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oee [108]
2 years ago
9

A steel wire of length 30.0 m and a copper wire of length 20.0m , both with 1.00 -mm diameters, are connected end to end and str

etched to a tension of 150 N . During what time interval will a transverse wave travel the entire length of the two wires?
Physics
1 answer:
Irina18 [472]2 years ago
6 0

The time interval it would take a transverse wave to travel the entire length of the two wires is equal to 0.3295 seconds.

<u>Given the following data:</u>

  • Length of steel wire = 30.0 m.
  • Length of copper wire = 20.0 m
  • Diameter of copper wire = 1.00 mm to m = 0.001 m.
  • Diameter of steel wire = 1.00 mm to m = 0.001 m.
  • Tension = 150 N.

<u>Scientific data:</u>

  • Density of steel = 7860 kg/m³.
  • Density of copper = 8920 kg/m³.

<h3>How to determine the time interval?</h3>

First of all, we would determine the speed of the wave of steel wire and copper wire respectively.

For the speed of the wave of steel wire, we have:

Mathematically, the speed of a wave of steel wire can be calculated by using this formula:

Vs = √[T/(ρπd²/4)]

Vs = √[150/(7860 × 3.142 × 0.001²)/4)]

Vs = √(150/0.0062)

Vs = √24,193.55

Speed, Vs = 155.54 m/s.

For the speed of the wave of copper wire, we have:

Mathematically, the speed of a wave of copper wire can be calculated by using this formula:

Vc = √[T/(ρπd²/4)]

Vc = √[150/(8920 × 3.142 × 0.001²)/4)]

Vc = √(150/0.0070)

Vc = √21,428.57

Speed, Vc = 146.39 m/s.

Now, we can determine the time interval:

Time = t₁ + t₂

Time = Ls/Vs + Lc/Vc

Time = 30.0/155.54 + 20.0/146.39

Time = 0.1929 + 0.1366

Time = 0.3295 seconds.

Read more on wave travel time here: brainly.com/question/13931407

#SPJ4

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Answer:

Average speed = 0.35 m/s

Explanation:

Given the following data;

Distance = 1.3 Km

Time = 62 minutes

To find the average speed in m/s;

First of all, we would convert the quantities to their standard unit (S.I) of measurement;

Conversion:

1.3 kilometres to meters = 1.3 * 1000 = 1300 meters

For time;

1 minute = 60 seconds

62 minutes = X

Cross-multiplying, we have;

X = 62 * 60

X = 3720 seconds

Now, we can calculate the average speed in m/s using the formula;

Speed = \frac {distance}{time}

Speed = \frac {1300}{3720}

Average speed = 0.35 m/s

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Explanation in File!
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A rock at rest falls off a tall cliff and hits the valley below after 3.5s. What is the rocks velocity as it hits the ground
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(a) Calculate the magnitude of the angular momentum of the earth in a circular orbit around the sun. Is it reasonable to model i
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Answer:

(a) L=2.6742\times 10^{40}\, kg.m^2.s^{-1}

(b) L_s=7.07 \times 10^{23} \, kg.m^2.s^{-1}

Explanation:

<u>For Earth we have:</u>

  • mass of earth, m_E=5.97\times 10^{24}\, kg
  • radius of earth, R_E=6.38\times 10^6m
  • orbital radius, r=1.5 \times 10^{11}m
  • period of rotation, t_{rot}=24h=86400\, s
  • period of revolution, t_{rev}= 1 yr=3.156\times 10^7 s

(a)

Angular momentum, L=?

∵L=I.\omega...............................(1)

For a particle of mass m moving in a circular path at a distance r from the axis,

I=m.r^2

& v=r.\omega

Putting respecstive values in eq. (1)

L=m_E\times r^{2}\times \omega

L=5.97\times 10^{24}\times (1.5 \times 10^{11})^2\times \frac{2\pi}{3.156\times 10^7}

L=2.6742\times 10^{40}\, kg.m^2.s^{-1}

To model earth as a particle is  reasonable because the distance between the sun and the earth is very large s compared to the radius if the earth.

(b)

For a uniform sphere of mass M and radius R and an axis through its center,  I=\frac{2}{5}M.R^2

using eq. (1)

L_s=(\frac{2}{5}m_E.R_E^2) \omega

L_s=\frac{2}{5} \times 5.97\times 10^{24} \times 6.38\times 10^6\times \frac{2\pi}{86400}

L_s=7.07 \times 10^{23} \, kg.m^2.s^{-1}

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3 years ago
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