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PolarNik [594]
3 years ago
6

Any help plz ? ………..

Physics
2 answers:
IrinaVladis [17]3 years ago
8 0

Answer:

-2+4-22--43

Explanation

This is the correct answer

Setler79 [48]3 years ago
7 0
1:2920103:93jaihuvsnall
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A 1.70 m long string has a standing wave with 2 loops at a frequency of 38.4 Hz. What is the speed of the waves in the string? (
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A wave pulse travels along a string at a speed of 200 cm/s. What will be the speed if:
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Answer:

a) v_f = \sqrt{\frac{2TL}{m}} = \sqrt{2} v = \sqrt{2}2m/s =2.83 m/s

The velocity increase by a factor of \sqrt{2}

b) v_f = \sqrt{\frac{TL}{4m}} = \frac{1}{2} v = \frac{1}{2} *2m/s =1 m/s

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c) v_f = \sqrt{\frac{4TL}{m}} = 2} v = 2 *2m/s =4m/s

The velocity increase by a factor of 2

d) v_f = \sqrt{\frac{4TL}{4m}} = v = 2m/s

The velocity not changes.

Explanation:

For this case we know that the velocity is v = 200 cm/s = 2m/s

v_f represent the final velocity after the changes specified,

Part a

The formula for the speed of a wave in a string is given by:

v = \sqrt{\frac{T}{\rho}}

And the linear density is defined as:

\rho = \frac{m}{L}

And if we replace this we got:

v = \sqrt{\frac{TL}{m}}

If the tension mass is doubled we have this:

v_f = \sqrt{\frac{2TL}{m}} = \sqrt{2} v = \sqrt{2}2m/s =2.83 m/s

The velocity increase by a factor of \sqrt{2}

Part b

If we mass is quadrupled we have this:

v_f = \sqrt{\frac{TL}{4m}} = \frac{1}{2} v = \frac{1}{2} *2m/s =1 m/s

The velocity decrease by a factor of 2.

Part c

If the length is quadrupled we have this:

v_f = \sqrt{\frac{4TL}{m}} = 2} v = 2 *2m/s =4m/s

The velocity increase by a factor of 2

Part d

For this case we know that the mass and the length are both quadrupled and we got:

v_f = \sqrt{\frac{4TL}{4m}} = v = 2m/s

The velocity not changes.

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