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DerKrebs [107]
3 years ago
7

Two tiny conducting spheres are identical and carry charges of -18.4 µC and +53.0 µC. They are separated by a distance of 2.73 c

m. (a) What is the magnitude of the force that each sphere experiences, and is the force attractive or repulsive?
Physics
1 answer:
melomori [17]3 years ago
4 0

Answer:

-11776.36 N

This force is attractive since both charges are of opposite sign

Explanation:

Given that

q_1=-18.4\mu C=-18.4\times 10^{-6}C\ and\\\ q_2=53\mu C=53\times 10^{-6}C

Distance between the spheres = 2.73 cm =0.0273 m

where K is Coulomb's constant = 9.10^ 9 [N.m^2 /C^2]

According to coulombs law we know that force between two charges is given by

F = \frac{kq_1q_2}{r^2}

F = \frac{(9\times10^9)\times (-18.4 \times 10^-^6)(53  \times 10^-^6}{0.0273^2)} \\= -11776.36N

This force is attractive since both charges are of opposite sign

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The frequency of the human ear canal is 2.92 kHz.

Explanation:

As the ear canal is like a tube with open at one end, the wavelength of sound passing through this tube will propagate 4 times its length of the tube. So wavelength of the sound wave will be equal to four times the length of the tube. Then the frequency can be easily determined by finding the ratio of velocity of sound to wavelength. As the velocity of sound is given as 339 m/s, then the wavelength of the sound wave propagating through the ear canal is  

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As length of the ear canal is given as 2.9 cm, it should be converted into meter as follows:

wavelength = 4*2.9*10^{-2} =0.116

Then the frequency is determined as

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