Answer:
a
b
Explanation:
From the question we are told that
The wavelength of the light is 
The distance of the slit separation is 
Generally the condition for two slit interference is
Where m is the order which is given from the question as m = 2
=> ![\theta = sin ^{-1} [\frac{m \lambda}{d} ]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20sin%20%5E%7B-1%7D%20%5B%5Cfrac%7Bm%20%5Clambda%7D%7Bd%7D%20%5D)
substituting values
Now on the second question
The distance of separation of the slit is
The intensity at the the angular position in part "a" is mathematically evaluated as
![I = I_o [\frac{sin \beta}{\beta} ]^2](https://tex.z-dn.net/?f=I%20%20%3D%20%20I_o%20%20%5B%5Cfrac%7Bsin%20%5Cbeta%7D%7B%5Cbeta%7D%20%5D%5E2)
Where 
is mathematically evaluated as
substituting values
So the intensity is
![I = I_o [\frac{sin (0.06581)}{0.06581} ]^2](https://tex.z-dn.net/?f=I%20%20%3D%20%20I_o%20%20%5B%5Cfrac%7Bsin%20%280.06581%29%7D%7B0.06581%7D%20%5D%5E2)
Well, almost any website can be, but if you are meaning for money, then google. if you are meaning reliable, then look for something with .org, its history and stuff
Answer:
Please see below as the answer is self-explanatory.
Explanation:
The low band of the VHF TV Spectrum, spans channels 2-6, from 54 to 88 Mhz.
In the analog TV, in the Americas, the total bandwidth of any channel is 6 Mhz, with the visual carrier modulated in VSS (Vestigial Side Band) at 1.25 Mhz from the lowest frequency of the channel.
The aural carrier is located at 4.5 Mhz from the visual carrier, and is FM modulated.
For Channel 6, which spans between 82 and 88 Mhz, the visual carrier is at 83.25 Mhz, so the aural carrier is at 87.75 Mhz, which falls within the FM Band, so it is possible to listen the audio part of this channel in a FM radio receiver, even at a lower volume, due to the FM radio has a greater deviation than TV aural carrier.
Answer:
P= 454.11 N
Explanation:
Since P is the only horizontal force acting on the system, it can be defined as the product of the acceleration by the total mass of the system (both cubes).
The friction force between both cubes (F) is defined as the normal force acting on the smaller cube multiplied by the coefficient of static friction. Since both cubes are subject to the same acceleration:
In order for the small cube to not slide down, the friction force must equal the weight of the small cube:
The smallest magnitude that P can have in order to keep the small cube from sliding downward is 454.11 N
