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.
Answer:
The distance is 3.1 m
Explanation:
The position vector of the fly relative to the corner of the wall is
r = (3.1, 0.5).
The distance of the fly from the corner will be calculated as the magnitude of the vector "r"
magnitude of vector 
Since the numbers to be added have only one decimal place 3.<u>1</u> and 0.<u>5</u>, the result of the sum will have to have one decimal place. The result of the square root will also have one decimal place.
The apple is not accelerating. That's a big tip off that the net force acting on it is zero.
The acceleration of the motion tells us about the net force acting on the body
Answer:
F = 0.78[N]
Explanation:
The given values correspond to forces, we must remember or take into account that the forces are vector quantities, that is, they have magnitude and direction. Since we have two X-Y coordinate axes (two-dimensional), we are going to decompose each of the forces into the X & y components.
<u>For F₁</u>
<u />
<u />
<u>For F₂</u>
![F_{x}=2*cos(60)\\F_{x}=1[N]\\F_{y}=-2*sin(60)\\F_{y}=-1.73[N]](https://tex.z-dn.net/?f=F_%7Bx%7D%3D2%2Acos%2860%29%5C%5CF_%7Bx%7D%3D1%5BN%5D%5C%5CF_%7By%7D%3D-2%2Asin%2860%29%5C%5CF_%7By%7D%3D-1.73%5BN%5D)
<u>For F₃</u>
<u />
<u />
Now we can sum each one of the forces in the given axes:
![F_{x}=1-0.866=0.134[N]\\F_{y}=2-1.73+0.5\\F_{y}=0.77[N]](https://tex.z-dn.net/?f=F_%7Bx%7D%3D1-0.866%3D0.134%5BN%5D%5C%5CF_%7By%7D%3D2-1.73%2B0.5%5C%5CF_%7By%7D%3D0.77%5BN%5D)
Now using the Pythagorean theorem we can find the total force.
![F=\sqrt{(0.134)^{2} +(0.77)^{2}}\\F= 0.78[N]](https://tex.z-dn.net/?f=F%3D%5Csqrt%7B%280.134%29%5E%7B2%7D%20%2B%280.77%29%5E%7B2%7D%7D%5C%5CF%3D%200.78%5BN%5D)