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inn [45]
3 years ago
15

A string of mass m and length L is under tension T. The speed of a wave in the string is v. What will be the speed of a wave in

the string if the mass of the string is increased to 2m, with no change in length?
Physics
1 answer:
Gekata [30.6K]3 years ago
3 0

Answer:

\frac{1}{\sqrt{2}}

Explanation:

The speed of a wave in a string is given by:

v=\sqrt{\frac{T}{m/L}}

where

T is the tension in the string

m is the mass of the string

L is the length

In this problem, the mass of the string is increased to 2m: m' = 2 m, while the length is not changed, L'=L. If the tension in the string is not changed, then the new speed of the wave in the string will be:

v'=\sqrt{\frac{T}{m'/L'}}=\sqrt{\frac{T}{2m/L}}=\frac{1}{\sqrt{2}}\sqrt{\frac{T}{m/L}}=\frac{v}{\sqrt{2}}

so, the speed of the wave decreases by a factor \frac{1}{\sqrt{2}}

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Answer: i) 2.356 × 10^-3 m = 2.356mm, ii) 4.712 × 10^-3 m = 4.712mm

Explanation: The formulae that relates the position of a fringe from the center to the wavelength, distance between slits and distance between slits and screen is given below as

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Where y = distance between nth fringes and the center fringe.

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R = distance between slits and screen = 1.0m

d = distance between slits = 0.25mm = 0.00025m

For distance between the first dark fringe and the center fringe.

This implies that m = 1

y = 1 × 589×10^-9 × 1/0.00025

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y = 2,356,000 × 10^-9

y = 2.356 × 10^-3 m = 2.356mm

For the second dark fringe, this implies that m = 2

y = 1 × 2 × 589×10^-9/0.00025

y = 1178 × 10^-9 /0.00025

y = 4,712,000 × 10^-9

y = 4.712 × 10^-3 m = 4.712mm

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Um comentarista de futebol certa vez comentou:"A bola bateu na trave e voltou duas vezes mais forte". Sabendo que quando a bola
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50 gram superball that strikes the wall at 1 m/s and bounces away at 0.8 m/s has the greater change in kinetic energy because the collision is elastic in nature that is bodies separates after collision and doesn't lose any kinetic energy.

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