Answer:
3 photons 
Explanation:
The energy of a photon E can be calculated using this formula:

Where  corresponds to Plank constant (6.626070x10^-34Js),
 corresponds to Plank constant (6.626070x10^-34Js),  is the speed of light in the vacuum (299792458m/s) and
 is the speed of light in the vacuum (299792458m/s) and  is the wavelength of the photon(in this case 800nm).
 is the wavelength of the photon(in this case 800nm).

Tranform the units

The band Gap is 4eV, divide the band gap between the energy of the photon:

Rounding to the next integrer: 3.
Three photons are the minimum to equal or exceed the band gap.
 
        
             
        
        
        
B. The voltage is the same across all resistors in the circuit.
        
                    
             
        
        
        
I believe an Atom is a very powerful source, the basic unit of a chemical element. An atom is a source of nuclear energy. 
But a molecule on the other hand isn't so different.
a group of atoms bonded together, representing the smallest fundamental unit of a chemical compound that can take part in a chemical reaction.
I hope that helps, have a fantastic day!
        
                    
             
        
        
        
Answer:
a) L=0. b) L = 262 k ^   Kg m²/s and c)  L = 1020.7 k^   kg m²/s
Explanation:
It is angular momentum given by
        L = r x p
Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum
One of the easiest ways to make this vector product is with the use of determinants
![{array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]](https://tex.z-dn.net/?f=%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%26y%26z%5C%5Cpx%26py%26pz%5Cend%7Barray%7D%5Cright%5D)
Let's apply this relationship to our case
Let's start by breaking down the speed
       v₀ₓ = v₀ cosn 45
       voy =v₀ sin 45
       v₀ₓ = 9 cos 45
       voy = 9 without 45
       v₀ₓ = 6.36 m / s
       voy = 6.36 m / s
a) at launch point r = 0 whereby L = 0
. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero
    vfy² = voy²- 2 g y
    y = voy² / 2g
    y = (6.36)²/2 9.8
    y = 2.06 m
Let's calculate the angular momentum
L= ![\left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%26y%260%5C%5Cpx%260%260%5Cend%7Barray%7D%5Cright%5D)
L = -px y k ^
L = - (m vox) (2.06) k ^
L = - 20 6.36 2.06 k ^
L = 262 k ^   Kg m² / s
The angular momentum is on the z axis
c) At the point of impact, at this point the height is zero and the position on the x-axis is the range
      R = vo² sin 2θ / g
      R = 9² sin (2 45) /9.8
      R = 8.26 m
L = ![\left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5Cx%260%260%5C%5Cpx%26py%260%5Cend%7Barray%7D%5Cright%5D) 
 
L = - x py k ^
L = - x m voy
L = - 8.26 20 6.36 k ^
L = 1020.7 k^   kg m² /s