The electric potential difference of a 150 µF capacitor is measured across the terminals of the capacitor and found to be 5 volts. What is the potential energy of the capacitor?

Answers

Answer 1

The potential energy stored in a capacitor is given by:
[tex]U= \frac{1}{2}CV^2 [/tex]
where
C is the capacitance of the capacitor
V is the potential difference across the plates of the capacitor

In our problem,
[tex]C=150 \mu F= 150 \cdot 10^{-6} F[/tex]
[tex]V=5 V[/tex]
Therefore, the potential energy of the capacitor is
[tex]U= \frac{1}{2}(150 \cdot 10^{-6}F)(5 V)^2 = 1.88 \cdot 10^{-3}J [/tex]

Answer 2

Answer:

it's 2 on edge

Explanation:

Related Questions

What is the correct formula for finding the frequency of an electromagnetic wave
A. F=c-A
B. F=A/c
C. F=A+c
D. F=c/A

Answers

Answer:
f = c / λ

Explanation:
Electromagnetic waves are types of periodic waves. They propagate with the same speed as light (3 * 10⁸ m/sec).
Therefore:
velocity of wave = c

Now, the equation that relates speed of wave and its frequency is as follows:
c = λf
where:
c is the speed of wave
f is the frequency of the wave
λ is the wavelength of the wave

Solve the above equation for frequency, we will end up with:
f = c / λ

is bread mold chemical or physical

Answers

Chemical change. Hope this helped!! ( and I hope I got it correct )

A bread mould is a chemical change and not physical because in a physical change no new substance is formed where as in a chemical change a new substance is formed and here he substance is the bread mould (a bacteria).
Thank you

A boat initially traveling at 10. meters per second accelerates uniformly at the rate of 5.0 meters per second2 for 10. seconds. How far does the boat travel during this time?

Answers

Final answer:

The boat would travel 350 meters during the given time period, assuming uniform acceleration. This is calculated using the equation of motion for distance.

Explanation:

The distance traveled by the boat can be calculated using the equation of motion: S = ut + 0.5at², where S is the distance traveled, u is the initial velocity, a is acceleration and t is time. In this case, u = 10 m/s, a = 5 m/s² and t = 10 s.

Substituting these values into the equation we get: S = 10*10 + 0.5*5*10². So, the total distance travelled by the boat = 100 m + 250 m = 350 m. Therefore, the boat travels 350 meters during this time.

Learn more about the Physics equation of motion here:

https://brainly.com/question/34615504

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How does paving parking lots and roads with concrete or asphalt affect surface water and groundwater?

Answers

Parking lots with roads and concrete reduce infiltration. Infiltration is the process by which water penetrates the soil. Reducing the amount of water that enters the soil can eventually impact groundwater levels in some areas by decreasing it over time. Paved roads lead to increased surface runoff which increases the possibility of flooding in periods of heavy rainfall. This is known as urban flooding.

I need help
Describe what happens to the energy of a wave if the frequency decreases. Enter your answer in the space provided.

Answers

The energy of the wave will decrease.

The energy of a wave is given as

E = h f

where E = energy of waver

h = plank's constant

f = frequency of the wave.

From the formula , we see that the energy of the wave is directly proportional to the frequency of the wave. hence as the frequency of the wave decrease, the energy of the wave will decrease.

What is the amount of displacement of a runner who runs exactly 2 laps around a 400 meter track?
A) 400 meters
B)0 meters
C)800 meters

Answers

Displacement is the distance and direction from the start point to the end point. Our runner finished exactly where he started. His displacement is zero.

The displacement is zero.

Hence, the correct option is B.

The displacement of an object is the change in its position.

In this case, the runner starts and ends at the same point, so the displacement is zero.

Even though the runner runs 800 meters (2 laps x 400 meters/lap), the runner does not change position.

The runner ends up where they started, so the displacement is zero.

Therefore, The displacement is zero.

Hence, the correct option is B.

To know more about displacement here

https://brainly.com/question/11934397

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If a car is traveling on the highway at a constant velocity, the force that pushes the car forward must be A. equal to the weight of the car. B. greater than the force of friction exerted upon the car. C. equal to the force of friction exerted upon the car. D. greater than the weight of the car.

Answers

c. equal to the force of friction exerted upon the car

Answer:

C. equal to the force of friction exerted upon the car.

Explanation:

We know according to Newton's First Law Of Motion , an object in rest or in uniform motion will be in same state unless or until an external force is not provided.

In other words , When no force apply on the body it being in the state of uniform motion or in rest.

In the case given, the car is travelling in the highway at a constant velocity. Therefore net force acting on car in horizontal direction is equal to zero.

It means all the forces acting on horizontal direction cancelling each other which cause net force equals to zero.

Here, to cancel force of friction the force that pushes the car in forward direction must be equals to it.

Therefore, C. option is correct.

A cable capable of pulling 4,500 N snapped while trying to drag a 20,000 N compressor across the street. What is the coefficient of static friction for this scenario?

Note: the 4500 N cable is used as Fs and is calculated the same as Fk
0.225
> 0.225
4,500 N
> 4,500 N
4.44
> 4.44
Please explain to me how you get your answer.

Answers

Answer:

[tex]\mu > 0.225[/tex]

Explanation:

The cable snapped because the frictional force [tex]f_s[/tex] was greater than the tension in the rope:

[tex]f_s > 4,500N[/tex]

Now,

[tex]f_s =\mu N[/tex]

where [tex]\mu[/tex] is the coefficient of static friction, and [tex]N[/tex] is the normal force. In our case, the normal force on the compressor is 20,000 N; therefore,

[tex]f_s =\mu (20,000N )[/tex].

Putting this into the condition [tex]f_s > 4,500N[/tex], we get:

[tex]\mu (20,000N )> 4,500N[/tex]

[tex]$\mu > \frac{4,500N}{20,000N} $[/tex]

[tex]\boxed{\mu > 0.225.}[/tex]

Answer:

>0.225 is correct if its not please let me know I'm willing to help!

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