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klemol [59]
3 years ago
9

If a power utility were able to replace an existing 500 kV transmission line with one operating at 1 MV, it would change the amo

unt of heat produced in the transmission line to
Physics
1 answer:
mestny [16]3 years ago
8 0

Answer:

It would change the amount of heat produced in the transmission line to four times the previous value.

Explanation:

Given;

initial voltage in the transmission line, V₁ = 500 kV = 500,000 V

Final voltage in the transmission line, V₂ = 1 MV = 1,000,000

The power lost in the transmission line due to heat is given by;

P = \frac{V^2}{R}

Power lost in the first wire;

P_1 = \frac{V_1^2}{R}

R = \frac{V_1^2}{P_1}

Power lost in the second wire

P_2 = \frac{V_2^2}{R}\\\\ R = \frac{V_2^2}{P_2}

Keeping the resistance constant, we will have the following equation;

\frac{V_2^2}{P_2} = \frac{V_1^2}{P_1} \\\\P_2 = \frac{V_2^2P_1}{V_1^2}\\\\

P_2 = \frac{(1,000,000)^2P_1}{(500,000)^2}\\\\P_2 =4P_1

Therefore, it would change the amount of heat produced in the transmission line to four times the previous value.

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Answer:

a) 2457J

b) 558W

c) 337N

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v=a*t\\t=\frac{a}{v}\\v=12\frac{km}{h}\frac{1000m}{km}*\frac{1h}{3600s}\\\\v=3.3m/s\\\\t=\frac{3.3m/s}{0.75m/s^2}\\t=4.4s

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d=\frac{1}{2}*a*t^2\\d=7.3m

Using the energy conservation formula:

K_i+U_i+W_d+W_f=K_f+U_f

Because the motion started from rest the initial kinetic energy is zero, the motion occurred in-ground level so the gravitational energy is zero too.

the work done by the friction force is given by:

W_f=F_f*d*cos(\theta)\\W_f=\µ*m*g*d*cos(180)\\W_f=0.08*220kg*9.8m/s^2*7.3m*(-1)\\W_f=-1259J

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W_d=\frac{1}{2}*220kg*(3.3m/s)^2+1259J\\W_d=2457J

The power is given by:

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