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Dahasolnce [82]
3 years ago
6

A heavy solid disk rotating freely and slowed only by friction applied at its outer edge takes 120 seconds to come to a stop.

Physics
1 answer:
alisha [4.7K]3 years ago
5 0

Answer:

The time is 16 min.

Explanation:

Given that,

Time = 120 sec

We need to calculate the moment of inertia

Using formula of moment of inertia

I=\dfrac{1}{2}MR^2

If the disk had twice the radius and twice the mass

The new moment of inertia

I'=\dfrac{1}{2}\times2M\times(2R)^2

I'=8I

We know,

The torque is

\tau=F\times R

We need to calculate the initial rotation acceleration

Using formula of acceleration

\alpha=\dfrac{\tau}{I}

Put the value in to the formula

\alpha=\dfrac{F\times R}{\dfrac{1}{2}MR^2}

\alpha=\dfrac{2F}{MR}

We need to calculate the new rotation acceleration

Using formula of acceleration

\alpha'=\dfrac{\tau}{I'}

Put the value in to the formula

\alpha=\dfrac{F\times R}{8\times\dfrac{1}{2}MR^2}

\alpha=\dfrac{2F}{8MR}

\alpha=\dfrac{\alpha}{8}

Rotation speed is same.

We need to calculate the time

Using formula angular velocity

\Omega=\omega'

\alpha\time t=\alpha'\times t'

Put the value into the formula

\alpha\times120=\dfrac{\alpha}{8}\times t'

t'=960\ sec

t'=16\ min

Hence, The time is 16 min.

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A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c
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Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = 71\times 10^{- 9} C

Q' = + 42 nC = 42\times 10^{- 9} C

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Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

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Now,

Electric field at the mid-point due to charge Q' is given by:

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The net Electric field is given by:

\vec{E_{net}} = \vec{E} - \vec{E'}

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