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finlep [7]
2 years ago
12

What is the speed of a transverse wave in a rope of length 3. 1 m and mass 86 g under a tension of 380 n?

Physics
1 answer:
Nitella [24]2 years ago
3 0

117 m/sec is the speed of a transverse wave in a rope of length 3. 1 m and mass 86 g under a tension of 380 n.

The wave speed v is given by

v= √τ/μ

​where τ is the tension in the rope and μ is the linear mass density of the rope.

The linear mass density is the mass per unit length of rope :

μ= m / L = (0.086 kg)/(3.1 m)=0.0277 kg/m.

v= \sqrt{ \frac{380 N}{0.0277 kg/m}}  = 117.125 m/sec (approx. 117 m/sec

In physics, a transverse wave is a wave whose oscillations are perpendicular to the direction of the wave's advance. This is in contrast to a longitudinal wave which travels in the direction of its oscillations. Water waves are an example of transverse wave.

Transverse waves commonly occur in elastic solids due to the shear stress generated; the oscillations in this case are the displacement of the solid particles away from their relaxed position, in directions perpendicular to the propagation of the wave. These displacements correspond to a local shear deformation of the material. Hence a transverse wave of this nature is called a shear wave. Since fluids cannot resist shear forces while at rest, propagation of transverse waves inside the bulk of fluids is not possible.

Learn more about Transverse waves here : brainly.com/question/13761336

#SPJ4

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Answer:

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Explanation:

We know that the length of the circular path, L the plane travels is

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Now, from θ = θ₀ + ωt where θ₀ = 0 rad, ω = angular speed  and t = time,

θ = θ₀ + ωt = 0 + ωt = ωt

So, v = rdθ/dt + θdr/dt

v = rω + ωtdr/dt

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v = (r + tdr/dt)v'/r

v = v' + tv'/r(dr/dt)

v = v'[1 + t(dr/dt)/r]

Given that v' = 110 m/s, t = 33.0s, r = 120 m and dr/dt = rate at which line is shortened = -0.80 m/s (negative since it is decreasing)

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v = 110 m/s[1 - 0.22]

v = 110 m/s(0.78)

v = 85.8 m/s

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