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1 year ago

Calculate the frequency if the number of revolutions is 300 and the paired poles are 50.

Physics

1 answer:

Andrei [34K]1 year ago

3 0

Answer: A

Explanation: We know that f=p*n

f=50*300=15000 Hz = 15kHz.

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Technician A says that power is the rate that energy is stored. Technician B says that power refers to the rate that energy is t

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Answer:

C. Both technicians A and B

Explanation:

From the physical definition, power is defined as the rate of a body doing work. It is expressed as

P = w/t watts

Where

w - is the work done or the energy of the system in joules

t - time

The unit of power is represented in watts.

Whenever there is a rate of change of energy in the system, it accounts for the efficiency of the power of the system.

Hence, the statements of both technicians are correct.

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3 years ago

A flywheel in the form of a uniformly thick disk of radius 1.33 m1.33 m has a mass of 70.6 kg70.6 kg and spins counterclockwise

ladessa [460]

Answer:

The constant torque required to stop the disk is 8.6 N-m in clockwise direction .

Explanation:

Let counterclockwise be positive direction and clockwise be negative direction .

Given

Radius of disk , r = 1.33 m

Mass of disc , m = 70.6 kg

Initial angular velocity , $\omega_i =217 rpm$

Final angular velocity , $\omega_f =0\, rpm$

Time taken to stop , t = 2.75 min

Let $\alpha$ be the angular acceleration

We know

$\omega _f=\omega _i+\alpha t$

=> $0=217+2.75\alpha =>\alpha = -78.9\frac{rev}{min^{2}}$

=> $\alpha =-\frac{78.9\times 2\pi}{60\times 60}\frac{rad}{s^{2}}=-0.138 \frac{rad}{s^{2}}$

Torque required to stop is given by

$\tau =I\alpha$

where moment of inertia , $I=\frac{mr^{2}}{2}=\frac{70.6\times 1.33^{2}}{2}kg.m^{2}=62.5 kg.m^{2}$

=> $\therefore \tau =-0.138\times 62.5\, N.m=-8.6\, N.m$

Thus the constant torque required to stop the disk is 8.6 N-m in clockwise direction .

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