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MakcuM [25]
3 years ago
7

2 * 1.5 * (.850/2)^2A small ball with mass 1.50 kg is mounted on one end of a rod 0.850 m long and of negligible mass. The syste

m rotates in a horizontal circle about the other end of the rod at 5060 rev/min.
Physics
1 answer:
grin007 [14]3 years ago
8 0

Answer

given,

mass of the rod = 1.50 Kg

length of rod = 0.85 m

rotational velocity = 5060 rev/min

now calculating the rotational inertia of the system.

I = m L^2        

where L is the length of road, we will take whole length of rod because mass is at  the end of it.      

I = 1.5 \times 0.85^2  

I = 1.084 kg.m²                        

hence, the rotational inertia the system is equal to I = 1.084 kg.m²

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Please help. I don’t understand this
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Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as

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According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so

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5 0
2 years ago
During which phase of the moon do neap tides occur?
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Where m1 and m2 are the masses of the two objects that are interacting and r is the distance between them.

For example, seeing the image below, point A is closer to the Moon than point b, and at the same time the center of mass of the Earth will feel more attracted to the Moon than point B. Therefore, that creates a tidal bulge in point A and point B.

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