I think it’s 10 kg since 20/2=10
The condition of earths atmosphere at a given time and place is the weather.
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Complete Question
Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute).Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light
Answer:
The diameter is
Explanation:
From the question we are told that
The best resolution is ![\theta = 0.3 \ arcsecond](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%200.3%20%5C%20%20arcsecond)
The wavelength is ![\lambda = 700 \ nm = 700 *10^{-9 } \ m](https://tex.z-dn.net/?f=%5Clambda%20%20%3D%20%20700%20%5C%20%20nm%20%3D%20%20700%20%2A10%5E%7B-9%20%7D%20%5C%20%20m)
Generally the
1 arcminute = > 60 arcseconds
=> x arcminute => 0.3 arcsecond
So
![x = \frac{0.3}{60 }](https://tex.z-dn.net/?f=x%20%3D%20%20%5Cfrac%7B0.3%7D%7B60%20%7D)
=> ![x = 0.005 \ arcminutes](https://tex.z-dn.net/?f=x%20%3D%200.005%20%5C%20%20arcminutes)
Now
60 arcminutes => 1 degree
0.005 arcminutes = > z degrees
=> ![z = \frac{0.005}{60 }](https://tex.z-dn.net/?f=z%20%3D%20%20%5Cfrac%7B0.005%7D%7B60%20%7D)
=> ![z = 8.333 *10^{-5} \ degree](https://tex.z-dn.net/?f=z%20%3D%20%208.333%20%2A10%5E%7B-5%7D%20%20%5C%20degree)
Converting to radian
![\theta = z = 8.333 *10^{-5} * 0.01745 = 1.454 *10^{-6} \ radian](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20z%20%3D%20%208.333%20%2A10%5E%7B-5%7D%20%20%2A%200.01745%20%3D%201.454%20%2A10%5E%7B-6%7D%20%5C%20%20radian)
Generally the resolution is mathematically represented as
![\theta = \frac{1.22 * \lambda }{ D}](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20%5Cfrac%7B1.22%20%2A%20%20%5Clambda%20%20%7D%7B%20D%7D)
=> ![D = \frac{1.22 * \lambda }{\theta }](https://tex.z-dn.net/?f=D%20%3D%20%20%5Cfrac%7B1.22%20%2A%20%5Clambda%20%7D%7B%5Ctheta%20%7D)
=>
=>
Perpendicular slope would be 1/3. so the equation will be Y=1/3x -4