Ask question

Ask question

All categories

4 years ago

7

Hydro-Quebec transmits power from hydroelectric dams in the far north of Quebec to the city of Montreal at 735kV. The lines are

935 km long and are 3.50 cm in diameter. Given the resistivity of copper is 1.68 x 10^-8 Ω.m. a) find the resistance of one of the lines, and b) the current carried by the wire.

1 answer:

Elanso [62]4 years ago

8 0

Answer:

a)

16.33 Ω

b)

45009.18 A

Explanation:

a)

L = length of the line = 935 km = 935000 m

d = diameter of the line = 3.50 cm = 0.035 m

ρ = resistivity of the line = 1.68 x 10⁻⁸ Ω.m

Area of cross-section of the line is given as

A = (0.25) πd²

A = (0.25) (3.14) (0.035)²

A = 0.000961625 m²

Resistance of the line is given as

$R=\frac{\rho L}{A}$

inserting the values

R = (1.68 x 10⁻⁸) (935000)/(0.000961625)

R = 16.33 Ω

b)

V = potential difference across the line = 735 kv = 735000 Volts

i = current carried by the wire

Using ohm's law, current carried by the wire is given as

$i=\frac{V}{R}$

i = 735000/16.33

i = 45009.18 A

You might be interested in

Can something have energy without having momentum? explain. can something have momentum without having energy? defend your answe

marta [7]

Momentum is a product mass and velocity. If a certain object posses a kinetic energy, then it should have a momentum since it is moving which has a velocity. However, if the object is at rest and only has potential energy, then it would not have momentum. So, for the first question the answer would be yes, an object can have energy without having any momentum. For the second question, every object whether it is moving or at rest, possess some energy, potential for an object at rest and kinetic for an object that is moving. Thus, the answer would be no, an object having momentum would always have energy.

8 0

3 years ago

What are the four system of measurement

Rudiy27

Answer:

ask google

Explanation:

5 0

4 years ago

In a football game, a defensive lineman exerts a force upon an offensive lineman, causing the offensive lineman to be accelerate

Grace [21]

the answer is B.twice as much

7 0

3 years ago

Read 2 more answers

A newspaper turns yellow in sunlight. Is this a physical or chemical change?

LekaFEV [45]

It is a chemical change. I always think of a chemical/physical change as if you could reverse it back as it started off, for example if you stepped on a can you can reverse the can back probably not exactly like it was before but you can still reverse it so this would be a physical change, and if you baked a pizza you could not reverse the dough and everything else back.

5 0

3 years ago

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.42 m. The mug

telo118 [61]

Answer:

a) $V_{x}=3.72m/s$ , b) ∠=-54.83°

Explanation:

In order to solve this problem, we must start with a drawing of the situation, this will help us visualize the problem better. (See picture attached).

a)

Now, the idea is that the beer mug has a horizontal speed and no vertical speed at initial conditions. So knowing this, we can start finding the initial velocity of the mug.

In order to do so, we need to find the time it takes for the mug to reach the ground. We can find it by using the following equation:

$y=y_{0}+V_{y0}t+\frac{1}{2}a_{y}t^{2}$

We can see from the drawing that y and the initial velocity in y are zero, so we can simplify our formula:

$0=y_{0}+\frac{1}{2}a_{y}t^{2}$

so we can solve for t, so we get:

$t=\sqrt{\frac{-(2)y_{0}}{a}}$

so now we can substitute the known values, so we get:

$t=\sqrt{\frac{-(2)(1.42)}{-9.8}}$

which yields:

t=0.538s

So we can use this value to find the velocity in x:

$V_{x}=\frac{x}{t}$

When substituting we get:

$V_{x}=\frac{2m}{0.538s}$

which yields:

$V_{x}=3.72m/s$

b)

In order to solve part b, we need to find the y-component of the velocity, for which we can use the following formula:

$\Delta y=\frac{V_{f}^{2}-V_{0}^{2}}{2a}$

We know that $V_{0}$ is zero, so we can simplify the expression:

$\Delta y=\frac{V_{yf}^{2}}{2a}$

So we can solve the equation for $V_{yf}^{2}$ so we get:

$V_{yf}=\sqrt{2\Delta y a}$

and when substituting the known values we get:

$V_{yf}=\sqrt{2(-1.42m)(-9.8m/s^{2})}$

which yields:

$V_{yf}=-5.28m/s$

Once we got the final velocity in y, we can use it together with the velocity in x to find the angle.

So we can use the following formula:

$tan \theta =\frac{V_{y}}{V_{x}}$

when solving for theta we get:

$\theta = tan^{-1}(\frac{V_{y}}{V_{x}})$

We can substitute so we get:

$\theta = tan^{-1}(\frac{-5.28m/s}{3.72m/s})$

which yields:

$\theta = -54.83^{o}$

7 0

3 years ago