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Doss [256]
3 years ago
13

A uniform, thin, solid door has height 2.20 m, width 0.870 m, and mass 23.0 kg. (a) find its moment of inertia for rotation on i

ts hinges. (b) is any piece of data unnecessary
Physics
1 answer:
vampirchik [111]3 years ago
8 0
The solution for this problem is:


a. The moment of inertia of the thin door rotating about the hinges axis is given by:

I = 1/3 mw^2 is to solve for the inertia, where: m is the mass and w is the width


= 23.0 x 0.870^2 / 3

= 17.4087 / 3

= 5.8029 kg m^2


b. The height of the door is not required for this calculation.
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Novay_Z [31]

Answer:

It will be cut in half

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Also, recall that

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Then we substitute in the previous equation

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Then we use trigonometry to find the width, we say

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Since the angle is small, we then have

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4 0
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Answer:

Following are the solution to this question:

Explanation:

Law:

\to \theta= 180^{\circ}- 50^{\circ}- 25^{\circ}

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Answer:

I believe D

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I hope this is correct good luck!

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