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lbvjy [14]
3 years ago
8

A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum e

mf generated is:
Physics
1 answer:
bogdanovich [222]3 years ago
3 0

Answer:

Maximum emf = 5.32 V

Explanation:

Given that,

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions per second

Magnetic field, B = 0.5 T

We need to find maximum emf generated in the loop. It is based on the concept of Faraday's law. The induced emf is given by :

\epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t

For maximum emf, \sin\omega t=1

So,

\epsilon=NB\pi r^2\omega \\\\\epsilon=NB\pi r^2\times 2\pi f\\\\\epsilon=10\times 0.5\times \pi (0.03)^2\times 2\pi \times 60\\\\\epsilon=5.32\ V

So, the maximum emf generated in the loop is 5.32 V.

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An object, experiencing no friction, keeps moving at a constant speed. What can we say about the net force on the object?
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Answer:

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The decomposition of ammonia to the elements is a first-order reaction with a half-life of 200 s at a certain temperature. How l
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Based on the half-life of the reaction, time taken for the partial pressure of ammonia to decrease from 0.100 atm to 0.00625 atm is 800 seconds.

<h3>What is the half-life of a substance?</h3>

The half-life of a substance is the time taken for half the amount of atoms present in that substance to decay.

The half-life of the reaction is 200 seconds.

After, one half-life, pressure reduces to 0.0500 atm

After, two half-lives, pressure reduces to 0.0250 atm

After, three half-lives, pressure reduces to 0.01250 atm

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Therefore, time taken for the partial pressure of ammonia to decrease from 0.100 atm to 0.00625 atm is 800 seconds.

Learn more about half-life at: brainly.com/question/26689704

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