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Genrish500 [490]
3 years ago
13

Electromagnetic induction is the production of a voltage across a conductor when it is exposed to a varying magnetic field. This

process is essential in which of these technologies? A)Laser technolgy. B)electric motors. C) Automobile manufacturing. D) Electric power distribution.
Physics
2 answers:
Darina [25.2K]3 years ago
6 0

Answer:

D) Electric power distribution.

Explanation:

Electric power distribution requires high voltages to efficiently transmit electric power. This requires use of a transformer which uses electromagnetic induction.

Airida [17]3 years ago
5 0

Answer:

D) Electric power distribution.

Explanation:

As we know that the Faraday's law of electromagnetic induction says that rate of change in magnetic flux linked with a closed conducting loop will induce EMF in the loop.

This induced EMF is used for power production so here when we use it for the production of voltage across conductor.

it is used in AC generator when a coil is rotated at high speed between strong magnetic field of magnets.

This is induced EMF is given by

EMF = NBA\omega sin(\omega t + \phi)

so here correct answer will be

D) Electric power distribution.

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3 years ago
1.5-m length of straight wire experiences a maximum force of 1.2 N when in a uniform magnetic field that is 1.8 T. 1) What curre
Iteru [2.4K]

Answer:

 I = 0.44 A

Explanation:

The magnetic force on a conductor is given by the expression

       F = I L x B

Where bold letters indicate vectors, I is the current, L is the vector in the direction of the current, and B is the magnetic field

Since the force is maximum, the wire must be perpendicular to the magnetic field, therefore

        F = I L B sin 90

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7 0
3 years ago
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Check My Work (1 remaining) Boron coated with SiC (or Borsic) reinforced aluminum containing aligned 20 vol% fibers is an import
Lubov Fominskaja [6]

Answer:

E = 12 * 10^{6} Psi

Explanation:

Modulus of Elasticity perpendicular to the fibre,

\frac{1}{E} = \frac{B_{B} }{E_{B} } + \frac{B_{Al} }{E_{Al} }

B_{B} = 20% = 0.2\\

B_{Al} = 80% = 0.8

E_{B} =55 * 10^{6} Psi

E_{Al} =10 * 10^{6} Psi

\frac{1}{E} = \frac{0.2 }{55 * 10^{6}  } + \frac{0.8 }{10 *10^{6} }

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3 years ago
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