Answer:
The length of the beam increasing is 9.64 ft/s.
Explanation:
Given that,
Height = 210 ft
Distance =290 ft
According to figure,
We need to calculate the angle
....(I)
Put the value of x in the equation
![\cos\theta=\dfrac{210}{290}](https://tex.z-dn.net/?f=%5Ccos%5Ctheta%3D%5Cdfrac%7B210%7D%7B290%7D)
![\cos\theta=\dfrac{21}{29}=0.72](https://tex.z-dn.net/?f=%5Ccos%5Ctheta%3D%5Cdfrac%7B21%7D%7B29%7D%3D0.72)
Now, ![\sin\theta=\dfrac{20}{29}](https://tex.z-dn.net/?f=%5Csin%5Ctheta%3D%5Cdfrac%7B20%7D%7B29%7D)
On differentiate of equation (I)
![-\sin\theta\dfrac{d\theta}{dt}=-\dfrac{-210}{x^2}\dfrac{dx}{dt}](https://tex.z-dn.net/?f=-%5Csin%5Ctheta%5Cdfrac%7Bd%5Ctheta%7D%7Bdt%7D%3D-%5Cdfrac%7B-210%7D%7Bx%5E2%7D%5Cdfrac%7Bdx%7D%7Bdt%7D)
![\sin\theta=\dfrac{210}{x^2}\dfrac{dx}{dt}](https://tex.z-dn.net/?f=%5Csin%5Ctheta%3D%5Cdfrac%7B210%7D%7Bx%5E2%7D%5Cdfrac%7Bdx%7D%7Bdt%7D)
Put the value in the equation
![\sin\dfrac{20}{29}\times2.0=\dfrac{210}{(290)^2}\dfrac{dx}{dt}](https://tex.z-dn.net/?f=%5Csin%5Cdfrac%7B20%7D%7B29%7D%5Ctimes2.0%3D%5Cdfrac%7B210%7D%7B%28290%29%5E2%7D%5Cdfrac%7Bdx%7D%7Bdt%7D)
![\dfrac{dx}{dt}=\sin\dfrac{20}{29}\times2.0\times\dfrac{290^2}{210}](https://tex.z-dn.net/?f=%5Cdfrac%7Bdx%7D%7Bdt%7D%3D%5Csin%5Cdfrac%7B20%7D%7B29%7D%5Ctimes2.0%5Ctimes%5Cdfrac%7B290%5E2%7D%7B210%7D)
![\dfrac{dx}{dt}=9.64\ ft/s](https://tex.z-dn.net/?f=%5Cdfrac%7Bdx%7D%7Bdt%7D%3D9.64%5C%20ft%2Fs)
Hence, The length of the beam increasing is 9.64 ft/s.
I would say its the last answer
C the correct but not sure?
Answer:
The width of the slit is 0.4 mm (0.00040 m).
Explanation:
From the Young's interference expression, we have;
(λ ÷ d) = (Δy ÷ D)
where λ is the wavelength of the light, D is the distance of the slit to the screen, d is the width of slit and Δy is the fringe separation.
Thus,
d = (Dλ) ÷ Δy
D = 3.30 m, Δy = 4.7 mm (0.0047 m) and λ = 563 nm (563 ×
m)
d = (3.30 × 563 ×
) ÷ (0.0047)
= 1.8579 ×
÷ 0.0047
= 0.0003951 m
d = 0.00040 m
The width of the slit is 0.4 mm (0.00040 m).