Which of the following provides evidence that geologic activity occurred in the past on Earth's Moon?
1 answer:
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The final kinetic energy of the ball is 2.45 J
Explanation:
We can solve this problem by using the law of conservation of energy.
In absence of frictional effect, the mechanical energy of the apple must be conserved during the fall. So we can write:
where :
is the initial potential energy, at the top
is the initial kinetic energy, at the top
is the final potential energy, at the bottom
is the final kinetic energy, at the bottom
By explicing the potential energy, we can rewrite the equation as:
where:
m = 0.5 kg is the mass of the apple
is the acceleration of gravity
is the initial height
is the final height
The initial kinetic energy is zero, since the ball starts from rest:
Therefore we can solve the equation for , the final kinetic energy of the ball:
Learn more about kinetic energy and potential energy:
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Answer:
Δθ₁ = 172.5 rev
Δθ₁h = 43.1 rev
Δθ₂ = 920 rev
Δθ₂h = 690 rev
Explanation:
- Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:
- Now, we need first to find the value of the angular acceleration, that we can get from the following expression:
- Since the machine starts from rest, ω₀ = 0.
- We know the value of ωf₁ (the operating speed) in rev/min.
- Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:
- Replacing by the givens in (2):
- Solving for α:
- Replacing (5) and Δt in (1), we get:
- in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:
- In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:
- First of all, we need to find the value of the angular acceleration during the second period.
- We can use again (2) replacing by the givens:
- ωf =0 (the machine finally comes to an stop)
- ω₀ = ωf₁ = 57.5 rev/sec
- Δt = 32 s
- Solving for α in (9), we get:
- Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:
- In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows: