Answer:
The answer is true
Explanation:
could you brainliest again they said I plagarized when I didn't
If you're moving, then you have kinetic energy.
If you're not at the bottom yet, then you still have
some potential energy left.
Answer:
a) ![\Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)](https://tex.z-dn.net/?f=%20%5CDelta%20%5Clambda%20%3D%20%5Clambda%27%20-%5Clambda_o%20%3D%20%5Cfrac%7Bh%7D%7Bm_e%20c%7D%20%281-cos%20%5Ctheta%29)
For this case we know the following values:
![h = 6.63 x10^{-34} Js](https://tex.z-dn.net/?f=%20h%20%3D%206.63%20x10%5E%7B-34%7D%20Js%20)
![m_e = 9.109 x10^{-31} kg](https://tex.z-dn.net/?f=%20m_e%20%3D%209.109%20x10%5E%7B-31%7D%20kg%20)
![c = 3x10^8 m/s](https://tex.z-dn.net/?f=%20c%20%3D%203x10%5E8%20m%2Fs)
![\theta = 37](https://tex.z-dn.net/?f=%20%5Ctheta%20%3D%2037)
So then if we replace we got:
![\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm](https://tex.z-dn.net/?f=%20%5CDelta%20%5Clambda%20%3D%20%5Cfrac%7B6.63x10%5E%7B-34%7D%20Js%7D%7B9.109%20x10%5E%7B-31%7D%20kg%20%2A3x10%5E8%20m%2Fs%7D%20%281-cos%2037%29%20%3D%204.885x10%5E%7B-13%7D%20m%20%2A%20%5Cfrac%7B1m%7D%7B1x10%5E%7B-15%7D%20m%7D%3D%20488.54%20fm)
b) ![\lambda_0 = \frac{hc}{E_0}](https://tex.z-dn.net/?f=%20%5Clambda_0%20%3D%20%5Cfrac%7Bhc%7D%7BE_0%7D)
With ![E_0 = 300 k eV= 300000 eV](https://tex.z-dn.net/?f=%20E_0%20%3D%20300%20k%20eV%3D%20300000%20eV)
And replacing we have:
![\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm](https://tex.z-dn.net/?f=%20%5Clambda_0%20%3D%20%5Cfrac%7B1240%20x10%5E%7B-9%7D%20eV%20m%7D%7B300000eV%7D%3D4.13%20x10%5E%7B-12%7Dm%20%3D%204.12%20pm)
And then the scattered wavelength is given by:
![\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm](https://tex.z-dn.net/?f=%20%5Clambda%20%27%20%3D%20%5Clambda_0%20%2B%20%5CDelta%20%5Clambda%20%3D%204.13%20%2B%200.489%20pm%20%3D%204.619%20pm)
And the energy of the scattered photon is given by:
![E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV](https://tex.z-dn.net/?f=%20E%27%20%3D%20%5Cfrac%7Bhc%7D%7B%5Clambda%27%7D%3D%20%5Cfrac%7B1240x10%5E%7B-9%7D%20eVm%7D%7B4.619x10%5E%7B-12%7D%20m%7D%3D268456.37%20eV%20-%20268.46%20keV)
c) ![E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV](https://tex.z-dn.net/?f=%20E_f%20%3D%20E_0%20-E%27%20%3D%20300%20-268.456%20kev%20%3D%2031.544%20keV)
Explanation
Part a
For this case we can use the Compton shift equation given by:
For this case we know the following values:
So then if we replace we got:
Part b
For this cas we can calculate the wavelength of the phton with this formula:
With
And replacing we have:
And then the scattered wavelength is given by:
And the energy of the scattered photon is given by:
Part c
For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:
A) ![2.03 m/s^2](https://tex.z-dn.net/?f=2.03%20m%2Fs%5E2)
Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:
Parallel:
(1)
where
m is the mass
g = 9.8 m/s^2 the acceleration of gravity
![\theta=22.5^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D22.5%5E%7B%5Ccirc%7D)
is the coefficient of friction
R is the normal reaction
a is the acceleration
Perpendicular:
(2)
From (2) we find
![R=mg cos \theta](https://tex.z-dn.net/?f=R%3Dmg%20cos%20%5Ctheta)
And substituting into (1)
![mg sin \theta - \mu_k mg cos \theta = ma](https://tex.z-dn.net/?f=mg%20sin%20%5Ctheta%20-%20%5Cmu_k%20mg%20cos%20%5Ctheta%20%3D%20ma)
Solving for a,
![a=g sin \theta - \mu_k g cos \theta = (9.8)(sin 22.5)-(0.19)(9.8)(cos 22.5)=2.03 m/s^2](https://tex.z-dn.net/?f=a%3Dg%20sin%20%5Ctheta%20-%20%5Cmu_k%20g%20cos%20%5Ctheta%20%3D%20%289.8%29%28sin%2022.5%29-%280.19%29%289.8%29%28cos%2022.5%29%3D2.03%20m%2Fs%5E2)
B) 5.94 m/s
We can solve this part by using the suvat equation
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
Here we have
v = ?
u = 0 (it starts from rest)
![a=2.03 m/s^2](https://tex.z-dn.net/?f=a%3D2.03%20m%2Fs%5E2)
s = 8.70 m
Solving for v,
![v=\sqrt{u^2+2as}=\sqrt{0+2(2.03)(8.70)}=5.94 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Bu%5E2%2B2as%7D%3D%5Csqrt%7B0%2B2%282.03%29%288.70%29%7D%3D5.94%20m%2Fs)
Answer:
Sorry this isn’t going to be any help. You don’t have any statement that I’m able to see.
Explanation: