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Svetradugi [14.3K]
3 years ago
14

Which phase happens immediately after a third quarter moon

Physics
2 answers:
damaskus [11]3 years ago
6 0

Answer:

waning gibbous

Explanation:

<u>new Moon. </u>

<u>waxing crescent Moon. </u>

<u>first quarter Moon. </u>

<u>waxing gibbous Moon. </u>

<u>full Moon. </u>

<u>waning gibbous Moon. </u>

<u>last quarter Moon. </u>

<u>waning crescent Moon</u>

These are the phases the moon goes through!

astra-53 [7]3 years ago
5 0

Answer:

Ur mom lol

Explanation:

Ur mom lol

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A cylinder with moment of inertia I1 rotates with angular speed ω0 about a frictionless vertical axle. A second cylinder, with m
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Answer:

Part(a): The final angular velocity is \omega_{f} = \dfrac{I_{1}\omega_{i}}{(I_{1} + I_{2})}

Part(b): The ratio of the rotational energies is \dfrac{k_{f}}{k_{i}}& = \dfrac{I_{1}}{(I_{1} + I_{2})},showing the the energy of th system will decrease.

Explanation:

Part(a):

If 'I' be the moment of inertia of an object and '\omega' be its angular velocity then the angular momentum 'L' of the object can  be written as

L = I \omega

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&& I_{1} \omega_{i} = (I_{1} + I_{2}) \omega_{f}\\&or,& \omega_{f} = \dfrac{I_{1}\omega_{i}}{(I_{1} + I_{2})}

Part(b):

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K_{i} = \dfrac{1}{2} I_{1} \omega_{i}^{2}

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K_{f} = \dfrac{1}{2}(I_{1} + I_{2}) \omega_{f}^{2}

Substituting the value of \omega_{f},

&& K_{f} = \dfrac{1}{2}(I_{1} + I_{2})\dfrac{I_{1}^{2}\omega_{i}^{2}}{(I_{1} + I_{2})^{2}} = \dfrac{1}{(I_{1} + I_{2})} \dfrac{1}{2}I_{1}\omega_{i}^{2} = \dfrac{1}{(I_{1} + I_{2})} K_{i}\\&\dfrac{k_{f}}{k_{i}}& = \dfrac{I_{1}}{(I_{1} + I_{2})}

The above expression shows that the ebergy of the system will decrease.

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