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Ivanshal [37]
3 years ago
7

A 10-kg disk-shaped flywheel of radius 9.0 cm rotates with a rotational speed of 320 rad/s. Part A Determine the rotational mome

ntum of the flywheel. Express your answer to two significant figures and include the appropriate units. Part B With what magnitude rotational speed must a 10-kg solid sphere of 9.0 cm radius rotate to have the same rotational momentum as the flywheel? Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Leya [2.2K]3 years ago
4 0

Answer:

(A). The rotational momentum of the flywheel is 12.96 kg m²/s.

(B). The rotational speed of sphere is 400 rad/s.

Explanation:

Given that,

Mass of disk = 10 kg

Radius = 9.0 cm

Rotational speed = 320 m/s

(A). We need to calculate the rotational momentum of the flywheel.

Using formula of momentum

L=I\omega

L=\dfrac{1}{2}mr^2\omega

Put the value into the formula

L=\dfrac{1}{2}\times10\times(9.0\times10^{-2})^2\times320

L=12.96\ kg m^2/s

(B). Rotation momentum of sphere is same rotational momentum of the  flywheel

We need to calculate the magnitude of the rotational speed of sphere

Using formula of rotational momentum

L_{sphere}=L_{flywheel}

I\omega_{sphere}=I\omega_{flywheel}

\omega_{sphere}=\dfrac{I\omega_{flywheel}}{I_{sphere}}

\omega_{sphere}=\dfrac{I\omega_{flywheel}}{\dfrac{2}{5}mr^2}

Put the value into the formula

\omega_{sphere}=\dfrac{12.96}{\dfrac{2}{5}\times10\times(9.0\times10^{-2})^2}

\omega_{sphere}=400\ rad/s

Hence, (A). The rotational momentum of the flywheel is 12.96 kg m²/s.

(B). The rotational speed of sphere is 400 rad/s.

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The momentum of an electron is 1.75 times larger than the value computed non-relativistically. What is the speed of the electron
FrozenT [24]

Answer:

<em>Speed of the electron is 2.46 x 10^8 m/s</em>

<em></em>

Explanation:

momentum of the electron before relativistic effect = M_{0} V

where M_{0} is the rest mass of the electron

V is the velocity of the electron.

under relativistic effect, the mass increases.

under relativistic effect, the new mass M will be

M = M_{0}/ \sqrt{1 - \beta ^{2}  }

where

\beta = V/c

c  is the speed of light = 3 x 10^8 m/s

V is the speed with which the electron travels.

The new momentum will therefore be

==> M_{0}V/ \sqrt{1 - \beta ^{2}  }

It is stated that the relativistic momentum is 1.75 times the non-relativistic momentum. Equating, we have

1.75M_{0} V = M_{0}V/ \sqrt{1 - \beta ^{2}  }

the equation reduces to

1.75 = 1/ \sqrt{1 - \beta ^{2}  }

square both sides of the equation, we have

3.0625 = 1/(1 - \beta ^{2} )

3.0625 - 3.0625\beta ^{2} = 1

2.0625 = 3.0625\beta ^{2}

\beta ^{2} = 0.67

β = 0.819

substitute for  \beta = V/c

V/c = 0.819

V = c x 0.819

V = 3 x 10^8 x 0.819 = <em>2.46 x 10^8 m/s</em>

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2001240Determine the specific kinetic energy of a mass whose velocity is 40 m/s, in kJ/kg.
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Answer:

The specific kinetic energy of a mass is 0.8 kJ/kg

Explanation:

Given that,

Velocity = 40 m/s

Specific kinetic energy is the kinetic energy per unit mass.

We need to calculate the specific kinetic energy

Using formula of specific kinetic energy

K.E=\dfrac{\dfrac{1}{2}mv^2}{m}

K.E=\dfrac{v^2}{2}

Put the value into the formula

K.E=\dfrac{40^2}{2}

K.E=800\ J/kg

We know that,

1 kJ = 1000 J

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The specific energy is

K.E=800\times0.001

K.E=0.8\ kJ/kg

Hence, The specific kinetic energy of a mass is 0.8 kJ/kg

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