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kati45 [8]
3 years ago
10

How long does it take electrons to get from a car battery to the starting motor? Assume the current is 300 A and the electrons t

ravel through a copper wire with cross-sectional area 0.21 cm2 and length 0.85 m.The number of charge carriers per unit volume is 8.49 1028 m3.
Engineering
1 answer:
yuradex [85]3 years ago
6 0

Answer:

t=13.49 min

Explanation:

Given that

I=300 A

A= 0.21 cm²

L= 0.85 m

Number of charge carriers per unit volume = 8.49 x 10²⁸ m⁻³.

We know speed of the electron given as

V=J/ne

J= I/A

So

V=I/neA

V=\dfrac{300}{8.49\times 10^{28}\times 1.6\times 10^{-19}\times 0.21\times 10^{-4} }

V=0.00105 m/s

lets take t is the time when electron travels from car battery to the starting motor

t = L/V

t=\dfrac{0.85}{0.00105}

t=809.52 s

t=13.49 min

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4 years ago
Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,
Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

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Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0
4 years ago
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