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Anuta_ua [19.1K]

2 years ago

11

A 310-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,010 A. If the conductor is copper

with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable? (Use 3.156 107 for the number of seconds in a year.)

1 answer:

Marizza181 [45]2 years ago

5 0

Answer:

t = 166 years

Explanation:

In order to calculate the amount of years that electrons take to cross the complete transmission line. You first calculate the drift speed of the electrons by using the following formula:

$v_d=\frac{I}{nqA}$ (1)

I: current on the wire = 1,010A

n: free charge density = 8.50*10^28 electrons/m^3

A: cross-sectional area of the transmission line = π*r^2

r: radius of the cross-sectional area = 2.00cm = 0.02m

You replace the values of the parameters in the equation (1):

$v_d=\frac{1,010A}{(8.50*10^{28}electron/m^3)(1.6*10^{-19}C)(\pi (0.02m)^2)}\\\\v_d=5.9*10^{-5}\frac{m}{s}$

Next, you use the following formula:

$t=\frac{x}{v_d}$ (2)

x: length of the line transmission = 310km = 310,000m

You replace the values of vd and x in the equation (2):

$t=\frac{310,000m}{5.9*10^{-5}m/s}=5.24*10^9s$

Finally, you convert the obtained t to seconds

$t=5.24*10^9s*\frac{1\ year}{3.156*10^7s}=166.03\ years$

The electrons take approximately 166 years to travel trough the complete transmission line

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To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after

As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

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⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

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