A 2.31-kg rope is stretched between supports that are 10.4 m apart, and has a tension in it of 49.2 N. If one end of the rope is slightly tweaked, how long will it take for the resulting disturbance to reach the other end?

Respuesta :

Answer:

The resulting disturbance takes approximately 0.70 seconds to reach the other end.

Explanation:

The velocity of a disturbance on a rope depends on the physical properties of the material and the tension exerted to it, the equation is:

[tex]v=\sqrt{\frac{T}{\mu}} [/tex]  (1)

With v the velocity of the disturbance, T the tension of the rope and μ the linear density of the rope. The linear density of the rope is the relation between mass M and length L so we have:

[tex]\frac{M}{L} [/tex] (2)

Using (2) on (1):

[tex]v=\sqrt{\frac{TL}{M}}=\sqrt{\frac{49.2*10.4}{2.31}}\simeq14.9\frac{m}{s} [/tex] (3)

If we assume the velocity of the disturbance is constant, it will take for the resulting disturbance to reach the other ends a time t:

[tex] t=\frac{L}{v}=\frac{10.4}{14.9}\simeq0.70\,s [/tex] (4)

Lanuel

The time time taken by the resulting disturbance to reach the other end is equal to 0.70 seconds.

Given the following data:

Mass = 2.31 kg.

Distance = 10.4 m.

Tension = 49.2 N.

How to calculate the time?

In order to determine the amount of time it would take the resulting disturbance to reach the other end, we would solve for the velocity by using the following formula:

V = √(TL/M)

V = √(49.2 × 10.4/2.31)

V = 14.9 m/s.

At constant velocity, the time time taken by the resulting disturbance to reach the other end is given by:

t = L/V

t = 10.4/14.9

Time, t = 0.70 seconds.

Read more on tension here: brainly.com/question/4080400

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