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3 years ago

The electric cell made _____ of currents possible.

Physics

2 answers:

Kitty [74]3 years ago

7 0

Electric cells are source of constant potential difference

these are also known as DC source.

they always follow ohm's law as per which

$V = iR$

since DC source are constant EMF source

it always produce constant current in it

so the correct answer must be

Steady Streams

baherus [9]3 years ago

4 0

Answer:

it would be steady streams

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The alternating current which crosses an apparatus of 600 W has a maximum value of 2.5 A. What is efficient voltage between its

azamat

Answer: Option (b) is correct.

Explanation:

Since we know that,

P = VI

where;

P = power

V= Voltage

I = Current

Since it's given that,

P = 600W

I = 2.5 A

equating these values in the above equation, we get;

V = $\frac{600}{2.5}$

V = 240 V

8 0

3 years ago

A light-year is a unit of a. time. c. mass. b. distance. d. density. please select the best answer from the choices provided a b

11111nata11111 [884]

Answer:

distance

Explanation:

it is the distance traveled by light in one year

3 0

2 years ago

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Mike says, I gave up guitar because I’ll never be good at it, some people seem to have a good ear music, not me. Which mindset i

AveGali [126]

A negative mind set

8 0

3 years ago

Based on the law of conservation of energy, which statement is false?

alexgriva [62]

I think it’s Energy is lost when machines don’t work right.

8 0

3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago