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malfutka [58]
1 year ago
12

An energy storage system based on a flywheel (a rotating disk) can store a maximum of 3.7 MJ when the flywheel is rotating at 16

000 revolutions per minute.
Physics
1 answer:
Likurg_2 [28]1 year ago
7 0

The moment of inertia of the flywheel is 2.63 kg-m^{2}

It is given that,

The maximum energy stored on the flywheel is given as

E=3.7MJ= 3.7×10^{6} J

Angular velocity of the flywheel is 16000\frac{rev}{min} = 1675.51\frac{rad}{sec}

So to find the moment of inertia of the flywheel. The energy of a flywheel in rotational kinematics is given by :

E = \frac{1}{2}Iw^{2}

By rearranging the equation:

I = \frac{2E}{w_{2} }

I = 2.63 kg-m^{2}

Thus the moment of inertia of the flywheel is 2.63 kg-m^{2}.

Learn more about moment of inertia here;

brainly.com/question/13449336

#SPJ4

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Answer:

The radius of the new planet is ~2.04 * 10⁶ m, or 2,041,752 m.

Explanation:

We can use Newton's Law of Universal Gravitation:

  • \displaystyle F_g=G\frac{Mm}{r^2}

Let's look at Newton's 2nd Law:

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We can set these equations equal to each other:

  • \displaystyle G\frac{Mm}{r^2} =ma

The mass of the second mass (astronaut) cancels out. We are left with:

  • \displaystyle G\frac{M}{r^2} =a

We are solving for the radius of the new planet, so we can rearrange the equation:

  • \displaystyle r=\sqrt{\frac{GM}{a} }

Substitute in our known values given in the problem (<u><em>G = 6.67 * 10⁻¹¹ </em></u><em> ; </em><u><em>M = 7.5 * 10²³</em></u><em> ; </em><u><em>a = 12</em></u>).

  • \displaystyle r =\sqrt{\frac{(6.67\times 10^{-11})(7.5 \times 10^{23}}{12} }
  • r=2.04 \times 10^6

The radius of the new planet is ~2.04 * 10⁶ m.

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