Answer:
Hi there, this image is your answer
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



Answer:
a)
125.6 rad/s
b)
25.12 rad/s²
Explanation:
a)
t = time required by the fan to get up to final operating speed = 5 sec
w = final operating rotational speed = 1200 rpm
we know that :
1 revolution = 2π rad
1 min = 60 sec
w = 
w = 
w = 125.6 rad/s
b)
w₀ = initial angular speed = 0 rad/s
α = angular acceleration
using the equation
w = w₀ + α t
125.6 = 0 + α (5)
α = 25.12 rad/s²
Answer:
1.4E-3J
Explanation:
Given that
Time = 8hrs = 28.8E3 seconds
Intensity= 90dB
D= 0.008m
Radius= 0.004m
So intensity is sound level Bis
10dBlog(I/Io)
I= 10 (B/10dB)Io
= 10( 90/10) x 10^-12
=0.001W/m²
But we know that
I = P/A
P= I πr²
= 5.02 x10^-8W
But energy is power x time
So E= 5.02E-8 x 28.8E3
= 1.4E-3J
Something is reproducing.