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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:

x=(0.088m)\cos(\sqrt{\frac{k}{m} }  t)

Explanation:

We first identify the elements of this simple harmonic motion:

The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.

The angular frequency ω can be calculated with the formula:

\omega =\sqrt{\frac{k}{m}}

Where k is the spring constant and m is the mass of the particle.

Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:

x=A\cos(\omega t)

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x=(0.088m)\cos(\omega t)

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x=(0.088m)\cos(\sqrt{\frac{k}{m} }  t)

7 0
4 years ago
A spelunker is surveying a cave. She follows a passage that takes her a distance 184 m straight west, then a distance 220 m in a
Sever21 [200]
Refer to the diagram shown below.

Define the unit vector i to point in the eastern direction, and the unit vector j to point in the northern direction.

The first distance is 184 m west. It is represented by
d₁ = -184 i

The second distance is 220 m at 30° south of east. It is
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The third distance is 104 m at 80 east of north. It is
d₃ = 104(sin 80° i + cos 80° j) =  102.42 i + 18.06 j

Let the fourth distance be 
d₄ = a i + b j

Because the traveler ends back at the original position, the vector sum of the distances is zero. It means that each component of the vector sum is zero.

The x-component yields
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a = -108.95

The y-component yields
0 - 110 + 18.06 + b = 0
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The magnitude of the fourth displacement is
√[(-108.95)² + 91.94² ] = 142.56 m

The direction is at an angle θ north of west, given by
θ = tan⁻¹ (91.94/108.95) = 40.2°

Answer:
The fourth displacement has a magnitude of 142.56 m. It is about 40° north of west.

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given that

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Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

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<h3>What is Kepler's third law?</h3>

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By using Kepler's third law, this can be written as,

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To know more about Kepler's third law, follow the link given below.

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