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Deffense [45]
3 years ago
10

An electric motor in a hair dryer is running at its normal constant operating speed and, thus, is drawing a relatively small cur

rent, as in part (b) of Example 12. The wire in the coil of the motor has some resistance. What happens to the temperature of the coil if the shaft of the motor is prevented from turning, so that the back emf is suddenly reduced to zero?
Physics
1 answer:
svlad2 [7]3 years ago
3 0

Answer:

The temperature of the coil will increase (over heating will occur)

Explanation:

This overheating generally occurs when the motor is overloaded, when a bearing seizes up, when something locks the motor shaft and prevents it from turning, or when the motor simply fails to start properly.

Back emf is zero when the motor is not turning, and it increases proportionally to the motor's angular velocity. As the motor turns faster and faster, the back emf grows, always opposing the driving emf, and reduces the voltage across the coil and the amount of current it draws.

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Explanation:

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A 1.65 kg mass stretches a vertical spring 0.260 m If the spring is stretched an additional 0.130 m and released, how long does
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Answer:

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

Explanation:

We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:

y(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right) (1)

Where:

y(t) - Position of the mass as a function of time, measured in meters.

A - Amplitude, measured in meters.

k - Spring constant, measured in newtons per meter.

m - Mass of the block, measured in kilograms.

t - Time, measured in seconds.

\phi - Phase, measured in radians.

The spring is now calculated by Hooke's Law, that is:

k = \frac{m\cdot g}{\Delta y} (2)

Where:

g - Gravitational acceleration, measured in meters per square second.

\Delta y - Deformation of the spring due to gravity, measured in meters.

If we know that m=1.65\,kg, g = 9.807\,\frac{m}{s^{2}} and \Delta y = 0.260\,m, then the spring constant is:

k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}

k = 62.237\,\frac{N}{m}

If we know that A = 0.130\,m, k = 62.237\,\frac{N}{m}, m=1.65\,kg, x(t) = 0\,m and \phi = 0\,rad, then (1) is reduced into this form:

0.130\cdot \cos (6.142\cdot t)=0 (1)

And now we solve for t. Given that cosine is a periodic function, we are only interested in the least value of t such that mass reaches equilibrium position. Then:

\cos (6.142\cdot t) = 0

6.142\cdot t = \cos^{-1} 0

t = \frac{1}{6.142}\cdot \left(\frac{\pi}{2} \right)\,s

t \approx 0.255\,s

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