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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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A piece of aluminum has a volume of 1.50 10-3 m3. the coefficient of volume expansion for aluminum is β = 69 ✕ 10-6 (°c)-1. the
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W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3  m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

β = 69*10^-6 (°C)^-1       V = 1.50*10^-3  m^3       ΔT = 298°C

So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

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So ΔV = 3.0843*10^-5 m^3

Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

To get work, multiply the air pressure and the volume change.

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W = 3.12 J

Hope this helps!

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3 years ago
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