• n = magnitude of normal force
• 1480 N = weight of the crate
• (370 N) sin(-20°) = magnitude of the vertical component of push
The crate doesn't move up or down, so it's also in equilibrium in this direction.
Solve for n :
n = 1480 N + (370 N) sin(-20°) ≈ 1606.55 N ≈ 1610 N
Then the coefficient of kinetic friction is µ such that
f = µn ⇒ µ = f/n ≈ 0.216
Answer:
The percentage of its mechanical energy does the ball lose with each bounce is 23 %
Explanation:
Given data,
The tennis ball is released from the height, h = 4 m
After the third bounce it reaches height, h' = 183 cm
= 1.83 m
The total mechanical energy of the ball is equal to its maximum P.E
E = mgh
= 4 mg
At height h', the P.E becomes
E' = mgh'
= 1.83 mg
The percentage of change in energy the ball retains to its original energy,
ΔE % = 45 %
The ball retains only the 45% of its original energy after 3 bounces.
Therefore, the energy retains in each bounce is
∛ (0.45) = 0.77
The ball retains only the 77% of its original energy.
The energy lost to the floor is,
E = 100 - 77
= 23 %
Hence, the percentage of its mechanical energy does the ball lose with each bounce is 23 %