Ask question

Ask question

All categories

3 years ago

11

Ask Your Teacher An electric utility company supplies a customer's house from the main power lines (120 V) with two copper wires

, each of which is 34.0 m long and has a resistance of 0.109 Ω per 300 m. (a) Find the potential difference at the customer's house for a load current of 116 A.

1 answer:

natta225 [31]3 years ago

4 0

Answer:

The potential difference at the customer's house is 117.1 V.

Explanation:

a) The potential difference at the customer's house can be calculated as follows:

$\Delta V_{h} = \Delta V_{p} - \Delta V_{l}$

<u>Where</u>:

$V_{h}$ : is the potential difference at the customer's house

$V_{p}$ : is the potential difference from the main power lines = 120 V

$V_{l}$ : is the potential difference from the lines

$\Delta V_{h} = \Delta V_{p} - IR$

The resistance, R, is:

$\frac{0.109 \Omega}{300 m}*2*34.0 m = 0.025 \Omega$

Now, the potential difference at the customer's house is:

$\Delta V_{h} = 120 V - 116A*0.025 \Omega = 117.1 V$

Therefore, the potential difference at the customer's house is 117.1 V.

I hope it helps you!

You might be interested in

1. A sprinter races in the 100 meter dash. It takes him 10 second to reach the finish line

poizon [28]

Answer:

v = 10 m/s

Explanation:

Given that,

Distance covered by a sprinter, d = 100 m

Time taken by him to reach the finish line, t = 10 s

We need to find his average velocity. We know that velocity is equal to the distance covered divided by time taken. So,

v = d/t

$v=\dfrac{100\ m}{10\ s}\\\\v=10\ m/s$

Hence, his average velocity is 10 m/s.

6 0

3 years ago

A wave traveling in water has a frequency of 500Hz and a wavelength of 3.00 meters. What is the speed of the wave?

jenyasd209 [6]

Answer:

1500 m/s

Explanation:

3 0

2 years ago

A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal

Setler [38]

Answer

given,

mass of the ball = 3 kg

swing in vertical circle with radius = 2 m

work done by the gravity = ?

work done by the tension = ?

Work done by the gravity = - m g Δh

Δ h = 2 + 2 = 4 m

Work done by the gravity = $- 3 \times 9.8 \times 4$

= -117.6 J

work done by gravity is equal to -117.6 J

Work done by tension will be equal to zero.

Zero because tension is always perpendicular to velocity

work done by tension is equal to 0 J

7 0

3 years ago

A heavy boy and a lightweight girl are balanced on a mass-less seesaw. If they both move forward so that they are one-half their

Rom4ik [11]

Answer:

b) Nothing will happen, the sea saw will still be balanced.

Explanation:

b) Nothing will happen, the sea saw will still be balanced.

Reason:-

When two kids are balanced, the sum of torques on the seesaw will be zero.

if each kid, reduces their distances by half, then the torque of each kid will be half and the sum of torque of each on the seesaw will be zero.

Therefore the seesaw is balanced

4 0

3 years ago

Read 2 more answers

A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of

sp2606 [1]

Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height $\Delta h$ depends on the square of the time:

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ,

where

$\Delta h$ is the change in the ball's height.
$g$ is the acceleration due to gravity.
$t$ is the time for which the ball is in the air.
$v_0$ is the initial vertical velocity of the ball.

The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. $\Delta h = -0.950\;\text{m}$ .
Gravity pulls objects toward the earth, so $g$ is also negative. $g \approx -9.81\;\text{m}\cdot\text{s}^{-2}$ near the surface of the earth.
Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, $v_0 = 0$ .

Solve for $t$ .

$\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t$ ;

$\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2}$ ;

$\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)}$ ;

$t \approx 0.440315\;\text{s}$ .

What's the initial horizontal velocity of the ball?

Horizontal displacement of the ball: $\Delta x = 0.352\;\text{m}$ ;
Time taken: $\Delta t = 0.440315\;\text{s}$

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

$\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}$ .

Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

3 0

3 years ago