Given data
*The given 4th harmonic frequency is 31.5 Hz
The fundamental frequency is calculated as

Hence, the fundamental frequency is 7.875 Hz
Answer:
a

b

Explanation:
From the question we are told that
The wavelength of the light is 
The distance of the slit separation is 
Generally the condition for two slit interference is

Where m is the order which is given from the question as m = 2
=> ![\theta = sin ^{-1} [\frac{m \lambda}{d} ]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20sin%20%5E%7B-1%7D%20%5B%5Cfrac%7Bm%20%5Clambda%7D%7Bd%7D%20%5D)
substituting values

Now on the second question
The distance of separation of the slit is

The intensity at the the angular position in part "a" is mathematically evaluated as
![I = I_o [\frac{sin \beta}{\beta} ]^2](https://tex.z-dn.net/?f=I%20%20%3D%20%20I_o%20%20%5B%5Cfrac%7Bsin%20%5Cbeta%7D%7B%5Cbeta%7D%20%5D%5E2)
Where
is mathematically evaluated as

substituting values


So the intensity is
![I = I_o [\frac{sin (0.06581)}{0.06581} ]^2](https://tex.z-dn.net/?f=I%20%20%3D%20%20I_o%20%20%5B%5Cfrac%7Bsin%20%280.06581%29%7D%7B0.06581%7D%20%5D%5E2)

Answer
given,
Side of copper plate, L = 55 cm
Electric field, E = 82 kN/C
a) Charge density,σ = ?
using expression of charge density
σ = E x ε₀
ε₀ is Permittivity of free space = 8.85 x 10⁻¹² C²/Nm²
now,
σ = 82 x 10³ x 8.85 x 10⁻¹²
σ = 725.7 x 10⁻⁹ C/m²
σ = 725.7 nC/m²
change density on the plates are 725.7 nC/m² and -725.7 nC/m²
b) Total change on each faces
Q = σ A
Q = 725.7 x 10⁻⁹ x 0.55²
Q = 219.52 nC
Hence, charges on the faces of the plate are 219.52 nC and -219.52 nC
we have to use newtons law of gravitation which is
F=GMm/r^2
G=6.67 x 10^<span>-11N kg^2/m^2
</span>M=<span>(15kg)
</span>m=15 kg
r=(3.0m)^2<span>
</span>putting values we have
<span>=(6.67 x 10^-11N kg^2/m^2)(15kg)(15kg)/(3.0m)^2 </span>
=1.67 x 10^-9N
You'd get an extra 40/60 of the energy, or 2/3. Multiply 5/3 by the required energy to get the actual consumption.