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3 years ago

8

This is a device used to change the flow of electricity through a circuit

2 answers:

Vinvika [58]3 years ago

8 0

Transformers or also called voltage transformers

Natalija [7]3 years ago

8 0

Answer:

Transformers are devices used in electrical circuits to change the voltage of electricity flowing in the circuit. Transformers can be used either to increase the voltage or decrease the voltage.

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Can someone help me

Illusion [34]

Thomas Edison is the answer im 100% sure of it.

5 0

3 years ago

what are the common elements that make up series and parallel circuits , and how are circuits diagrammed?

Ahat [919]

<h2>Answer:</h2>

A series circuit occurs when the elements are connected along a simple path so the same current flows through all the elements. On the other hand, a parallel circuit occurs when there are two or more paths for the electricity to flow. The diagram are shown in the Figure below. We have chosen a source and resistors to illustrate this problem.

3 0

3 years ago

Viewed moving objects from pond water under his microscope and named them “animalcules”.

galben [10]

Anton Von Leeuwenhoek, in the early 1600s, saw these tiny microbes and called them "animalcules" and "wee beasties".

5 0

3 years ago

Dimension equation of work

kkurt [141]

Answer:

Explanation:

Work

Other units Foot-pound, Erg

In SI base units 1 kg⋅m2⋅s−2

Derivations from other quantities W = F ⋅ s W = τ θ

Dimension M L2 T−2

Idk if this is what u are looking for but i hope this help.:)

3 0

3 years ago

The surface charge density on an infinite charged plane is - 2.10 ×10−6C/m2. A proton is shot straight away from the plane at 2.

inn [45]

Explanation:

Formula to calculate electric field because of the plate is as follows.

E = $\frac{\sigma}{2 \times \epsilon_{o}}$

= $\frac{2.10 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}$

= $1.18 \times 10^{5} N/C$

Now, we will consider that equilibrium of forces are present there. So,

ma = qE

a = $\frac{1.6 \times 10^{-19} \times 1.18 \times 10^{5}}{1.67 \times 10^{-27}}$

= $1.13 \times 10^{13} m/s^2$

According to the third equation of motion,

$v^{2} = 2 \times a \times d$

or, d = $\frac{v^{2}}{2d}$

= $\frac{(2.4 \times 10^{6})^{2}}{2 \times 1.13 \times 10^{13}}$

= 0.254 m

Thus, we can conclude that the proton will travel 0.254 m before reaching its turning point.

7 0

3 years ago