Use this media to help you complete the question.

An energy plant produces an output potential of 1500 kV and serves a city 143 km away. A high-voltage transmission line carries

jekas [21]

Answer:

2123.55 $/hr

Explanation:

Given parameters are:

$V_{plant} = 1500$ KV

L = 143 km

I = 500 A

$\rho = 2.4$ $\Omega / km$

So, we will find the voltage potential provided for the city as:

$V_{wire} =IR = I\rho L = 1500*2.4*143 = 514.8$ kV

$V_{city} = V_{plant}- V_{wire} = 1500-514.8 = 985.2$ kV

Then, we will find dissipated power because of the resistive loss on the transmission line as:

$P = I^2R = I^2\rho L=500^2*2.4*143 = 8.58*10^7$ W

Since the charge of plant is not given for electric energy, let's assume it randomly as $x = \frac{\dollar 0.081}{kW.hr}$

Then, we will find the price of energy transmitted to the city as:

$Cost = P * x = 8.58*10^7 * 0.081 * 0.001 = 6949.8$ $/hr

To calculate money per hour saved by increasing the electric potential of the power plant:

Finally,

$I_{new} = P/V_{new} = I/1.2\\P_{new} = I_{new}^2R_{wire}\\Cost = P_{new}/1.44=6949.8/1.44 = 4826.25$ $/hr

The amount of money saved per hour = $6949.8 - 6949.8/1.44 = 2123.55$ $/hr

Note: For different value of the price of energy, it just can be substituted in the equations above, and proper result can be found accordingly.

3 0

3 years ago

An Olympic runner complete the 200 m sprint and 23 seconds. What is the runners average speed (round your answer to the nearest

kipiarov [429]

Should be 8.7 m/s. Just divide 200 by 23 and round from there.

3 0

3 years ago

Cheryl is riding on the edge of a merry-go-round, 2m from the center, which is rotating with an increasing angular speed. Cheryl

WINSTONCH [101]

In circular motion we know that there is two type of acceleration that Cheryl experience

1. Tangential acceleration

2. Centripetal acceleration

here given that

$a_t = 3 m/s^2$

for centripetal acceleration we know that

$a_c = \frac{v^2}{R}$

$a_c = \frac{4^2}{2} = 8 m/s^2$

now we know that both centripetal acceleration and tangential acceleration is perpendicular to each other

so total acceleration is vector sum of both and given as

$a_{net} = \sqrt{8^2 + 3^2} = 8.54 m/s^2$

so total acceleration is 8.54 m/s^2

8 0

3 years ago

A lightweight string is wrapped several times around the rim of a small hoop. If the free end of the string is held in place and

MariettaO [177]

Answer:

Explanation:

Let T be the tension

For linear motion of hoop downwards

mg -T = ma , m is mass of the hoop . a is linear acceleration of CG of hoop .

For rotational motion of hoop

Torque by tension

T x R , R is radius of hoop.

Angular acceleration be α,

Linear acceleration a = α R

So TR = I α

= I a / R

a = TR² / I

Putting this value in earlier relation

mg -T = m TR² / I

mg = T ( 1 + m R² / I )

T = mg / ( 1 + m R² / I )

mg / ( 1 + R² / k² )

Tension is less than mg or weight because denominator of the expression is more than 1.

5 0

3 years ago

If your speed slows down you will have what acceleration?

vitfil [10]

When speed slows down you have kinetic energy.

8 0

4 years ago

Read 2 more answers