Answer is C is the correct answer
Answer:
First Question

Second Question
The wavelength is for an X-ray
Explanation:
From the question we are told that
The width of the wall is 
The first excited state is
The ground state is 
Gnerally the energy (in MeV) of the photon emitted when the proton undergoes a transition is mathematically represented as
![E = \frac{h^2 }{ 8 * m * l^2 [ n_1^2 - n_0 ^2 ] }](https://tex.z-dn.net/?f=E%20%20%20%3D%20%20%20%5Cfrac%7Bh%5E2%20%7D%7B%208%20%2A%20m%20%20%2A%20%20l%5E2%20%5B%20n_1%5E2%20-%20n_0%20%5E2%20%5D%20%7D)
Here h is the Planck's constant with value 
m is the mass of proton with value 
So
![E = \frac{( 6.626*10^{-34})^2 }{ 8 * (1.67 *10^{-27}) * (10 *10^{-15})^2 [ 2^2 - 1 ^2 ] }](https://tex.z-dn.net/?f=E%20%20%3D%20%20%20%5Cfrac%7B%28%206.626%2A10%5E%7B-34%7D%29%5E2%20%7D%7B%208%20%2A%20%281.67%20%2A10%5E%7B-27%7D%29%20%20%2A%20%20%2810%20%2A10%5E%7B-15%7D%29%5E2%20%5B%202%5E2%20-%201%20%5E2%20%5D%20%7D)
=> 
Generally the energy of the photon emitted is also mathematically represented as

=> 
=> 
=> 
Generally the range of wavelength of X-ray is 
So this wavelength is for an X-ray.
There's little gravity so your weight would change but not your mass
Answer:
22 revolutions
Explanation:
2 rev/s = 2*(2π rad/rev) = 12.57 rad/s
The angular acceleration when it starting

The angular acceleration when it stopping:

The angular distance it covers when starting from rest:


The angular distance it covers when coming to complete stop:


So the total angular distance it covers within 22 s is 62.8 + 75.4 = 138.23 rad or 138.23 / (2π) = 22 revolutions
To solve this problem it is necessary to apply the concepts related to Dopler's Law. Dopler describes the change in frequency of a wave in relation to that of an observer who is in motion relative to the Source of the Wave.
It can be described as

c = Propagation speed of waves in the medium
= Speed of the receiver relative to the medium
= Speed of the source relative to the medium
Frequency emited by the source
The sign depends on whether the receiver or the source approach or move away from each other.
Our values are given by,
Velocity of car
velocity of motor
Velocity of sound
Frequency emited by the source
Replacing we have that



Therefore the frequency that hear the motorcyclist is 601.7Hz