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3 years ago

Calculate the electric potential energy in a capacitor that stores 4.0 x 10-10 C

Physics

2 answers:

serious [3.7K]3 years ago

5 0

5.0•10^-8J

Ape.x .........-_-

ioda3 years ago

4 0

The electric potential energy is $1\cdot 10^{-7} J$

Explanation:

The electric potential energy stored in a capacitor is given by the equation

$PE=Q \Delta V$

where

Q is the charge stored in the capacitor

$\Delta V$ is the potential difference between the plates

For the capacitor in this problem, we have

$Q=4.0\cdot 10^{-10}C$ is the charge

$\Delta V=250.0 V$ is the potential difference

Substituting, we find the potential energy:

$PE=(4.0\cdot 10^{-10})(250.0)=1\cdot 10^{-7} J$

Learn more about capacitors:

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3 years ago

A truck pushes a pile of dirt horizontally on a frictionless road with a net force of 20\, \text N20N20, start text, N, end text

meriva

Answer:

300 Nm ; 300 J

Explanation:

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3 years ago

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Reptile [31]

Answer:

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6 0

3 years ago

An alternating source drives a series RLC circuit with an emf amplitude of 6.04 V, at a phase angle of +30.3°. When the potentia

Vinvika [58]

Answer:

-8.56V

Explanation:

Our values are given by,

e = 6.04 V

Φ = 30.3

VC = 5.32

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$e(t) = e* sin(wt)$

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$e(t) = 6.04 * sin(32.5 + 180)$

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Now, Using Kirchoff Voltage Law,

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5 0

3 years ago

A hoop (I = MR2) of mass 3 kg and radius 1.1 m is rolling at a center-of-mass speed of 11 m/s. An external force does 842 J of w

Scilla [17]

Answer:

$v_f = 20 m/s$

Explanation:

Since the hoop is rolling on the floor so its total kinetic energy is given as

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now for pure rolling condition we will have

$v = R\omega$

also we have

$I = mR^2$

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$W = KE_f - KE_i$

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3 0

3 years ago