Please help in desperate

1. Consider three objects: Object A is a hoop of mass m and radius r; Object B is a sphere

Zarrin [17]

a) Object C (the disk) has the greatest rotational inertia ( $\frac{3}{2}mr^2$ )

b) Object B (the sphere) has the smallest rotational inertia ( $\frac{4}{5}mr^2$ )

Explanation:

The moments of inertia of the three objects are the following:

1) For a hoop of negligible thickness, it is

$I=MR^2$

where M is its mass and R its radius. For the hoop in this problem,

M = m

R = r

Therefore, its moment of inertia is

$I=(m)(r)^2=mr^2$

2) For a solid sphere, the moment of inertia is

$I=\frac{2}{5}MR^2$

where M is its mass and R its radius. For the sphere in this problem,

M = 2m

R = r

Therefore, its moment of inertia is

$I=\frac{2}{5}(2m)(r)^2=\frac{4}{5}mr^2$

3) For a disk of negligible thickness, the moment of inertia is

$I=\frac{1}{2}MR^2$

where M is its mass and R its radius. For the disk in this problem,

M = 3m

R = r

Therefore, its moment of inertia is

$I=\frac{1}{2}(3m)(r)^2=\frac{3}{2}mr^2$

So now we can answer the two questions:

a) Object C (the disk) has the greatest rotational inertia ( $\frac{3}{2}mr^2$ )

b) Object B (the sphere) has the smallest rotational inertia ( $\frac{4}{5}mr^2$ )

Learn more about inertia:

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#LearnwithBrainly

7 0

3 years ago

I shared a picture of the problem. It’s a basic Physics question and an Algebra question.

julia-pushkina [17]

Hence the expression of ω in terms of m and k is

$\omega = \sqrt{\frac{k}{m}$

Given the expressions;

$T_s = 2 \pi \sqrt{\frac{m}{k} } \ and \ T_s = \frac{2 \pi}{\omega}$

Equating both expressions we will have;

$2 \pi \sqrt{\frac{m}{k} } = \frac{2 \pi}{\omega}$

Divide both equations by 2π

$\frac{2 \pi\sqrt{\frac{m}{2 \pi} } }{2 \pi}=\frac{\frac{2 \pi}{\omega} }{2\pi}\\\sqrt{\frac{m}{2 \pi} } = \frac{1}{\omega}\\$

Square both sides

$(\sqrt{\frac{m}{k} } )^2 = (\frac{1}{\omega} )^2\\\frac{m}{k} = \frac{1}{\omega ^2} \\\omega ^2 = \frac{k}{m}$

Take the square root of both sides

$\sqrt{\omega ^2} =\sqrt{\frac{k}{m} } \\\omega = \sqrt{\frac{k}{m}$

Hence the expression of ω in terms of m and k is

$\omega = \sqrt{\frac{k}{m}$

3 0

3 years ago

15 points Plz help me no links

wolverine [178]

Answer:

alternating mountain ranges and valley.

Hope u find this helpful

7 0

3 years ago

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tiny-mole [99]

Answer:

When you release the opening of the balloon, gas quickly escapes to equalize the pressure inside with the air pressure outside of the balloon. The escaping air exerts a force on the balloon itself. ... That opposing force—called thrust, in this case—propels the rocket forward.

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3 years ago

How much heat is released to freeze 47.30 grams of copper at its freezing point of 1,085°C? The latent heat of fusion of copper

krek1111 [17]

the answer is -9,697

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3 years ago

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