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qaws [65]
2 years ago
14

A copper wire has a diameter of 2.05 mm and carries a current of 5 A due solely to electrons. Given that the free (capable to mo

ve) electron concentration in copper is 1023 electrons per cm3, find the average velocity of the electrons in the wire. Give the velocity in mm/s and in miles/hour.
Physics
1 answer:
LekaFEV [45]2 years ago
6 0

To solve this problem we will apply the concepts related to the average velocity in an electron, defined as the value of the current on the product between the charge, the number of electrons, the constancy pi and the squared radius. Our values are given as,

D = 2.05 mm

I = 5A

e = -1.6 x 10^{-19} C

Q = 10^{23} e/cm^3 = 10^{29}e/m^3

The radius would b,

R = \frac{D}{2} = \frac{2.05*10^{-3}m}{2}

R = 1.025*10^{-3}m

Now the average velocity

V = \frac{I}{(Q e \pi R^2)}

Here,

I = Current

Q = Electron concentration

e = Charge of electron

R = Radius

V = \frac{5}{(10^{29})(-1.6*10^{-19})(\pi)(1.025*10^{-3})^2}}

V = 9.46*10^{-5}m/s

Converting to the values required we have that,

V = 9.46*10^{-5} m/s (\frac{1000mm}{1m})

V = 9.46*10^{-2}mm/s

And,

V = 9.46*10^{-5}m/s (\frac{3600s}{1hour})(\frac{0.000621371miles}{1m})

V = 2.116*10^{-4}mph

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<span>The correct answer is: Mechanical Energy

Explanation:
As the guitar strings are plunked, the potential energy stored in the strings has an ability to make them vibrate. When the strings are vibrating, that potential energy is actually converted to the kinetic energy. Hence, the whole phenomena contains both the kinetic energy and the potential energy. The sum of kinetic energy and the potential energy is called Mechanical energy. Therefore, the correct answer is Mechanical Energy.</span>
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MArishka [77]

Answer:

Same magnitude of the 10 nc charge cause the electric field is external.

Explanation:

To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.

As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.

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You didn't give us anything other than the question. No options are provided so I cannot answer. Nobody can.

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