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3 years ago

Which of the following would describe a length that is 2.0×10^-3 of a meter? a: 2.0 kilometers

Physics

1 answer:

natita [175]3 years ago

5 0

Answer:

Option (c) $2\times 10^{-3}\ m=2\ millimeters$

Explanation:

Here, it is required to describe the given length in a particular unit. Firstly, we need to see the following conversions as :

1 meter = 0.001 kilometers

1 meter = 10⁻⁶ megameters

1 meter = 1000 millimeters

1 meter = 1000000 micrometers

From the given option, the correct one is (c) because, 1 meter = 1000 millimeters

So, $2\times 10^{-3}\ m=2\ millimeters$

Hence, the correct option is (c). Hence, this is the required solution.

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

alisha [4.7K]

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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3 years ago