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Georgia [21]
3 years ago
7

A single coil of wire, with a radius of 0.13 m is rotated in a uniform magnetic field such that the angle between the field vect

or and the area vector obeys θ=ωt. If the strength of the field is 3.746 T, and the angular frequency is 524.7 rad/s, what is the induced emf in the loop at t=1.16 s?
Physics
1 answer:
Vedmedyk [2.9K]3 years ago
5 0

The magnitude of emf induced in the single coil of wire rotated in the uniform magnetic field is 0.171 V.


<h3>Induced emf</h3>

The emf induced in the loop is determined by applying Faraday's law of electromagnetic induction.

emf = N(dФ/dt)

where;

  • N is number of turns of the wire
  • Ф is magnetic flux

Ф = BA

where;

  • B is magnetic field strength
  • A is the area of the loop

emf = NBA/t

A = πr²

A = π x (0.13)²

A = 0.053 m²

emf = NBA/t

emf = (1 x 3.746 x 0.053)/(1.16)

emf = 0.171 V

Thus, the magnitude of emf induced in the single coil of wire is 0.171 V.

Learn more about induced emf here: brainly.com/question/13744192

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There are several information's already given in the question. Based on those given information's the answer can be easily deduced. 
First angle of depression = 18°33'
                                        = [18 + (33/60)] degree
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Second angle of depression = 51°33'
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                                              = 51.55 degrees
Height of the lighthouse = 200 feet
Then
Distance traveled by the boat from when it
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                                                   = (200/0.3356) - (200/1.2594)
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I hope the procedure is clear enough for you to understand.
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3 years ago
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395,000 meters in 9000
marta [7]

Answer:

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Explanation:

The computation of the speed is shown below:

As we know that

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Distance is 395,000 meters

Time is 9,000 seconds

Now placing these values to the formula

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As speed shows the relation between the distance and time and the same is to be considered i.e by applying the formula

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3 years ago
a 2.4*10^2 N force is pulling an 85 kg refrigerator across a horizontal surface. the force acts at an angle of 20.0 above the su
shepuryov [24]

Answer:

a) The work done by pulling force, W = 1804.21 joules

b) The work done by the kinetic frictional force is, w = 1332.8 J

Explanation:

Given,

The pulling force, Fₐ = 2.4 x 10² newton

The mass of the refrigerator, m = 85 Kg

The angle of pulling force acting to the surface, ∅ = 20°

The coefficient of kinetic friction is, μₓ = 0.200

The distance covered by the refrigerator, s = 8 m

The force acting in the direction parallel to the surface is given by

                                      F = Fₐ cos∅

∴                                     F = 2.4 x 10² x cos 20°

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The work done by the pulling force is given by

                                  W = F · S

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The kinetic friction force is given by the formula

                                  Fₓ = μₓ · η

Where

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                              η - Normal force, (mg)

∴                             Fₓ = 0.200 x 85 x 9.8

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The work done by the kinetic frictional force is given by

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                                     = 1332.8 J

Hence, the work done by the kinetic frictional force is, w = 1332.8 J

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