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Natali [406]
2 years ago
13

when electric power is transmitted over long distance, line losses can be reduced by generating AC rather than DC voltage and by

transmitting?
Engineering
1 answer:
Tanya [424]2 years ago
3 0

When electric power is transferred over a long distance then, line loss can be reduced by high voltage and low current. Thus, option A is correct.

<h3>What is electric power?</h3>

The complete question is attached to the image below.

The electric circuit transfers the electric current then it is called electric power. The loss in the line over a larger distance is reduced by transferring AC instead of DC voltage.

The AC transmitted is of high voltage, and the power is low. The equation for current is given,

P = VI

The increase in the voltage causes the decrease in the current and in turn the resistance also reduces resulting in the loss of the power of the circuit to decrease.

Therefore, the loss is reduced by transferring high voltage and, low current AC.

Learn more about high volatge AC here:

brainly.com/question/18069892

#SPJ1

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An engineer is considering time of convergence in a new Layer 3 environment design. Which two attributes must be considered? (Ch
mestny [16]

Answer:

Options B and E

Explanation:

To give sustainable environmental design when considering time of convergence in layer 3, an engineer must consider the loss of a valid forwarding path and the table updates since these will determine whether the design becomes fit or not.

7 0
3 years ago
There are two questions about SolidWorks.
Nataliya [291]

Answer:

1. It is a good practice to fully define a sketch to avoid having erroneous dimensions on the faces of a solid, this avoids that when it is required to make an assembly with the drawn part appear assembly errors.

2. The 2D sketch should always be done on a plane, so solidworks would ask you to select a plane on which you want to make the sketch, on the other hand, if it is a 3D sketch, solidworks allows you to do it without the need for Select any plane.

4 0
3 years ago
Two metallic terminals protrude from a device. The terminal on the left is the positive reference for a voltage called vx (the o
tresset_1 [31]

Answer:

Vx = 6.242 x 10raised to power 15

Vy = -6.242 x 10raised to power 15

Explanation:

from E = IVt

but V = IR from ohm's law and Q = It from faraday's first law

I = Q/t

E = Q/t x V x t = QV

hence, E =QV

V = E/Q

3 0
3 years ago
Read 2 more answers
To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105
GenaCL600 [577]

Answer:

62.14\ \text{miles}

6213727.37\ \text{miles}

Explanation:

The distance of the chain would be the product of the dislocation density and the volume of the metal.

Dislocation density = 10^5\ \text{mm}^{-2}

Volume of the metal = 1000\ \text{mm}^3

10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}

1\ \text{mile}=1609.34\ \text{m}

\dfrac{10^5}{1609.34}=62.14\ \text{miles}

The chain would extend 62.14\ \text{miles}

Dislocation density = 10^{10}\ \text{mm}^{-2}

Volume of the metal = 1000\ \text{mm}^3

10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}

\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}

The chain would extend 6213727.37\ \text{miles}

3 0
3 years ago
Using the results of the Arrhenius analysis (Ea=93.1kJ/molEa=93.1kJ/mol and A=4.36×1011M⋅s−1A=4.36×1011M⋅s−1), predict the rate
uysha [10]

Answer:

k = 4.21 * 10⁻³(L/(mol.s))

Explanation:

We know that

k = Ae^{-E/RT} ------------------- euqation (1)

K= rate constant;

A = frequency factor = 4.36 10^11 M⁻¹s⁻¹;

E = activation energy = 93.1kJ/mol;

R= ideal gas constant = 8.314 J/mol.K;

T= temperature = 332 K;

Put values in equation 1.

k = 4.36*10¹¹(M⁻¹s⁻¹)e^{[(-93.1*10^3)(J/mol)]/[(8.314)(J/mol.K)(332K)}

k = 4.2154 * 10⁻³(M⁻¹s⁻¹)

here M =mol/L

k = 4.21 * 10⁻³((mol/L)⁻¹s⁻¹)

 or

k = 4.21 * 10⁻³((L/mol)s⁻¹)

or

k = 4.21 * 10⁻³(L/(mol.s))

3 0
3 years ago
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