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marissa [1.9K]
3 years ago
5

A constant voltage of 12.00 V has been observed over a certain time interval across a 1.20 H inductor. The current through the i

nductor, measured as 3.00 A at the beginning of the time interval, was observed to increase at a constant rate to a value of 8.00 A at the end of the time interval. How long was this time interval?
Physics
1 answer:
kiruha [24]3 years ago
5 0

Answer:

The time is 0.5 sec.

Explanation:

Given that,

Voltage V= 12.00 V

Inductance L= 1.20 H

Current = 3.00 A

Increases rate = 8.00 A

We need to calculate change in current

\Delta A = 8.00-3.00= 5.00\ A

We need to calculate the time interval

Using formula of inductor

V=L\dfrac{\Delta A}{\Delta t}

\Delta t =\dfrac{L\Delta A}{V}

Where, \Delta A = change in current

V = voltage

L = inductance

Put the value into the formula

\Delta t=\dfrac{1.20\times5.00}{12.00}

\Delta t=0.5\ sec

Hence, The time is 0.5 sec.

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Answer:

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Explanation:

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A 70 ft rope hangs from a helicopter above this room. The rope has a mass per unit length of 2 lb/ft. In order to be rescued fro
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Answer:

The work done to get you safely away from the test is  2.47 X 10⁴ J.

Explanation:

Given;

length of the rope, L = 70 ft

mass per unit length of the rope, μ = 2 lb/ft

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mass of the 70 ft rope  = 2 lb/ft x 70 ft

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Total mass to be pulled to the helicopter, M = 120 lbs  + 140 lbs  

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The work done is calculated from work-energy theorem as follows;

W = Mgh

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g is acceleration due gravity = 32.17 ft/s²

h is height the total mass is raised = length of the rope = 70 ft

W = 260 Lb x 32.17 ft/s²  x 70 ft

W = 585494 lb.ft²/s²

1 lb.ft²/s² = 0.0421 J

W = 585494 lb.ft²/s²  = 2.47 X 10⁴ J.

Therefore, the work done to get you safely away from the test is  2.47 X 10⁴ J.

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