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3 years ago

May someone help me with this question, please and thank you :3

Mathematics

2 answers:

gizmo_the_mogwai [7]3 years ago

6 0

Answer:

Step-by-step explanation:

factor $4^{3}$ out of $4^{5}$

$\frac{4^{3}(4^{2}) }{4^{3} }$

cancel common factors

$\frac{4^{3}(4^{2}) }{4^{3}(1) }=\frac{4^{2} }{1}$

simplify

$4^{2}$

solve

9966 [12]3 years ago

3 0

Answer:

4^2 or 16

4^5 / 4^ 3

*subtract the exponents*

5 - 3 = 2

4^2 = 16

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<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

$\rm I_x = \int y^2 dA\\\\I_y = \int x^2 dA$

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

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

#SPJ4

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