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deff fn [24]
3 years ago
5

A 3.47-m rope is pulled tight with a tension of 106 N. A wave crest generated at one end of the rope takes 0.472 s to propagate

to the other end of the rope. What is the mass of the rope (in kg)
Physics
1 answer:
konstantin123 [22]3 years ago
6 0

Answer:

The mass of the rope is 1.7 kg.

Explanation:

Given;

length of the rope, L = 3.47 m

tension on the rope, T = 106 N

period of the wave, t = 0.472 s

frequency of the is calculated as;

f = \frac{1}{t} \\\\f = \frac{1}{0.472} \\\\f = 2.1186 \ Hz

the speed of the wave is calculated as;

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

where;

v is speed of the wave = fλ

λ is the wavelength

μ is mass per unit length

f\lambda = \sqrt{\frac{T}{\mu} } \\\\f(2l) = \sqrt{\frac{T}{\mu} } \\\\2fl =  \sqrt{\frac{T}{\mu} } \\\\(2fl)^2 = \frac{T}{\mu}\\\\4f^2l^2 =\frac{T}{\mu}\\\\\mu = \frac{T}{4f^2l^2}\\\\\frac{m}{l} =  \frac{T}{4f^2l^2}\\\\m = \frac{T}{4f^2l}\\\\m = \frac{106}{4(2.1186)^2(3.47)}\\\\m = 1.7 \ kg

Therefore, the mass of the rope is 1.7 kg.

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0.15 s

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From the question given above, the following data were obtained:

Speed of sound (v) = 330 m/s

Distance (x) = 25 m

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The time taken for the echo of the sound to the bat can be obtained as follow:

v = 2x / t

330 = 2 × 25 / t

330 = 50 / t

Cross multiply

330 × t = 50

Divide both side by 330

t = 50 / 330

t = 0.15 s

Thus, it will take 0.15 s for the echo of the sound to the bat

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What is the mass of the other block?
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Let us consider the tension produced on both the sides of the rope is T.

We have been given that the rope is mass less and the rope is passing through a pulley which is stationary.

Let\ m_{1} and\ m_{2}\ are\ the\ masses\ of\ two\ blocks

Let\ m_{1}\ is\ moving\ in\ vertical\ upward\ direction\ while\ m_{2}\ is\ in\ downward\ \ direction

Hence\ m_{2} =4.5\ kg

The body is moving downward with an acceleration of  \frac{3}{4} g

As the rope is the same one which is passing over a mass less pulley and connected by two masses,hence, acceleration of each block will be the same in magnitude.

For\ body\ m_{1}

Here the tension is acting in vertical upward direction and the  weight is acting in vertical downward direction. Here,the body is moving in vertical upward direction. Hence, the net force acting on it is-

                           T-m_{1} g=m_{1}a       [1]

    For\ body\ m_{2}

Here the tension is acting in vertical upward direction while weight is in vertical downward direction. The body is moving in downward direction. Hence the net force acting on it will be-

                            m_{2} g-T=m_{2} a   [2]

Combing 1 and 2 we get-

                          T-m_{1}g=m_{1}a

                          m_{2} g-T=m_{2} a

                      -------------------------------------------------

                      [m_{2} -m_{1} ]g=a[m_{1}+ m_{2}]

                      [4.5-m_{1}]g =\frac{3}{4}g[4.5+ m_{1}]

                      4[4.5-m_{1}] =3[4.5+m_{1} ]

                      7m_{1} =4.5 kg

                      m_{1} = 0.64286 kg    [ans]

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