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melamori03 [73]
3 years ago
9

a solid sphere and a hollow sphere with equal mass are rotated about an axis through their centers. both spheres experience equa

l torque. which sphere will reach a speed of 10 rad/s first? Explain your answer.​
Physics
1 answer:
adelina 88 [10]3 years ago
4 0

Answer:

Solid sphere

Explanation:

The solid sphere will have a lower moment of inertia as it carries more mass closer to the axis of rotation.

A lower moment of inertia results in a higher angular acceleration under a given torque.

α = τ/I

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3 is 3, 4 is 3, 5 is 3, 6 is false, 7 is true, 8 is false, 9 is false, 10 is 3
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Which of the following statements about dark matter and dark energy is/are true? Choose one or more: Both exceed the matter-ener
cricket20 [7]

Answer:

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Explanation:

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Liquid nitrogen boils at 77.2 k. what is this temperature in degrees celsius
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<span>77.2 k is -195.95 C
       
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Explain how carbon is cycled between the hydrosphere and geosphere. Use specific examples.
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4 years ago
Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du
Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time t is

\theta=\dfrac\alpha2t^2

It rotates 44.5 rad in this time, so we have

44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}

b. Since acceleration is constant, the average angular velocity is

\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2

where \omega_f is the angular velocity achieved after 6.00 s. The velocity of the disk at time t is

\omega=\alpha t

so we have

\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}

making the average velocity

\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}

Another way to find the average velocity is to compute it directly via

\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}

c. We already found this using the first method in part (b),

\omega=14.8\dfrac{\rm rad}{\rm s}

d. We already know

\theta=\dfrac\alpha2t^2

so this is just a matter of plugging in t=12.0\,\mathrm s. We get

\theta=179\,\mathrm{rad}

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2

Then for t=6.00\,\rm s we would get the same \theta=179\,\rm rad.

7 0
4 years ago
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