The railroad tracks will expand because the heat waves make them bigger
The question for this problem would be the minimum headphone delay, in ms, that will cancel this noise.
The 200 Hz. period = (1/200) = 0.005 sec. It will need to be delayed by 1/2, so 0.005/2, that is = 0.0025 sec. So converting sec to ms, will give us the delay of:Delay = 2.5 ms.
Answer:D
Explanation:
It was right o khan academy
Answer:
820 nm
Explanation:
We are given that
Wavelength=![\lambda=410 nm](https://tex.z-dn.net/?f=%5Clambda%3D410%20nm)
![\lambda=410\times 10^{-9} m](https://tex.z-dn.net/?f=%5Clambda%3D410%5Ctimes%2010%5E%7B-9%7D%20m)
![1nm=10^{-9} m](https://tex.z-dn.net/?f=1nm%3D10%5E%7B-9%7D%20m)
![\theta=14.5^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D14.5%5E%7B%5Ccirc%7D)
For first minimum therefore
m=0
We know that for destructive interference
![(m+\frac{1}{2})\lambda=dsin\theta](https://tex.z-dn.net/?f=%28m%2B%5Cfrac%7B1%7D%7B2%7D%29%5Clambda%3Ddsin%5Ctheta)
Substitute the values
![(0+\frac{1}{2})\times 410\times 10^{-9}=dsin 14.5](https://tex.z-dn.net/?f=%280%2B%5Cfrac%7B1%7D%7B2%7D%29%5Ctimes%20410%5Ctimes%2010%5E%7B-9%7D%3Ddsin%2014.5)
![d=\frac{410\times 10^{-9}}{2\times sin 14.5}](https://tex.z-dn.net/?f=d%3D%5Cfrac%7B410%5Ctimes%2010%5E%7B-9%7D%7D%7B2%5Ctimes%20sin%2014.5%7D)
![d=820\times 10^{-9} m=820 nm](https://tex.z-dn.net/?f=d%3D820%5Ctimes%2010%5E%7B-9%7D%20m%3D820%20nm)
Hence, the distance between two slits that produces the first minimum=820 nm