Atmospheric refraction is the deviation of light or other electromagnetic wave from a straight line as it passes through the atmosphere due to the variation in air density as a function of height. ... Refraction not only affects visible light rays, but all electromagnetic radiation, although in varying degrees.
So in short, the answer is D.
(My answer got deleted because it didnt explain which is dumb)
The answer is B tell me if I am wrong.
Let us consider two bodies having masses m and m' respectively.
Let they are separated by a distance of r from each other.
As per the Newtons law of gravitation ,the gravitational force between two bodies is given as -
where G is the gravitational force constant.
From the above we see that F ∝ mm' and 
Let the orbital radius of planet A is
= r and mass of planet is
.
Let the mass of central star is m .
Hence the gravitational force for planet A is 
For planet B the orbital radius
and mass
Hence the gravitational force 
![f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }](https://tex.z-dn.net/?f=f_%7B2%7D%20%3DG%5Cfrac%7Bm%2A3m_%7B1%7D%20%7D%7B%5B2r_%7B1%7D%5D%20%5E%7B2%7D%20%7D)

Hence the ratio is 
[ ans]
Answer:
6.88 mA
Explanation:
Given:
Resistance, R = 594 Ω
Capacitance = 1.3 μF
emf, V = 6.53 V
Time, t = 1 time constant
Now,
The initial current, I₀ = 
or
I₀ = 
or
I₀ = 0.0109 A
also,
I = ![I_0[1-e^{-\frac{t}{\tau}}]](https://tex.z-dn.net/?f=I_0%5B1-e%5E%7B-%5Cfrac%7Bt%7D%7B%5Ctau%7D%7D%5D)
here,
τ = time constant
e = 2.717
on substituting the respective values, we get
I = ![0.0109[1-e^{-\frac{\tau}{\tau}}]](https://tex.z-dn.net/?f=0.0109%5B1-e%5E%7B-%5Cfrac%7B%5Ctau%7D%7B%5Ctau%7D%7D%5D)
or
I =
or
I = 0.00688 A
or
I = 6.88 mA
Answer:
1.63366
Explanation:
I got this answer from calculator soups physics calculators. I really recommend their website for formulas.