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anygoal [31]
3 years ago
15

The wavelength of a wave on a string is 1.2 meters. If the speed of the wave is 60 meters/second, what is its frequency? A. 0.20

hertz B. 2.0 hertz C. 50 hertz D. 10 hertz E. 15 hertz
Physics
1 answer:
anastassius [24]3 years ago
8 0

Answer:

50 Hz

Explanation:

The frequency of a wave is given by

f=\frac{v}{\lambda}

where

v is the speed of the wave

\lambda is the wavelength

In this problem,

v = 60 m/s

\lambda=1.2 m

So the frequency is

f=\frac{60 m/s}{1.2 m}=50 Hz

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There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel
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Answer:

The total electric potential at mid way due to 'q' is \frac{q}{4\pi\epsilon_{o}d}

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V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}

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Now,

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Now, if the charge q is mid way between the field, then distance is \frac{d}{2}.

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\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}

Similarly, for plate B:

\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

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