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Inga [223]
3 years ago
15

(a) Calculate the self-inductance of a 42.2 cm long, 10.0 cm diameter solenoid having 1000 loops. (b) How fast can it be turned

off if the induced emf cannot exceed 3.00 V and a current of 18.6 A flows through this inductor?
Physics
1 answer:
kaheart [24]3 years ago
6 0

Answer:

self-inductance is 23.4 mH

0.145 seconds it be turned off

Explanation:

given data

length L = 42.2 cm

diameter d = 10 cm

no of loop N = 1000 loops

induced emf = 3 V

current I = 18.6 A

to find out

self-inductance and How fast it turned off emf

solution

we apply here formula of self-inductance that is

self-inductance = u×N²×A / L      .......................1

put here value

self-inductance = 4π×1000²×(π×r² / 0.422)

self-inductance = 4π×1000²×(π×0.05² / 0.422)

self-inductance = 0.0234 H

so self-inductance is 23.4 mH

and

we know emf = Ldi /dt

so time = 0.0234 × 18.6

time = 0.145 seconds

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The steel would expand by 4. 8 * 10^-3 cm

<h3>How to determine the linear expansion</h3>

The change in length  ΔL is proportional to length  L. It is dependent on the temperature, substance, and length.

Using the formula:

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ΔL = 1.2 * 10^-5 * 10 * 40

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From a hot air balloon 2 km​ high, a person looks east and sees one town with angle of depression of 16 degrees. He then looks w
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In angle of depression, there is assumption that object is closer to the person observing it, so there is parallel horizontal for both observing and object been observed.

hot air balloon 2 km​ high,

there exist two triangles

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g A projectile of mass 3 kg is launched horizontally from an initial height 3 m with an initial velocity 10 m/s. This velocity i
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Answer:

The kinetic energy at ground will be "238.2 J".

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h = 3 m

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By using the conservation of energy at points A and B,

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3 years ago
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