This diagram shows half of the magnetic field of a bar magnet. Carefully draw in the other

Find the energy u of the capacitor in terms of c and q by using the definition of capacitance and the formula for the energy in

gtnhenbr [62]

The formula for the energy in a capacitor , u in terms of q and c is q²/2c

<h3>What is the energy of a capacitor?</h3>

The energy of a capacitor u = 1/2qv where

q = charge on capacitor and
v = voltage across capacitor.

<h3>What is the capacitance of a capacitor?</h3>

Also, the capacitance of a capacitor c = q/v where

q = charge on capacitor and
v = voltage across capacitor.

So, v = q/c

<h3>The formula for energy of the capacitor in terms of q and c</h3>

Substituting v into u, we have

u = 1/2qv

= 1/2q(q/c)

= q²/2c

So, the formula for the energy in a capacitor , u in terms of q and c is q²/2c

Learn more about energy in a capacitor here:

brainly.com/question/10705986

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3 0

2 years ago

What is the attractive force between all matter in the universe?

kvv77 [185]

The attractive force between all matter in the universe is gravity.

4 0

3 years ago

Read 2 more answers

Does current in Helmholtz coils flow in the same or opposite direction through each coil? Explain.

zhenek [66]

To get a uniform field in the central region between the coils, current flows in the same direction in each.

5 0

3 years ago

Please answer it is urgent <br>I will make u brainlist

adelina 88 [10]

For B, it is because water is a really good conductor of electricity, so the electrician will get shocked

6 0

4 years ago

A tuning fork labeled 392 Hz has the tip of each of its two prongsvibrating with an amplitude of 0.600mma) What is the maximum s

DanielleElmas [232]

a) 1.48 m/s

The tuning fork is moving by simple harmonic motion: so, the maximum speed of the tip of the prong is related to the frequency and the amplitude by

$v_{max}=\omega A$

where

$v_{max}$ is the maximum speed

$\omega$ is the angular frequency

A is the amplitude

For the tuning fork in the problem, we have

$\omega=2\pi f=2 \pi(392 Hz)=2462 rad/s$ , where f is the frequency

$A=0.600 mm=6\cdot 10^{-4} m$ is the amplitude

Therefore, the maximum speed is

$v_{max}=(2462 rad/s)(6\cdot 10^{-4}m)=1.48 m/s$

b) $3.0\cdot 10^{-5} J$

The fly's maximum kinetic energy is given by

$K=\frac{1}{2}mv_{max}^2$

where

$m=0.0270 g=2.7\cdot 10^{-5} kg$ is the mass of the fly

$v_{max}=1.48 m/s$ is the maximum speed

Substituting into the equation, we find

$K=\frac{1}{2}(2.7\cdot 10^{-5}kg)(1.48 m/s)^2=3.0\cdot 10^{-5} J$

7 0

3 years ago