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3 years ago

A _____ increases or decreases voltage in a power line

2 answers:

4 0

A transformer is an electrical device that transfers energy between two or more circuits through electromagnetic induction. Transformers are mainly used to increase or decrease voltage in power lines or electric power applications. So your answer would be D. transformer

IceJOKER [234]3 years ago

4 0

Answer:

The correct answer is 4.

Explanation:

A transformer is an electrical element that allows increasing or decreasing the voltage in an alternating current electrical circuit, maintaining the power that enters the equipment, in the case of an ideal transformer, constant at the output. Basically, it converts the alternating electric energy of a certain voltage level into alternating energy of another voltage level, based on the phenomenon of electromagnetic induction.

Have a nice day!

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A certain dam generates 120 MJ of mechanical (hydroelectric) energy each minute. If the conversion from mechanical to electrical

Karolina [17]

Answer:

electric energy ( power ) = 300000 W

Explanation:

given data

mechanical (hydroelectric) energy = 120 MJ/min = 2000000 J/s

efficiency = 15 % = 0.15

solution

we know that Efficiency of electric engine is expression as

Efficiency = Mechanical energy ÷ electric energy ......................1

and here dam electrical power output is

put here value in equation 1

electric energy ( power ) = Efficiency × Mechanical energy ( power )

electric energy ( power ) = 0.15 × 2000000 J/s

electric energy ( power ) = 300000 W

8 0

3 years ago

Which material will heat up the most quickly if placed near a heat source?

Thepotemich [5.8K]

Answer:

Metal

Explanation:

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7 0

3 years ago

Read 2 more answers

Science question. What can cause a change in the direction of a magnetic field?

ivann1987 [24]

Answer:

- Whenever current travels through a conductor, a magnetic field is generated. ... Depending on the shape of the conductor, the contour of the magnetic field will vary. ... Click the Reverse button to change the direction of the current flow ... to make a fist (or to wrap around the wire in question) is the direction of ...

Explanation:

4 0

3 years ago

Residential building codes typically require the use of 12-gauge copper wire (diameter 0.2053 cm) for wiring receptacles. Such c

AysviL [449]

Answer:

a) E = 4.26 W

b) E' = 6.724 W

c) copper wire is the safer option to use.

Explanation:

Given:

Diameter of the 12 gauge copper wire, d = 0.2053 cm

Thus, Radius of the 12 gauge copper wire, r = 0.2053 cm

/ 2 = 0.10265 cm = 0.10265 × 10⁻² m

Now.

the area (A) comes out as

A = π × (0.10265 × 10⁻²)²

A = 3.3103 × 10⁻⁶ m²

Length of the copper wire, L = 2.10 m

a) The resisitivity (ρ) of copper = 1.68 × 10⁻⁸ ohm m

Now,

the resistance of the copper , R = ρL/A

R = (1.68 × 10⁻⁸ × 2.1) / ( 3.3103 × 10⁻⁶)

R = 0.01065 ohm

The Energy (E) is given as,

E = I²R

where, I is the current

I = 20.0 A

on substituting the values, we get

E = 20.0² × 0.01065

E = 4.26 W

(b) For the aluminium

Resisitivity, ρ' = 2.65 × 10⁻⁸ ohm m

Now, the resistance of the aluminium wire, R' = (ρ' × L) / A

Since the cross-section of the aluminium wire is same as the copper wire

thus,

R = (2.65 × 10⁻⁸ × 2.1) / ( 3.3103 × 10⁻⁶)

R = 0.0168 ohm

Therefore,

The Rate of energy produced by the aluminium wire, E' = I²R'

E' = 20.0² × 0.0168

E' = 6.724 W

(c) From the above results, we can conclude that the power consumed or the rate of energy produced by the aluminium wire is more.

Hence, copper wire is the safer option to use.

8 0

4 years ago

if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu

RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

$F = \frac{k|q_1||q_2|}{r^2}$

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

$F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N$

Therefore, the mutual force between the two point charges is 319.64 N

4 0

3 years ago