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myrzilka [38]
3 years ago
8

Some special vehicles have spinning disks (flywheels) to store energy while they roll downhill. They use that stored energy to l

ift themselves uphill later on. Their flywheels have relatively small rotational masses but spin at enormous angular speeds. How would a flywheel’s kinetic energy change if its rotational mass were 7 times larger but its angular speed were 7 times smaller?
Physics
1 answer:
Evgesh-ka [11]3 years ago
4 0

Answer:

Smaller by 7 times

Explanation:

Rotational mass is called moment of inertia

So, initial moment of inertia, I1 = I

initial angular velocity, ω1 = ω

Final moment of inertia, I2 = 7I

Final angular velocity, ω2 = ω/7

The kinetic energy in rotational motion is given by

K = \frac{1}{2}I\omega ^{2}

So, initial kinetic energy of rotation

K_{1} = \frac{1}{2}I_{1}\omega_{1} ^{2}= \frac{1}{2}I\omega ^{2}

So, final kinetic energy of rotation

K_{2} = \frac{1}{2}I_{2}\omega_{2} ^{2}= \frac{1}{2}7I \left (\frac{\omega}{7}  \right ) ^{2}

K_{2} = \frac{K_{1}}{7}

Thus, teh kinetic energy becomes smaller by 7 times.

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An object located near the surface of Earth has a weight of a 245 N
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Answer:

The mass of the object is 24.5 kg and weight of the object on Mars is 91.14 N.

Explanation:

Weight of the object on the surface of Earth, W = 245 N

On the surface of Earth, acceleration due to gravity, g = 10 m/s²

Weight of an object is given by :

W = mg

m is mass

m=\dfrac{W}{g}\\\\m=\dfrac{245\ N}{10\ m/s^2}\\\\=24.5\ kg

So, the mass of the object is 24.5 kg

Acceleration due to gravity on Mars, g' = 3.72 m/s²

Weight of the object on Mars,

W' =mg'

W' = 24.5 kg × 3.72 m/s²

= 91.14 N

So, the weight of the object on Mars is 91.14 N.

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