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DerKrebs [107]
3 years ago
13

While playing her guitar , karen plucks one string with increasing levels of force. What effect does this have on the sound prod

uced?
Physics
2 answers:
Lena [83]3 years ago
8 0

Answer:

The amplitude of vibration of string will increase due to which loudness of sound will increase

Explanation:

As we know that the guitar is based on the principle of Resonance. When string of the guitar vibrates at a given frequency then the sound produced in the hollow part of the guitar will also be at same frequency.

This is known as resonance condition, so guitar will produce same frequency sound as that of frequency of string.

Now if the string is plucked with increasing level of force then it will increase the amplitude of vibrations of the string due to which the sound produced in the guitar will also be of same level.

So here we can say that amplitude and intensity of sound related as

I = kA^2

so on increasing amplitude the intensity will increase and hence it will produce loud sound

Nikitich [7]3 years ago
8 0

Answer:

B. The amplitude of the sound keeps increasing.

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Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2
Alinara [238K]

Answer:

a)E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

b)E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

Explanation:

Given that

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q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

E=K\dfrac{q}{r^2}

E_1=K\dfrac{q_1}{r_1^2}

E_2=K\dfrac{q_2}{r_2^2}

Now by putting the va;ues

a)

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C

E_1=72.11\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C

E_2=21.58\times 10^{6}\ N/C

The net electric field

E=E_1-E_2

E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C

E_1=120.3\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C

E_2=147.92\times 10^{6}\ N/C

The net electric field

E=E_1+E_2

E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

   

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