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3 years ago

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Ask Your Teacher A coil has an inductance of 3.00 mH, and the current in it changes from 0.200 A to 1.50 A in a time interval of

0.200 s. Find the magnitude of the average induced emf in the coil during this time interval.

1 answer:

Pepsi [2]3 years ago

8 0

Answer:

Magnitude of the average induced emf in the coil $= 0.0195$ Volts

Explanation:

Induced emf in a coil with current flowing through it is can be calculated using the relation given below

$emf = -L\frac{dI}{dt}$ ------------ Equation (1)

Where "L" represents the inductance in H

dI represents the change in current in Ampere

and dt represents the change in time or time period for which the current is flowing in seconds

Given

$L = 3.00$ mH $= 3 * 10^ {-3}$ H

$I_0 = 0.200$ Ampere

$I_t = 1.50$ Ampere

$t = 0.200$ seconds

Substituting the given values in equation (1) we get -

$emf = - (3 * 10^{-3}) * \frac{1.500 - 0.200}{0.200} \\emf = 0.0195$

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3 years ago

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AysviL [449]

Answer:

The impulse transferred to the nail is 0.01 kg*m/s.

Explanation:

The impulse (J) transferred to the nail can be found using the following equation:

$J = \Delta p = p_{f} - p_{i}$

Where:

$p_{f}$ : is the final momentum

$p_{i}$ : is the initial momentum

The initial momentum is given by:

$p_{i} = m_{1}v_{1_{i}} + m_{2}v_{2_{i}}$

Where 1 is for the hammer and 2 is for the nail.

Since the hammer is moving down (in the negative direction):

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And because the nail is not moving:

$v_{2_{i}}= 0$

$p_{i} = m_{1}v_{1_{i}} = 0.25 kg*(-10 m/s) = -2.5 kg*m/s$

Now, the final momentum can be found taking into account that the hammer remains in contact with the nail during and after the blow:

$p_{f} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

Since the hammer and the nail are moving in the negative direction:

$v_{1_{f}}$ = $v_{2_{f}}$ = -9.7 m/s

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Finally, the impulse is:

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Therefore, the impulse transferred to the nail is 0.01 kg*m/s.

I hope it helps you!

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3 years ago