Answer:
the wavelength λ of the light when it is traveling in air = 560 nm
the smallest thickness t of the air film = 140 nm
Explanation:
From the question; the path difference is Δx = 2t (since the condition of the phase difference in the maxima and minima gets interchanged)
Now for constructive interference;
Δx= 
replacing ;
Δx = 2t ; we have:
2t =
Given that thickness t = 700 nm
Then
2× 700 = --- equation (1)
For thickness t = 980 nm that is next to constructive interference
2× 980 = ----- equation (2)
Equating the difference of equation (2) and equation (1); we have:'
λ = (2 × 980) - ( 2× 700 )
λ = 1960 - 1400
λ = 560 nm
Thus; the wavelength λ of the light when it is traveling in air = 560 nm
b)
For the smallest thickness 
∴ 
Thus, the smallest thickness t of the air film = 140 nm
Answer:
The graphs are attached
Explanation:
We are told that he starts with a constant speed of 25 m/s for a distance of 100 m.
At constant velocity, v = distance/time
time(t) = distance(d)/velocity(v)
t1 = 100/25
t1 = 4 s
Now, we are told that he applies his brakes and accelerates uniformly to a stop just as he reaches a wall 50m away.
It means, he decelerate and final velocity is zero.
Thus;
v² = u² + 2as
0² = 25² + 2a(50)
25² = - 100a
625 = - 100a
a = - 625/100
a = - 6.25 m/s²
v = u + at
0 = 25 + (-6.25t)
25 = 6.25t
t = 25/6.25
t = 4 s
With the values gotten, kindly find attached the distance-time and velocity-time graphs.
Answer:
specific purpose statement
Explanation:
It is a specific purpose statement made for a persuasive speech on the question of fact.
A specific purpose statement helps to build on the general purpose (that is to inform) and to make it more specific to the audience. So if the first speech is an informative speech, our general purpose is to inform our audience about a very specific realm of knowledge.
A specific purpose statement is given to audience to persuade on specific information.
Answer:
<em>0.45 mm</em>
Explanation:
The complete question is
a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?
A) 0.45 mm
B) 0.63 mm
C.) 0.68 mm
D) 0.91 mm
Current in the fuse is 1.0 A
Current density of the fuse when it melts is 620 A/cm^2
Area of the wire in the fuse = I/ρ
Where I is the current through the fuse
ρ is the current density of the fuse
Area = 1/620 = 1.613 x 10^-3 cm^2
We know that 10000 cm^2 = 1 m^2, therefore,
1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2
Recall that this area of this wire is gotten as
A = 
where d is the diameter of the wire
1.613 x 10^-7 = 
6.448 x 10^-7 = 3.142 x 
=
d = 4.5 x 10^-4 m = <em>0.45 mm</em>
Calculate q* E * d
<span>Put q = 1.6 x 10^-19 </span>
<span>E = 325 </span>
<span>d = 4.5
I hope this helps!</span>
