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pickupchik [31]
3 years ago
8

If the total mass is m, find the moment of inertia about an axis through the center and perpendicular to the plane of the square

. use the parallel-axis theorem.
Physics
1 answer:
My name is Ann [436]3 years ago
5 0
Question: A thin, uniform rod is bent into a square<span> of... A thin, uniform rod is bent into a </span>square<span> of side length a. </span>If the total mass is M<span>, </span>find the moment of inertia about an axis through the center and perpendicular to the plane of the square<span>. </span>Use the parallel-axis theorem<span>.</span>
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If I'm not mistaken it should be the digestive system due to the fact that our mouths and stomachs break down food and our intestines absorb any water and nutrients
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During the fission reaction shown, how did the target nucleus change ?
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A. The target nucleus split into two nuclei, each with fewer nucleons than the original.

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A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
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Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

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Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

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What type of compound forms Molecules?
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