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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:

R= 602 .11 N

Explanation:

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2 years ago
A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 1.26 x 10^4 s. What is
Soloha48 [4]

Answer: V=5839.051m/s  

Explanation:

According to the <u>Third Kepler’s Law</u> of Planetary motion:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;:

T=1.26(10)^{4}s is the period of the satellite

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=5.98(10)^{24}kg is the mass of the Earth

a  is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).

On the other hand, the orbital velocity V is given by:

V=\sqrt{\frac{GM}{a}}   (2)

Now, from (1) we can find a, in order to substitute this value in (2):

a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}   (3)

a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}   (4)

a=11705845.57m   (5)

Substituting (5) in (2):

V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{11705845.57m}}   (6)

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6 0
3 years ago
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Explanation:

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Nataly_w [17]

Answer:

a) 15.78 mi/h/s

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