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Anna007 [38]
2 years ago
8

Although the evidence is weak, there has been a concern in recent years over possible health effects from the magnetic fields ge

nerated by electric transmission lines. A typical high-voltage transmission line is 20 m above the ground and carries a 200 A current at a potential of 110 kV.
a. What is the magnetic field strength on the ground directly under such a transmission line?
b. What percentage is this of the earth's magnetic field of 50 ?T?
Physics
1 answer:
sergey [27]2 years ago
7 0

Answer:

a

The magnetic field strength is  B = 2 \mu T

Explanation:

From the question we are told that

             The  length line   above the ground  is  R = 20m

              The current of the line is  I  = 200A

              The voltage of the line is V = 110kV

Generally magnetic field strength is mathematically represented as

              B = \frac{\mu_o I}{2 \pi R}

Where  \mu_0 is the permeability of free space  = 4\pi * 10^{-7} N/A^2

             B = \frac{(4\pi * 10^{-7} N/A^2) *200}{2 \pi *20}

                = (2.0*10^{-7})[\frac{200}{20} ]

               = 2*10^{-6}T

              = 2 \mu T

Earths magnetic field is approximately given as 50 \mu T

   So the percentage would be

                           = \frac{Magnetic \ Field \ Intensity \ Of  \ Line}{Earth's \ Magnetic \ Field} * 100

                          = \frac{2 \mu T }{50 \mu T } * 100

                          =4%            

         

           

             

               

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For the adverage speed, we should first add the speed of each family in mph , so 90+60, which equals 150 and divide that by 2 because there are 2 speeds so the average speed is 75mph.

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Normally, jet engines push air out the back of the engine, resulting in forward thrust, but commercial aircraft often have thrus
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Answer:

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Uses of thrust reversal in practice:

When the ejected air is moving forward direction then the thrust force moving backward direction due to reversal thrust the speed of the craft slows down.

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Answer:

a) x = 8.8 cm * cos (9.52 rad/s * t)

b) x = 8.45 cm

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This is a Simple Harmonic Motion, and most Simple Harmonic Motion equations start from the equilibrium point. In this question however, we are starting from the max displacement the equations, and thus, it ought to be different.

From the question, we are given that

A = 8.8 cm = 0.088 m

t = 0.66 s

Now, we need to find the angular speed w, such that

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w = (2 * 3.142) / 0.66

w = 6.284 / 0.66

w = 9.52 rad/s

The displacement equation of Simple Harmonic Motion is usually given as

x = A*sin(w*t)

But then, the equation starts from the equilibrium point at 0 sec, i.e x = 0 m

When you have to start from the max displacement, then the equation would be

x = A*cos(w*t).

So when t = 0 the cos(0) = 1, and then x = A which is max displacement.

Thus, the equation is

x = 8.8 cm * cos (9.52 rad/s * t)

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Answer:

<h2>The velocity is 7.82 m/s</h2>

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We then substitute the given data into the formula to solve for velocity.

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velocity= 7.82 m/s

<h2>The velocity is 7.82 m/s</h2>
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