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drek231 [11]
3 years ago
9

A 0.35-kgkg cord is stretched between two supports, 7.4 mm apart. When one support is struck by a hammer, a transverse wave trav

els down the cord and reaches the other support in 0.88 ss .What is the tension in the cord? (Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Agata [3.3K]3 years ago
4 0

Answer:

The tension in the string is T = 1.49*10^{-6}N.

Explanation:

For a string with tension T and linear density \mu_d carrying a transverse wave at speed v it is true that

v = \sqrt{\dfrac{T}{\mu_d} }

solving for T we get:

T = \dfrac{v^2}{\mu_d}.

Now, the transverse wave covers the distance of 7.4mm in 0.88s, which means it's speed is

v =\dfrac{7.4*10^{-3}m}{0.88s} \\\\v = 8.4*10^{-3}s

And it's linear density (mass per unit length) is

\mu_d = \dfrac{0.35kg}{7.4*10^{-3}m} \\\\\mu_d = 47.3kg/m

Therefore, the tension in the cord is

T = \dfrac{(8.4*10^{-3}m/s^2)^2}{47.3kg/m}.

\boxed{T = 1.5*10^{-6}N}

or in micro newtons

T =1.5\mu N

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Answer:

Part a)

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Part b)

T_R = 379 N

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

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Part a)

so from above equations we have

514 = 0.906T_L + 0.985(\frac{T_L}{0.41})

514 = 3.3 T_L

T_L = 155.4 N

Part b)

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T_R = \frac{T_L}{0.41}

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Answer:

Below

Explanation:

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Rounding to 3 sig figs

     = 81.7 m

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