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Andru [333]
2 years ago
7

In a model generator, a 310-turn rectangular coil 0.075 m by 0.18 m rotates with an angular frequency of 12.2 rad/s in a uniform

magnetic field of 0.64 T. What is the maximum emf induced in the coil
Physics
1 answer:
Novosadov [1.4K]2 years ago
4 0

This question involves the concepts of the induced emf, magnetic field, and angular frequency.

The maximum emf induced in the coil is "32.68 volts".

The following formula can be used to find out the induced emf in the coil.

E =N\omega AB\ Sin(\omega t)

But, for maximum induced emf sin function should have a value of 1. Therefore,

E = N\omega AB

where,

E = induced emf = ?

N = No. of turns in the coil = 310 turns

ω = angular frequency = 12.2 rad/s

A = Area of coil = (0.075 m)(0.18 m) = 0.0135 m²

B = magnetic field = 0.64 T

Therefore,

E=(310)(12.2\ rad/s)(0.0135\ m^2)(0.64\ T)

<u>E = 32.68 volts</u>

Learn more about induced emf here:

brainly.com/question/18441805?referrer=searchResults

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love history [14]

Answer:

F = 39.2 N

Explanation:

Since, the object is in uniform motion. Therefore, the frictional force on object will be:

Frictional Force = μk N = μk mg

where,

μk = coefficient of kinetic friction = 0.2

m = mass of crate = 10 kg

g = 9.8 m/s²

Therefore,

Frictional Force = (0.2)(10 kg)(9.8 m/s²)

Frictional Force = 19.6 N

The horizontal component of force must be equal to this frictional force to continue the uniform motion:

F Sin 30° = 19.6 N

F = 19.6 N/Sin 30°

<u>F = 39.2 N</u>

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The answer would be C
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The psychologist known for latent learning and cognitive maps is _________. A. Robert Rescorla B. Edward Tolman C. William James
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Answer:

B

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B. Edward Tolman

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Read 2 more answers
Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts
Nata [24]

(a) The system of interest if the acceleration of the child in the wagon is to be calculated are the wagon and the children outside the wagon.

(b) The acceleration of the child-wagon system is 0.33 m/s².

(c) Acceleration of the child-wagon system is zero when the frictional force is 21 N.

<h3>Net force on the third child</h3>

Apply Newton's second law of motion;

∑F = ma

where;

  • ∑F is net force
  • m is mass of the third child
  • a is acceleration of the third child

∑F = 96 N - 75 N - 12 N = 9 N

Thus, the system of interest if the acceleration of the child in the wagon is to be calculated are;

  • the wagon
  • the children outside the wagon

<h3>Free body diagram</h3>

           →                 →              Ф                         ←

         1st child      friction       wagon                2nd child

<h3>Acceleration of the  child and wagon system</h3>

a = ∑F/m

a = 9 N / 27 kg

a = 0.33 m/s²

<h3>When the frictional force is 21 N</h3>

∑F = 96 N - 75 N - 21 N = 0 N

a = ∑F/m

a = 0/27 kg

a = 0 m/s²

Learn more about net force here: brainly.com/question/14361879

#SPJ1

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2 years ago
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