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
Impulse = Integral of F(t) dt from 0.012s to 0.062 s
Given that you do not know the function F(t) you have to make an approximation.
The integral is the area under the curve.
The problem suggest you to approximate the area to a triangle.
In this triangle the base is the time: 0.062 s - 0.012 s = 0.050 s
The height is the peak force: 35 N.
Then, the area is [1/2] (0.05s) (35N) = 0.875 N*s
Answer> 0.875 N*s
Answer:
Tension in the cable is T = 16653.32 N
Explanation:
Give data:
Cross section Area A = 1.3 m^2
Drag coefficient CD = 1.2
Velocity V = 4.3 m/s
Angle made by cable with horizontal =30 degree
Density 
Drag force FD is given as
Drag force = 14422.2 N acting opposite to the motion
As cable made angle of 30 degree with horizontal thus horizontal component is take into action to calculate drag force
TCos30 = F_D
T = 16653.32 N
Answer:
0.12 K
Explanation:
height, h = 51 m
let the mass of water is m.
Specific heat of water, c = 4190 J/kg K
According to the transformation of energy
Potential energy of water = thermal energy of water
m x g x h = m x c x ΔT
Where, ΔT is the rise in temperature
g x h = c x ΔT
9.8 x 51 = 4190 x ΔT
ΔT = 0.12 K
Thus, the rise in temperature is 0.12 K.
Answer:
v_average = 500 m / min
Explanation:
Average speed is defined
v = (x_{f} -x₀) / Δt
let's look in each section
section 1
the variation of the distance is 800 in a time of 1.4 min
v₁ = 800 / 1.4
v₁ = 571.4 m / min
section 2
distance interval 500 in a 1.6 min time interval
v₂ = 500 / 1.6
v₂ = 312.5 m / min
section 3
distance interval 1200 m in a time 2 min
v₃ = 1200/2
v₃ = 600 m / min
taking the speed of each section we can calculate the average speed
the distance traveled
Δx = 800 + 500 + 1200
Δx = 2500 m
the time spent
Δt = 1.4 + 1.6+ 2
Δt = 5 min
v_average = Δx / Δt
v_average = 2500/5
v_average = 500 m / min
