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mrs_skeptik [129]
3 years ago
16

Rohit is trying to tune an instrument. He strikes a tuning fork and then plays a note that almost matches it. Rohit hears the vo

lume of the two waves beating, which sounds like "wah-wah-wah-wah."
Physics
2 answers:
stellarik [79]3 years ago
5 0

The number of "wah"s every second is the difference between the frequencies of the tuning fork and the note he is playing.

uranmaximum [27]3 years ago
4 0

Answer:

The two waves produced have different frequencies.

Explanation:

Rohit is trying to tune an instrument. He strikes a tuning fork and then plays a note that almost matches it. Rohit hears the volume of the two waves beating, which sounds like "wah-wah-wah-wah."

The options can be the following

The two waves produced have different speeds.

The two waves produced have different amplitudes.

The two waves produced have different wavelengths.

The two waves produced have different frequencies.

A wave is a disturbance which  transfers energy through a medium without displacing the medium itself.

The two waves have different frequencies to be able to produce such a sound. That's called a beat frequency. the waves interfere constructively.

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Given:

Radius of the ring (r) = 10.0 cm = 0.10 m           [1 cm = 0.01 m]

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Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

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(a)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2

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(b)

Given:

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Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C

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Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2

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(d)

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Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2

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