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2 years ago

What is the amount of charge a capacitor can store per volt of potential difference?

2 answers:

4 0

The amount of charge a capacitor can store per volt of potential difference is called the capacitance of the capacitor.

<h3>What is the capacitance?</h3>

The capacitance has to do with the amount of charge that is stored by a capacitor. We know that a capacitor is a device that can be used to store electric charges. We have in it, two capacitors separated by a dielectric material.

Hence, the amount of charge a capacitor can store per volt of potential difference is called the capacitance of the capacitor.

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Vlad1618 [11]2 years ago

3 0

Capacitance is the amount of charge a capacitor can store per volt of potential difference.

<h3>What is capacitance?</h3>

Capacitance is defined as the capacitors ability to store an electrical charge on its plates. Capacitance is derived from capacity which means ability to store.

So we can conclude that Capacitance is the amount of charge a capacitor can store per volt of potential difference.

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Suppose you are in a moving car and the motor stops running. You step on the brakes and slow the car to half speed. If you relea

In-s [12.5K]

Answer:

See the answer below

Explanation:

If you step on the brake of a car while driving, the frictional force between the tires of the car and the surface of the road increases in opposition to the motion of the car. Consequently, the car slows down.

If you release your foot from the brake pedal when the car is still at half speed, the frictional force reduces and the car speeds up a bit even without pressing the throttle. Eventually, the frictional force will slow down and stop the car if the throttle is not pressed.

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3 years ago

What type of wave is shown below?

Elena-2011 [213]

Answer:

B. Transverse wave.

Explanation:

Because it has troughs and crests.

5 0

3 years ago

A battery has an emf of 15.0 V. The terminal voltage of the battery is 12.2 V when it is delivering 14.0 W of power to an extern

solong [7]

Answer:

The value of R and the internal resistance of the battery are 10.6 ohm and 2.45 ohm

Explanation:

Given that,

Emf of battery = 15.0 V

Voltage = 12.2 V

Power = 14.0 W

(a). We need to calculate the value of R

Using formula of power

$P=IV$

$P=\dfrac{V^2}{R}$

$R=\dfrac{V^2}{P}$

Where, R = resistance

P = power

V = voltage

Put the value into the formula

$R =\dfrac{(12.2)^2}{14.0}$

$R=10.6\ \Omega$

(b). We need to calculate the internal resistance of the battery

Firstly we calculate the current

Using formula of current

$I=\dfrac{V}{R}=\dfrac{P}{V}$

Put the value of P and V into the formula

$I =\dfrac{14}{12.2}$

$I=1.14\ A$

We calculate the internal resistance

Using formula of emf

$e-V=Ir$

$r =\dfrac{e-V}{I}$

Put the value into the formula

$r=\dfrac{15.0-12.2}{1.14}$

$r = 2.45\ \Omega$

Hence, The value of R and the internal resistance of the battery are 10.6 ohm and 2.45 ohm

3 0

3 years ago

Can you please solve this for me urgently want to make sure if my answers are correct?

MA_775_DIABLO [31]

Answer:

Resolving horizontally. :

$\sum d_{x} = - (17.0 \cos 20.0) - (11.0 \cos 35.0) + (30.0 \cos 50.0) + 0 \\ { \underline{d _{x} = - 5.702 \: m}} \\ \\ \sum d _{y} = (17.0 \sin 20.0) + 12.0 - (11.0 \sin 35.0) - (30.0 \sin 50.0) \\ { \underline{d _{y} = - 11.476 \: m}}$

therefore, for resultant:

$d = \sqrt{ {d _{y} }^{2} + d _{x} {}^{2} }$

substitute:

$d = \sqrt{ {( - 5.702)}^{2} + {( - 11.476)}^{2} } \\ \\ d = \sqrt{164.211} \\ \\ { \boxed{ \boxed{ \bf{d = 12.8 \: m}}}}$

6 0

3 years ago

Briefly explain how an electric motor works.

NeTakaya

You put electricity into it at one end and an axle (metal rod) rotates at the other end giving you the power to drive a machine of some kind.

3 0

3 years ago