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SpyIntel [72]
4 years ago
11

What is the moment of inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a

final velocity of 6.00 m/s?
Physics
1 answer:
Paha777 [63]4 years ago
7 0
E = mgh +  \frac{1}2} m v^{2} + \frac{1}{2} I \omega^{2} = mgh +  \frac{1}2} m  r ^{2}   \omega ^{2}  + \frac{1}{2} I \omega^{2}

for a solid cylinder:  I =  \frac{1}{2} m r^{2}
for a hollow cylinder: I = mr^{2}

I will look at the case of a hollow cylinder:

E = mgh + I \omega ^{2} = constant \\ \\ I =  \frac{mgh}{  \omega^{2} }

That is as far as i get.


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The collision between two gas molecules or billiard balls can be approximated as elastic collisions. Elastic collisions are exchanges of kinetic energy between two bodies having different reference frames in which the total kinetic energy of the two bodies after collision is equal to the energy before collision.'

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Mandarinka [93]

The period of the orbit would increase as well

Explanation:

We can answer this question by applying Kepler's third law, which states that:

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Mathematically,

\frac{T^2}{a^3}=const.

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In this problem, the question asks what happens if the distance of the Earth from the Sun increases. Increasing this distance means increasing the semi-major axis of the orbit, a: but as we saw from the previous equation, the orbital period of the Earth is proportional to a, therefore as a increases, T increases as well.

Therefore, the period of the orbit would increase.

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