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liraira [26]
2 years ago
11

If the tension is 300 N, what is the difference in the speed of a wave traveling from one wire to the other?

Physics
1 answer:
RUDIKE [14]2 years ago
6 0

Answer:

   v₁ / v₂ = √μ ₂/μ ₁

Explanation:

The speed of a traveling wave on a string is given by

         v = √ T /μ

Where v is the speed of the wave, T the tension and μ  is the linear density of the string

Let's apply this equation to our case

               v₁ = √ T / μ ₁

               v2 = √ T /μ ₂

If the voltage of the two systems is the same

              T = v₁² μ₁ = v₂² μ₂

We cleared

            v₁² / v₂² = μ₂ / μ₁

             v₁ / v₂ = √μ ₂/μ ₁

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4 0
2 years ago
Mechanical energy is conserved in the presence of which of the following types of forces?magnetic
Tju [1.3M]
<h2>Answer: electrostatic and gravitational force </h2><h2 />

Mechanical energy remains constant (conserved) if only <u>conservative forces</u> act on the particles.  

In this sense, the following forces are conservative:  

-Gravitational  

-Elastic

-Electrostatics  

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According to this, mechanical energy is conserved in the presence of electrostatic and gravitational forces.

7 0
3 years ago
Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

5 0
3 years ago
According to the graph, how many atoms would remain after two half-lives?
Fudgin [204]

Answer:

Let No be initial no of atoms

N = N0 / 2      after 1 half-life

N = N0 / 4     after 2 half-lives

So after 2 half-lives 20 of the 80 atoms remain

4 0
2 years ago
A train traveled 350km in 2.5h. what was the average speed of the train?
MaRussiya [10]
Speed equals distance divided by time, so 350 divided by 2.5 equals 140 kilometers per hour.
3 0
3 years ago
Read 2 more answers
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