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Darya [45]
3 years ago
13

A piano wire has a tension of 650 N and a mass per unit length of 0.060 g/cm. What is the speed of waves on this wire

Physics
1 answer:
Trava [24]3 years ago
4 0

Answer:

The speed of waves on this wire is 329.14 m/s

Explanation:

Given;

tension of the wire, T = 650 N

mass per unit length, μ = 0.06 g /cm  = 0.006 kg/m

(convert the unit of mass per length in g/cm to kg/m by dividing by 10 = 0.06 / 10 = 0.006 kg/m)

The speed of waves on this wire is given as;

v = \sqrt{\frac{T}{\mu} } \\\\v = \sqrt{\frac{650}{0.006} }\\\\v = 329.14 \ m/s

Therefore, the speed of waves on this wire is 329.14 m/s

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Each separate weight has a 'moment'.
The moment of each weight is: 

             (the weight of it) x (its distance from the pivot/fulcrum) .

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The lever (or the see-saw) is balanced when (the sum of all the moments
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When is the absorption spectrum of an electron generated
serious [3.7K]
A
Excitation to a higher energy state requires energy which is absorbed from the electromagnetic waves applied.
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Which situation is an example of transferring heat by means of convection?
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3 years ago
A blacksmith heats a 1.5 kg iron horseshoe to 500 C, containing 20 kg of water at 18 C then plunges it into a bucket What is the
Tema [17]

Answer:

21.85 C

Explanation:

mass of iron = 1.5 kg, initial temperature of iron, T1 = 500 C

mass of water = 20 kg, initial temperature of water, T2 = 18 C

let T be the equilibrium temperature.

Specific heat of iron = 449 J/kg C

specific heat of water = 4186 J/kg C

Use the principle of caloriemetry

heat lost by the hot body = heat gained by the cold body

mass of iron x specific heat of iron x decrease in temperature = mass of water  x specific heat of water x increase in temperature

1.5 x 449 x (500 - T) = 20 x 4186 x (T - 18)

336750 - 673.5 T = 83720 T - 1506960

1843710 = 84393.5 T

T = 21.85 C

8 0
2 years ago
When a car of mass 1167 kg accelerates from 10.0 m/s to some final speed, 4.00  10 5J of work are done. Find this final speed.
Akimi4 [234]
  • Mass=1167kg
  • Initial velocity=u=10m/s
  • Acceleration=a=4m/s^2
  • Work done=105J=W
  • Final velocity=v=?
  • Force=F
  • Distance=d

Apply Newton's second law

\\ \tt\hookrightarrow F=ma

\\ \tt\hookrightarrow F=1167(4)=4668N

Now

\\ \tt\hookrightarrow W=Fd

\\ \tt\hookrightarrow d=\dfrac{W}{F}

\\ \tt\hookrightarrow d=\dfrac{105}{4668}

\\ \tt\hookrightarrow d=0.022m

Now

  • d be s

According to third equation of kinematics

\\ \tt\hookrightarrow v^2=u^2+2as

\\ \tt\hookrightarrow v^2=10^2+2(4)(0.022)

\\ \tt\hookrightarrow v^2=100+8(0.022)

\\ \tt\hookrightarrow v^2=100+0.176

\\ \tt\hookrightarrow v^2=100.176

\\ \tt\hookrightarrow v=10.001m/s

8 0
2 years ago
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