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Yuri [45]
4 years ago
13

An inductor is connected in series to a fully charged capacitor. Which of the following statements are true? Check all that appl

y.- As the capacitor is charging, the current is increasing.- The stored electric field energy can be greater than the stored magnetic field energy.- As the capacitor is discharging, the current is increasing.- The stored electric field energy can be less than the stored magnetic field energy.- The stored electric field energy can be equal to the stored magnetic field energy.
Physics
1 answer:
Luba_88 [7]4 years ago
3 0

Answer:

As the capacitor is discharging, the current is increasing

Explanation:

Lets take

C= Capacitance

L=Inductance

V=Voltage

I= Current

The total energy E given as

E=\dfrac{IL^2}{2}+\dfrac{CV^2}{2}

We know that total energy E is conserved so when electric energy 1/2 CV² decreases then magnetic energy 1/2 IL²  will increases.

It means that when charge on the capacitor decreases then the current will increase.

As the capacitor is discharging, the current is increasing

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According to Newton’s law of universal gravitation, which statements are true? As we move to higher altitudes, the force of grav
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According to Newton’s law of universal gravitation, as we move to higher altitudes, the force of gravity on us decreases and as we gain mass, the force of gravity on us increases both are the true statement.  

<u>Explanation:  </u>

Newton law of universal gravity extends gravity beyond the earth's surface. This gravity depends on the masses directly and inverse to the distance square between their centers.

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A string of length L, mass per unit length \mu, and tension T is vibrating at its fundamental frequency. What effect will the fo
viva [34]

The fundamental frequency on a vibrating string is given by:

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}

where

L is the length of the string

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Keeping this equation in mind, we can now answer the various parts of the question:

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In this case, the length of the string is doubled:

L' = 2L

Substituting into the expression of the fundamental frequency, we find the new frequency:

f'=\frac{1}{2(2L)}\sqrt{\frac{T}{\mu}}=\frac{1}{2}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\frac{f}{2}

So, the fundamental frequency will halve.

(b) the fundamental frequency will decrease by a factor \sqrt{2}

In this case, the mass per unit length is doubled:

\mu'=2\mu

Substituting into the expression of the fundamental frequency, we find the new frequency:

f'=\frac{1}{2L}\sqrt{\frac{T}{2 \mu}}=\frac{1}{\sqrt{2}}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\frac{f}{\sqrt{2}}

So, the fundamental frequency will decrease by a factor \sqrt{2}.

(c) the fundamental frequency will increase by a factor \sqrt{2}

In this case, the tension is doubled:

T'=2T

Substituting into the expression of the fundamental frequency, we find the new frequency:

f'=\frac{1}{2L}\sqrt{\frac{2T}{\mu}}=\sqrt{2}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\sqrt{2}f

So, the fundamental frequency will increase by a factor \sqrt{2}.

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