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Alexeev081 [22]
3 years ago
7

Calculate the moment of inertia and the rotational kinetic energy for the following objects spinning about a central axis (in un

its of Joules): a solid sphere with a mass of 200 grams and a radius of 5.0 cm rotating with an angular speed of 2.5 rad/sec. a thin spherical shell with a mass of 200 grams and a radius of 5.0 cm rotating with an angular speed of 2.5 rad/sec. a solid cylinder with a mass of 200 grams and a radius of 5.0 cm rotating with an angular speed of 2.5 rad/sec. a hoop with a mass of 200 grams and a radius of 5.0 cm rotating with an angular speed of 2.5 rad/sec.
Physics
1 answer:
Helga [31]3 years ago
6 0

Answer:

a) K = 6.25 10⁻⁴ J , b)  K = 10.41  10⁻⁴ J , c)  K = 7.81 10⁻⁴ J , d)  K = 15.625 10⁻⁴ J

Explanation:

The moment of inertia is

         I = ∫ r² dm

For high symmetry bodies it is tabulated.

The rotational kinetic energy is

         K = ½ I w²

Let's calculate the values ​​of the given bodies

a) solid sphere

         I = 2/5 M R²

         I = 2/5 0.200 0.05²

         I = 2 10⁻⁴ kg m²

        K = ½ 2 10⁻⁴ 2.5²

        K = 6.25 10⁻⁴ J

b) spherical shell

        I = 2/3 M R²

        I = 2/3 0.200 0.05²

        I = 3.33 10⁻⁴ kg m²

        K = ½ 3.33 10-4 2.5²

        K = 10.41  10⁻⁴ J

c) solid cylinder

        I = ½ M R²

        I = ½ 0.200 0.05²

        I = 2.5 10⁻⁴ kg m²

       K = ½ 2.5 10-4 2.5²

       K = 7.81 10⁻⁴ J

d) hoop

       I = M R²

       I = 0.200 0.05²

       I = 5 10⁻⁴ kg m²

      K = ½ 5 10⁻⁴ 2.5²

      K = 15.625 10⁻⁴ J

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5 0
2 years ago
An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p
igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

Explanation:

Given:

Distance from the wire to the field point r = 2.83 \times 10^{-2} m

Speed of electron v = 35.5 \%c

Current I = 17.7 A

For finding the acceleration,

First find the magnetic field due to wire,

  B = \frac{\mu _{o}I }{2\pi r }

Where \mu_{o} = 4\pi   \times 10^{-7}

  B = \frac{4\pi \times 10^{-7}  \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }

  B = 12.50 \times 10^{-5} T

The magnetic force exerted on the electron passing through straight wire,

  F = qvB  

  F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}

  F = 21.3 \times 10^{-16} N

From the newton's second law

  F = ma

Where m = mass of electron = 9.1 \times 10^{-31} kg

So acceleration is given by,

   a = \frac{F}{m}

   a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }

   a = 2.34 \times 10^{15} \frac{m}{s^{2} }

Therefore, the magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

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