All categories

2 years ago

To reduce inductive reactance, what devices are normally placed on transmission or distribution lines?

Physics

1 answer:

UkoKoshka [18]2 years ago

4 0

Answer : Capacitors

Explanation : Capacitors are normally placed on transmission or distribution lines when to reduce inductive reactance.

This is because it enhances electromechanical and voltage stability , limit voltage dips at network nodes and reduces the power loss.

So, we can say that inductive reactance normally replace by the capacitors.

You might be interested in

How are vibration waves and energy related to sounds

Lesechka [4]

In electromagnetic waves, energy is transferred through vibrations of electric and magnetic fields. ... In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels.

3 0

2 years ago

Read 2 more answers

Why is an element considered a pure substance?

anastassius [24]

Answer:

An element is a pure substance and is made of only one type of atom.

Explanation:

It cannot be broken down into a simpler substance.

3 0

2 years ago

A girl and a boy are riding on a merry go round that is turning at a constant rate. The girl is near the outer edge, and the boy

galina1969 [7]

Answer:

The girl has greater tangential acceleration

Explanation:

The angular acceleration ( $\alpha$ ) of the merry go round is equal to the rate of the change of the angular velocity, $\omega$ :

$\alpha = \frac{d\omega}{dt}$

Since all the points of the merry go round complete 1 circle in the same time, the angular velocity of each point of the merry go round is the same, and so all the points also have the same angular acceleration.

The tangential acceleration instead is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the distance from the centre of the merry go round

Since the girl is near the outer edge and the boy is closer to the centre, the value of r for the girl is larger than for the boy, so the girl has greater tangential acceleration.

5 0

2 years ago

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

yawa3891 [41]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

$E_{pe}\:(elastic\:potential\:energy) = 5184\:J$

$K\:(constant) = 16200\:N/m$

$x\:(displacement) =\:?$

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

$E_{pe} = \dfrac{k*x^2}{2}$

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:

$E_{pe} = \dfrac{k*x^2}{2}$

$5184 = \dfrac{16200*x^2}{2}$

$5184*2 = 16200*x^2$

$10368 = 16200\:x^2$

$16200\:x^2 = 10368$

$x^{2} = \dfrac{10368}{16200}$

$x^{2} = 0.64$

$x = \sqrt{0.64}$

$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark$

Answer:

The displacement of the spring = 0.8 m

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

3 0

3 years ago

Two identical loudspeakers, speaker 1 and speaker 2, are 2.0 m apart and are emitting 1700-hz sound waves into a room where the

Debora [2.8K]

Given the following in the problem:

Distances : 2.0 m and 4.0 m
Sound waves : 1700 hz
Speed of sound : 340 m/s

Get the wavelength of the sound by using the formula:
Lambda = speed of sound/sound waves
Lambda = 340 m/s / 1700 hz
Lambda = 0.2
Get the path length difference to the point from the two speakers
L1 = 4mL2 = sqrt (42+ 22) m
Delta = 4.47

x = delta / lambda

If the outcome is nearly an integer, the waves strengthen at the point. If it is nearly an integer +0.5 the waves interfere destructively at the point. If it is neither the point is somewhat in in the middle.

Solving x = (4.47 – 4) / (0.2) = 2.35 an integer +0.5 so it’s a point of destructive interference.

3 0

2 years ago