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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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If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t seconds is given by
fgiga [73]

Answer:

a) v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s

v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s

v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s

v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s

b) v=65-32(2)=1ft/s

Explanation:

From the exercise we got the ball's equation of position:

y=65t-16t^{2}

a) To find the average velocity at the given time we need to use the following formula:

v=\frac{y_{2}-y_{1}  }{t_{2}-t_{1}  }

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

y_{t=2}=65(2)-16(2)^{2} =66ft

y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft

v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s

--

y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft

v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s

--

y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft

v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s

--

y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft

v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s

b) To find the instantaneous velocity we need to derivate the equation

v=\frac{df}{dt}=65-32t

v=65-32(2)=1ft/s

7 0
3 years ago
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