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shutvik [7]
3 years ago
9

While punting a football, a kicker rotates his leg about the hip joint. The moment of inertia of the leg is 3.75 kg⋅m² and its r

otational kinetic energy is 175 J.
(a) What is the angular velocity of the leg?
(b) What is the velocity of the tip of the punter's shoe if it is 1.05 m from the hip joint?
Physics
1 answer:
forsale [732]3 years ago
4 0

Answer:

9.6609 rad/s

10.143945 m/s

Explanation:

I = Moment of inertia = 3.75 kgm²

K = Kinetic energy = 175 J

r = Radius = 1.05 m

Kinetic energy is given by

K=\dfrac{1}{2}I\omega^2\\\Rightarrow \omega=\sqrt{\dfrac{2K}{I}}\\\Rightarrow \omega=\sqrt{\dfrac{2\times 175}{3.75}}\\\Rightarrow \omega=9.6609\ rad/s

The angular velocity of the leg is 9.6609 rad/s

Velocity is given by

v=r\omega\\\Rightarrow v=1.05\times 9.6609\\\Rightarrow v=10.143945\ m/s

The velocity of the tip of the punters shoe is 10.143945 m/s

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What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally h
zysi [14]

Complete question:

What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)

Answer:

The peak emf generated by the coil is 15.721 kV

Explanation:

Given;

Radius of coil, r = 0.250 m

Number of turns, N = 500-turn

time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s

magnetic field strength, B = 0.425 T

Induced peak emf = NABω

where;

A is the area of the coil

A = πr²

ω is angular velocity

ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s =  60 rev/s

Induced peak emf = NABω

                               = 500 x (π x 0.25²) x 0.425 x 376.738

                               = 15721.16 V

                               = 15.721 kV

Therefore, the peak emf generated by the coil is 15.721 kV

5 0
3 years ago
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