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Sladkaya [172]
2 years ago
10

A 20 cm square frame that can rotate about the 00' axis is placed in a homogeneous magnetic field with 0.5T induction directed v

ertically downwards.
a) Draw vectors of forces acting on the sides of the frame.
b) calculate the values of these forces and the resultant force acting on the frame.
c) calculate the torque of the frame.

Physics
1 answer:
IceJOKER [234]2 years ago
3 0

Answer:

Hello my friend. You are sh+t, your parents are sh+t and if you had some pet it is sh+t too.

Explanation:

Goodbay :)

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A 70.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 15.0 m
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Answer:

1.3823 rad/s

20.7345 m/s

28.66129935 m/s²

a=2.92164g

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Explanation:

r = Radius = 15 m

m = Mass of person = 70 kg

g = Acceleration due to gravity = 9.81 m/s²

Angular velocity is given by

\omega=13.2\times \dfrac{2\pi}{60}\\\Rightarrow \omega=1.3823\ rad/s

Angular velocity is 1.3823 rad/s

Linear velocity is given by

v=r\omega\\\Rightarrow v=15\times 1.3823\\\Rightarrow v=20.7345\ m/s

The linear velocity is 20.7345 m/s

Centripetal acceleration is given by

a_c=r\omega^2\\\Rightarrow a_c=15\times 1.3823^2\\\Rightarrow a_c=28.66129935\ m/s^2

The centripetal acceleration is 28.66129935 m/s²

Acceleration in terms of g

\dfrac{a}{g}=\dfrac{28.66129935}{9.81}\\\Rightarrow a=2.92164g

a=2.92164g

Centripetal force is given by

F_c=ma_c\\\Rightarrow F_c=70\times 28.66129935\\\Rightarrow F_c=2006.29095\ N

The centripetal force is 2006.29095 N radially outward

The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.

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