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Vika [28.1K]
3 years ago
12

A fireman is sliding down a fire pole. As he speeds up, he tightens his grip on the pole, thus increasing the vertical frictiona

l force that the pole exerts on the fireman. When the force on his hands equals his weight, what happens to the fireman?
A. The fireman comes to a stop.
B. The fireman descends with slower and slower speed.
C. The fireman descends with a smaller but non-zero acceleration.
D. The fireman continues to descend, but with constant speed.
E. The acceleration of the fireman is now upward.
Physics
1 answer:
Gnesinka [82]3 years ago
6 0

Answer:

The fireman continues to descend but with constant speed.

Explanation:  

According to Newton first law of motion, The object which is rest continues to remain at rest or object moving with constant speed move with constant velocity when acted upon by an unbalanced force.

When the net force acting on the object becomes zero the object moves with constant speed.

Hence when the frictional force acting on the fireman hands equals to his body weight , the net force acting on the fireman becomes zero and thus fireman continues to descend with constant speed.

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Answer:

Part a)

Average EMF for half cycle is

E_{avg} = 2.64 V

Part b)

For one complete cycle we will have

E_{avg} = 0

Part c)

Maximum induced EMF will be at

t = 0.025 s and 0.075 s

minimum induced EMF is at

t = 0.05s and 0.1 s

Explanation:

As we know that magnetic field is oscillating in direction as well as magnitude

so induced EMF is given as

E = NBA\omega sin(\omega t)

Part a)

For average value of EMF from positive maximum to negative maximum which is equal to half cycle

so we have

E_{avg} = NBA\omega \frac{2}{T}\int_0^{T/2} sin(\omega t) dt

E_{avg} = \frac{2NBA\omega}{\pi}

E_{avg} = \frac{2(10)(0.12)(0.055)(2\pi (10))}{\pi}

E_{avg} = 2.64 V

Part b)

For one complete cycle we will have

E_{avg} = NBA\omega \frac{1}{T}\int_0^T sin(\omega t) dt

E_{avg} = 0

Part c)

Maximum induced EMF will be at

t = \frac{T}{4} and \frac{3T}{4}

here we know

T = \frac{1}{f} = 0.1 s

t = 0.025 s and 0.075 s

minimum induced EMF is at

t = \frac{T}{2} and T

so it is

t = 0.05s and 0.1 s

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A thin aluminum meter stick hangs from a string attached to the 50.0 cm mark of the stick. From the 0.00 cm mark on the meter st
devlian [24]

Answer:

Stop cheating in exam

Explanation:

Shame!!!!

I am sorry but I will have to refer you to the student conduct at UTA.

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muminat
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8 0
2 years ago
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Answer:

B splits and goes through two components

Explanation:

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So, the correct answer is

B splits and goes through two components

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3 years ago
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