Answer:
If the frequency of the motion of a simple harmonic oscillator is doubled , then maximum speed of the oscillator changes by the factor 2
Explanation:
We know that in a simple harmonic oscillator the maximum speed is given by
= ![Aw](https://tex.z-dn.net/?f=Aw)
Here A is amplitude which is constant , so from above equation we see that maximum speed is directly proportional to
of the oscillation .
Since ![w = 2 \pi f](https://tex.z-dn.net/?f=w%20%3D%202%20%5Cpi%20f)
= 2
Where
is the maximum speed when frequency is doubled .
Part A)
First of all, let's convert the radii of the inner and the outer sphere:
![r_A = 11.0 cm = 0.110 m](https://tex.z-dn.net/?f=r_A%20%3D%2011.0%20cm%20%3D%200.110%20m)
![r_B = 16.5 cm=0.165 m](https://tex.z-dn.net/?f=r_B%20%3D%2016.5%20cm%3D0.165%20m)
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
![C=4 \pi \epsilon _0 \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=](https://tex.z-dn.net/?f=C%3D4%20%5Cpi%20%5Cepsilon%20_0%20%20%5Cfrac%7Br_A%20r_B%7D%7Br_B-%20r_A%7D%3D4%5Cpi%288.85%20%5Ccdot%2010%5E%7B-12%7DC%5E2m%5E%7B-2%7DN%5E%7B-1%7D%29%20%5Cfrac%7B%280.110m%29%280.165m%29%7D%7B0.165m-0.110m%7D%3D)
![=3.67\cdot 10^{-11}F](https://tex.z-dn.net/?f=%3D3.67%5Ccdot%2010%5E%7B-11%7DF)
Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
![Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C](https://tex.z-dn.net/?f=Q%3DCV%3D%283.67%5Ccdot%2010%5E%7B-11%7DF%29%28100%20V%29%3D3.67%5Ccdot%2010%5E%7B-9%7DC)
Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
![E\cdot (4 \pi r^2) = \frac{Q}{\epsilon _0}](https://tex.z-dn.net/?f=E%5Ccdot%20%284%20%5Cpi%20r%5E2%29%20%3D%20%20%5Cfrac%7BQ%7D%7B%5Cepsilon%20_0%7D%20)
from which
![E(r) = \frac{Q}{4 \pi \epsilon_0 r^2}](https://tex.z-dn.net/?f=E%28r%29%20%3D%20%20%5Cfrac%7BQ%7D%7B4%20%5Cpi%20%5Cepsilon_0%20r%5E2%7D)
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
![E(0.111m)=2680 N/C](https://tex.z-dn.net/?f=E%280.111m%29%3D2680%20N%2FC)
And so, the energy density at r=0.111 m is
![U= \frac{1}{2} \epsilon _0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3](https://tex.z-dn.net/?f=U%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Cepsilon%20_0%20E%5E2%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%288.85%5Ccdot%2010%5E%7B-12%7DC%5E2m%5E%7B-2%7DN%5E%7B-1%7D%29%282680%20N%2FC%29%5E2%3D3.17%20%5Ccdot%2010%5E%7B-5%7DJ%2Fm%5E3)
Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor:
![Q=3.67 \cdot 10^{-9}C](https://tex.z-dn.net/?f=Q%3D3.67%20%5Ccdot%2010%5E%7B-9%7DC)
. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
![E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C](https://tex.z-dn.net/?f=E%280.164%20m%29%3D%20%5Cfrac%7BQ%7D%7B4%20%5Cpi%20%5Cepsilon_0%20r%5E2%7D%3D1228%20N%2FC%20)
And therefore, the energy density at this distance from the center is
Answer:
Given:
Thermal Kinetic Energy of an electron, ![KE_{t} = \frac{3}{2}k_{b}T](https://tex.z-dn.net/?f=KE_%7Bt%7D%20%3D%20%5Cfrac%7B3%7D%7B2%7Dk_%7Bb%7DT)
= Boltzmann's constant
Temperature, T = 1800 K
Solution:
Now, to calculate the de-Broglie wavelength of the electron,
:
![\lambda_{e} = \frac{h}{p_{e}}](https://tex.z-dn.net/?f=%5Clambda_%7Be%7D%20%3D%20%5Cfrac%7Bh%7D%7Bp_%7Be%7D%7D)
(1)
where
h = Planck's constant = ![6.626\times 10^{- 34}m^{2}kg/s](https://tex.z-dn.net/?f=6.626%5Ctimes%2010%5E%7B-%2034%7Dm%5E%7B2%7Dkg%2Fs)
= momentum of an electron
= velocity of an electron
= mass of electon
Now,
Kinetic energy of an electron = thermal kinetic energy
![\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dm_%7Be%7Dv_%7Be%7D%5E%7B2%7D%20%3D%20%5Cfrac%7B3%7D%7B2%7Dk_%7Bb%7DT)
![}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}](https://tex.z-dn.net/?f=%7Dv_%7Be%7D%20%3D%20%5Csqrt%7B2%5Cfrac%7B%5Cfrac%7B3%7D%7B2%7Dk_%7Bb%7DT%7D%7Bm_%7Be%7D%7D%7D)
![}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}](https://tex.z-dn.net/?f=%7Dv_%7Be%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B3%5Ctimes%201.38%5Ctimes%2010%5E%7B-%2023%7D%5Ctimes%201800%7D%7B9.1%5Ctimes%2010_%7B-%2031%7D%7D%7D)
(2)
Using eqn (2) in (1):
![\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm](https://tex.z-dn.net/?f=%5Clambda_%7Be%7D%20%3D%20%5Cfrac%7B6.626%5Ctimes%2010%5E%7B-%2034%7D%7D%7B9.1%5Ctimes%2010_%7B-%2031%7D%5Ctimes%202.86%5Ctimes%2010%5E%7B5%7D%7D%20%3D%202.55%20nm)
Now, to calculate the de-Broglie wavelength of proton,
:
![\lambda_{p} = \frac{h}{p_{p}}](https://tex.z-dn.net/?f=%5Clambda_%7Bp%7D%20%3D%20%5Cfrac%7Bh%7D%7Bp_%7Bp%7D%7D)
(3)
where
= mass of proton
= velocity of an proton
Now,
Kinetic energy of a proton = thermal kinetic energy
![\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dm_%7Bp%7Dv_%7Bp%7D%5E%7B2%7D%20%3D%20%5Cfrac%7B3%7D%7B2%7Dk_%7Bb%7DT)
![}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}](https://tex.z-dn.net/?f=%7Dv_%7Bp%7D%20%3D%20%5Csqrt%7B2%5Cfrac%7B%5Cfrac%7B3%7D%7B2%7Dk_%7Bb%7DT%7D%7Bm_%7Bp%7D%7D%7D)
![}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}](https://tex.z-dn.net/?f=%7Dv_%7Bp%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B3%5Ctimes%201.38%5Ctimes%2010%5E%7B-%2023%7D%5Ctimes%201800%7D%7B1.6726%5Ctimes%2010_%7B-%2027%7D%7D%7D)
(4)
Using eqn (4) in (3):
![\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm](https://tex.z-dn.net/?f=%5Clambda_%7Bp%7D%20%3D%20%5Cfrac%7B6.626%5Ctimes%2010%5E%7B-%2034%7D%7D%7B1.6726%5Ctimes%2010_%7B-%2027%7D%5Ctimes%206.674%5Ctimes%2010%5E%7B3%7D%7D%20%3D%205.94%5Ctimes%2010%5E%7B-%2011%7D%20m%20%3D%200.0594%20nm)
Answer:
![\Phi_{E} = E\pi r^2 \omega t](https://tex.z-dn.net/?f=%5CPhi_%7BE%7D%20%3D%20E%5Cpi%20r%5E2%20%5Comega%20t)
Explanation:
The electric flux is defined as the multiple of electric field and the area that the electric field passes through, such that
![\Phi_{E} = \vec{E}\vec{A}](https://tex.z-dn.net/?f=%5CPhi_%7BE%7D%20%3D%20%5Cvec%7BE%7D%5Cvec%7BA%7D)
When calculating the electric flux, the angle between the directions of electric field and the area becomes important, especially if the angle is changing with time.
The above formula can be rewritten as follows
![\Phi_{E} = EA\cos(\theta)](https://tex.z-dn.net/?f=%5CPhi_%7BE%7D%20%3D%20EA%5Ccos%28%5Ctheta%29)
where θ is the angle between the electric field and the area of the loop. Note that, the direction of the area of the loop is perpendicular to the plane of the loop.
If the loop is rotating with constant angular velocity ω, then the angle can be written as follows
![\theta = \omega t](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Comega%20t)
At t = 0, cos(0) = 1 and the electric flux through the loop is at its maximum value.
Therefore the electric flux can be written as a function of time
![\Phi_{E} = E\pi r^2 \omega t](https://tex.z-dn.net/?f=%5CPhi_%7BE%7D%20%3D%20E%5Cpi%20r%5E2%20%5Comega%20t)
A solar eclipse will be visible over a wide area of the north polar region
on Friday, March 20.
England is not in the path of totality, but it's close enough so that a large
part of the sun will be covered, and it will be a spectacular sight.
For Londoners, the eclipse begins Friday morning at 8:25 AM,when the
moon just begins to eat away at the sun's edge. It advances slowly, as more
and more of the sun disappears, and reaches maximum at 9:31 AM. Then
the obscured part of the sun begins to shrink, and the complete disk is
restored by the end of the eclipse at 10:41AM, after a period of 2 hours
16 minutes during which part of the sun appears to be missing.
The catch in observing the eclipse is:
<em><u>YOU MUST NOT LOOK AT THE SUN</u></em>.
Staring at the sun for a period of time can cause permanent damage to
your vision, even though <em><u>you don't feel it while it's happening</u></em>.
This is not a useful place to try and give you complete instructions or
suggestions for observing the sun over a period of hours. Please look
in your local newspaper, or search online for phrases like "safe eclipse
viewing".