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2 years ago

Need help with two of these! Thank you

Mathematics

1 answer:

astraxan [27]2 years ago

6 0

Step-by-step explanation:

17) 3x+5-12+4x+2x

3x+4x+2x +5-12

8x-7

if x =6

8*6-7 =48-7

= 41

18.) 9(3x-4y+8)

27x-36y+72

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2 years ago

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Find the moment of inertia about the y-axis of the thin semicirular region of constant density

arlik [135]

The moment of inertia about the y-axis of the thin semicircular region of constant density is given below.

$\rm I_y = \dfrac{1}{8} \times \pi r^4$

<h3>What is rotational inertia?</h3>

Any item that can be turned has rotational inertia as a quality. It's a scalar value that indicates how complex it is to adjust an object's rotational velocity around a certain axis.

Then the moment of inertia about the y-axis of the thin semicircular region of constant density will be

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x = r cos θ

y = r sin θ

dA = r dr dθ

Then the moment of inertia about the x-axis will be

$\rm I_x = \int _0^r \int _0^{\pi} (r\sin \theta )^2 \ r \ dr \ d\theta\\\\\rm I_x = \int _0^r \int _0^{\pi} r^3 \sin ^2\theta \ dr \ d\theta$

On integration, we have

$\rm I_x = \dfrac{1}{8} \times \pi r^4$

Then the moment of inertia about the y-axis will be

$\rm I_y = \int _0^r \int _0^{\pi} (r\cos\theta )^2 \ r \ dr \ d\theta\\\\\rm I_y = \int _0^r \int _0^{\pi} r^3 \cos ^2\theta \ dr \ d\theta$

On integration, we have

$\rm I_y = \dfrac{1}{8} \times \pi r^4$

Then the moment of inertia about O will be

$\rm I_o = I_x + I_y\\\\I_o = \dfrac{1}{8} \times \pi r^4 + \dfrac{1}{8} \times \pi r^4\\\\I_o = \dfrac{1}{4} \times \pi r^4$

More about the rotational inertia link is given below.

brainly.com/question/22513079

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