Let us start from considering monochromatic light as an incidence on the film of a thickness t whose material has an index of refraction n determined by their respective properties.
From this point of view part of the light will be reflated and the other will be transmitted to the thin film. That additional distance traveled by the ray that was reflected from the bottom will be twice the thickness of the thin film at the point where the light strikes. Therefore, this relation of phase differences and additional distance can be expressed mathematically as

We are given the second smallest nonzero thickness at which destructive interference occurs.
This corresponds to, m = 2, therefore


The index of refraction of soap is given, then

Combining the results of all steps we get

Rearranging, we find



The gravitational force is inversely proportional to the
square of the distance between their centers. So the
force is greatest when the distance is zero.
To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

Where,
F = Force
r = Radius
Replacing we have that,



The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore


Finally, angular acceleration is a result of the expression of torque by inertia, therefore



PART B)
The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians
, therefore



Answer:
t = 0.657 s
Explanation:
First, let's use the appropiate equations to solve this:
V = √T/u
This expression gives us a relation between speed of a disturbance and the properties of the material, in this case, the rope.
Where:
V: Speed of the disturbance
T: Tension of the rope
u: linear density of the rope.
The density of the rope can be calculated using the following expression:
u = M/L
Where:
M: mass of the rope
L: Length of the rope.
We already have the mass and length, which is the distance of the rope with the supports. Replacing the data we have:
u = 2.31 / 10.4 = 0.222 kg/m
Now, replacing in the first equation:
V = √55.7/0.222 = √250.9
V = 15.84 m/s
Finally the time can be calculated with the following expression:
V = L/t ----> t = L/V
Replacing:
t = 10.4 / 15.84
t = 0.657 s