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Sauron [17]
2 years ago
14

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 12.0 cm ,

and the outer sphere has radius 14.0 cm . A potential difference of 100 V is applied to the capacitor.
A)What is the energy density at r= 12.1 cm , just outside the inner sphere?
B)What is the energy density at r = 13.9 cm , just inside the outer sphere?
C)For a parallel-plate capacitor the energy density is uniform in the region between the plates, except near the edges of the plates. Is this also true for a spherical capacitor?

There are formulas related to energy density but i couldn't figure out the solution.I would be thankful for your helps.
Physics
1 answer:
iren [92.7K]2 years ago
5 0
Energy Density = 1/2 × ε(0) × (V/d)^2

V = 100, d = 0.01, ε(0) = 8.85 x 10^-12
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A physics student walks 100 meters in 80 seconds. The student stops for 30 seconds, and then walks 200 meters farther in 90 seco
Elan Coil [88]

Answer:

B i think is the answer

Explanation:

i feel like it is B because if you put them together and the answer is 1.5 so it is B

8 0
2 years ago
Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the crit
Dennis_Churaev [7]

Complete Question:

A beam of white light is incident on the surface of a diamond at an angle \theta_{a},  since the index of refraction depends on the light's wavelength, the different colors that comprise white light will spread out as they pass through the diamond. For example, the indices of refraction in diamond are n_{red} = 2.410 for red light and n_{blue} = 2.450 for blue light. Thus, blue light and red light are refracted at different angles inside the diamond. The surrounding air has n_{air} = 1.000.

Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the critical angle θcrit, the light will not be refracted out into the air, but instead it will be totally internally reflected back into the diamond. Find θcrit. Express your answer in degrees to four significant figures.

Answer:

\theta_{crit} = 24.09^{0}

Explanation:

Only the blue refracted ray is related to the critical angle in this question

n_{air} = 1.000

n_{blue} = 2.450

The relationship between the critical angle(\theta_{crit}), n_{air} and n_{blue} can be given as sin \theta_{crit} = \frac{n_{air} }{n_{blue} }

sin \theta_{crit} = \frac{1 }{2.450 }\\\theta_{crit} = sin^{-1} \frac{1 }{2.450 }\\\theta_{crit} = sin^{-1} 0.4082^{0}\\  \theta_{crit} = 24.09^{0}

6 0
2 years ago
A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, th
ziro4ka [17]

Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

A.

In the first question, we have to caculate the constant of the spring with this equation:

m*g=k*x

Getting the k:

k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\

The energy of a spring:

E=\frac{1}{2}*k*x^{2}

For the first case:

E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]

For the second case:

E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]

If you take the relation E2/E1 = 4.

B.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

E=\frac{1}{2}*k*x^{2}⇒k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]

C.

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

E=W*h=500[N]*10[m]=5000[J]

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

D.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J

7 0
2 years ago
Which statement correctly describes the relationship between frequency and wavelength?
Len [333]
The relationship between the frequency and wavelength of a wave is given by the equation:

v=λf, where v is the velocity of the wave, λ is the wavelength and f is the frequency. 

If we divide the equation by f we get:

λ=v/f

From here we see that the wavelength and frequency are inversely proportional. So as the frequency increases the wavelength decreases. 

So the second statement is true: As the frequency of a wave increases, the shorter the wavelength is.  
3 0
2 years ago
Read 2 more answers
A damped harmonic oscillator consists of a block of mass 2.5 kg attached to a spring with spring constant 10 N/m to which is app
Cerrena [4.2K]

Answer:

0.5% per oscillation

Explanation:

The term 'damped oscillation' means an oscillation that fades away with time. For Example; a swinging pendulum.

Kinetic energy, KE= 1/2×mv^2-------------------------------------------------------------------------------------------------------------(1).

Where m= Mass, v= velocity.

Also, Elastic potential energy,PE=1/2×kX^2----------------------------------------------------------------------------------------------------------------------(2).

Where k= force constant, X= displacement.

Mechanical energy= potential energy (when a damped oscillator reaches maximum displacement).

Therefore, we use equation (3) to get the resonance frequency,

W^2= k/m--------------------------------------------------------------------------------------(3)

Slotting values into equation (3).

= 10/2.5.

= ✓4.

= 2 s^-1.

Recall that, F= -kX

F^2= (-0.1)^2

Potential energy,PE= 1/2 ×0.01

Potential energy= 0.05 ×100

= 0.5% per oscillation.

6 0
2 years ago
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