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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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bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. what is the veloc
USPshnik [31]

Answer:

The final velocity of the car A is -1.053 m/s.

Explanation:

For an elastic collision both the kinetic energy and the momentum of the system are conserved.

Let us call

m_A = mass of car A;

v_{A1} = the initial velocity of car A;

v_{A2} = the final velocity of car A;

and

m_B = mass of car B;

v_{B1} = the initial velocity of car B;

v_{B2} = the final velocity of car B.

Then, the law of conservation of momentum demands that

m_Av_{A1}+m_Bv_{B1} =m_Av_{A2}+m_Bv_{B2}

And the conservation of kinetic energy says that

\dfrac{1}{2} m_Av_{A1}^2+\dfrac{1}{2}m_Bv_{B1}^2=\dfrac{1}{2}m_Av_{A2}^2+\dfrac{1}{2}m_Bv_{B2}^2

These two equations are solved for final velocities  v_{A2} and v_{B2} to give

$v_{A2} =\frac{m_A-m_B}{m_A+m_B} v_{A1}+\frac{2m_B}{m_A+m_B} v_{B1}$

$v_{B2} =\frac{2m_A}{m_A+m_B} v_{A1}+\frac{m_B-m_A}{m_A+m_B} v_{B1}$

by putting in the numerical values of the variables we get

$v_{A2} =\frac{281-209}{281+209} (2.82)+\frac{2*209}{281+209} (-1.72)$

\boxed{v_{A2} = -1.05m/s}

and

$v_{B2} =\frac{2*281}{281+209} (2.82)+\frac{209-281}{281+209} (-1.72)$

\boxed{v_{B2} = 3.49m/s}

Thus, the final velocity of the car A is -1.053 m/s and of car B is 3.49 m/s.

4 0
4 years ago
Brainliest for whoever is first and right.
devlian [24]

The answer is 9.8, did this last year in AP Science

3 0
3 years ago
A bicycle racer is going downhill at 11.0 m/s when, to his horror, one of his 2.25 kg wheels comes off when he is 75.0 m above t
alex41 [277]

Answer:

a.) Speed V = 29.3 m/s

b.) K.E = 1931.6 J

Explanation: Please find the attached files for the solution

4 0
4 years ago
A material that provides little or no resistance to the flow of electric current is called a/an
GarryVolchara [31]
Conductor .......................
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A ball weighing 2.0 N falls through a distance of 4 m.
Naddik [55]

Answer:

This question is incomplete, a possible question would be to calculate the work done by the ball.

Work done = 8J

Explanation:

Work done by an object is calculated my multiplying the force by the distance.

The weight of this ball is 2.0N, which also represents the force exerted by the ball. The ball travels through a distance of 4m

Hence, work done = F × d

Work done = 2N × 4m

Work done = 8Nm or 8J

4 0
3 years ago
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