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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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Explanation:

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Unlike poles attract each other. There is magnetic force of attraction between the opposite poles. South- North poles attract each other.

Mono poles cannot exist.

From the given statements, the magnetic poles are described by:

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3 years ago
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If an automobile engine delivers a power of 50.0 hp, how much time will it take for the engine to do 6.40 x 10^4 j of work? (1 h
aleksandrvk [35]

Given data

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                = 50 × 746,  we know that 1 hp = 746 W.

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Power is defined as rate of doing work and the unit of power is<em> Watt.</em>

Mathematically,

         Power = (Work / time)   Watts

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                      = 6.40 ×10⁴/37300

                      = <em>1.74 seconds  </em>

<em>  </em><em>Therefore , the engine need 1.74 seconds to do 6.40 6.40 ×10⁴ J of work </em>

<em> </em>


7 0
4 years ago
The total charge a battery can supply is rated in mA⋅h , the product of the current (in mA ) and the time (in h ) that the batte
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Answer: 0.2  hours

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Fron this we can obtain:

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then considering that this battery can supply 1800 mA in one hour we have this battery can supply 350 mA  in x time in the form:

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x hour--------350 mA

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