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Lady_Fox [76]
3 years ago
12

Suppose a piano tuner hears 3 beats per sec-

Physics
1 answer:
vova2212 [387]3 years ago
8 0

Answer: 3 The string should be loosened because  the frequency of the string is 5 Hz above the  correct frequency.

Explanation:  In order to explain this problem we have to consider the relationship bewteen the Tension and the frequency, this is given by:

We know that:

v=\sqrt{T/\mu where μ and T are the mass linear density and Tension of teh string, respectively.

We also know that v=λ*ν the speed is related to the wavelength (λ) and the frequency (ν).

Then we have that frequency depents  directely of the Tension (T) on the string.

In our case the frequency to be tuned is 3 Hz, firstly the piano tuner slightly tightening then he  hears 5 beats per second (5 Hz). Then it is neccesary loose the tension on the string because the actual frequency is above to the correct frequency (3 Hz).

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h = 0 + 0.5 * 9.81 m/s^2 * (4.9 s)^2

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Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at
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Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

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