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qaws [65]
3 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]3 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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Answer:

natural is the answer

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Andrea's near point is 20.0 cm and her far point is 2.0 m. Her contact lenses are designed so that she can see objects that are
Shalnov [3]

Answer:

the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

Explanation:

Given the data in the question,

near point = 20 cm

far point = 2 m = 200 cm

Now, for an object that is infinitely far away, the image is at is its far point.

so using the following expression, we can determine the focal length

1/f = 1/i + 1/o

where f is the focal length, i is the image distance and o is the object distance.

here, far point i = 2 m = 200 cm  and v is ∞

so we substitute

1/f = 1/(-200 cm)  +  1/∞

f = -200 cm

Also, for object at its closest point, the image appear at near point,

so

1/f = 1/i + 1/o

we make o the subject of formula

o = ( i × f ) / ( i - f )

given that near point i = 20 cm

we substitute

o = ( -20 × -200 ) / ( -20 - (-200) )

o = 4000 / 180

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4 0
3 years ago
What happens to the kinetic energy of a body when: a) the mass of the body is doubled at constant velocity? b) the velocity of t
blagie [28]
Using the formula KE=1/2mv^2

a: The kinetic energy doubles.
b: The kinetic energy quadruples.
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Hopefully it’s clear how the formula can show you this.
3 0
3 years ago
The SI system uses three base units. Question 6 options: True False
GalinKa [24]

Answer:

The answer is false

Explanation:

Though the mostly used SI unit of measurement or the most popular units are the

Length,

Time and

Mass

i.e meter (m), seconds (s), kilogram (kg)

Aside all the above stated units for measurements there are other four basic units which are itemized  bellow.

they are

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6 0
3 years ago
Starting from rest, a 2.1x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts
nirvana33 [79]

Answer:

1.327363 m/s

0.00090243 m

Explanation:

u = Initial velocity

v = Final velocity

m = Mass of flea

Energy

E=\frac{1}{2}m(v^2-u^2)\\\Rightarrow 3.7\times 10^{-4}=2.1\times 10^{-4}(v^2-0)\\\Rightarrow v=\sqrt{\frac{3.7\times 10^{-4}}{2.1\times 10^{-4}}}\\\Rightarrow v=1.32736\ m/s

The velocity of the flea when leaving the ground is 1.327363 m/s

W=F\times s\\\Rightarrow s=\frac{W}{F}\\\Rightarrow s=\frac{3.7\times 10^{-4}}{0.41}\\\Rightarrow s=0.00090243\ m

The flea will travel 0.00090243 m upward

8 0
3 years ago
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