Answer:
Regions near rivers have water surfaces that rapidly change in temperature from cold to hot.
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The correct answer is: wavelength =
4562 nm
Explanation:Rydberg's formula is given as:
![\frac{1}{\lambda} = R[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ]](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%5Clambda%7D%20%3D%20R%5B%20%5Cfrac%7B1%7D%7Bn_1%5E2%7D%20%20-%20%5Cfrac%7B1%7D%7Bn_2%5E2%7D%20%5D%20)
--- (1)
Where
R = Rydberg's constant = 1.096 * 10^7 per meter

= 5

= 7
λ = Wavelength
Plug in the values in (1):
(1)=>
![\frac{1}{\lambda} = (1.096 * 10^7)[ \frac{1}{5^2} - \frac{1}{7^2} ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Clambda%7D%20%3D%20%281.096%20%2A%2010%5E7%29%5B%20%5Cfrac%7B1%7D%7B5%5E2%7D%20-%20%5Cfrac%7B1%7D%7B7%5E2%7D%20%5D)
Answer:
Liquid's index of refraction, n₁ = 1.27
Explanation:
It is given that,
The critical angle for a liquid in air is, 
We have to find the refractive index of the liquid. Critical angle of a liquid is defined as the angle of incidence in denser medium for which the angle of refraction is 90°.
Using Snell's law as :

Here, 

Where
n₂ = Refractive index of air = 1
n₁ = refractive index of liquid
So,


n₁ = 1.269
or n₁ = 1.27
Hence, the refractive index of liquid is 1.27
Well,
arctan is a bijection from R into (-pi/2 , pi/2)*and
pi is a period of tangent function:
so
as we have : tan(7pi/4) = tan(pi - pi/4) = - tan(pi/4)
we finally get :
<span>arctan(tan(7pi/4)) = artan(tan(- pi/4)) = - pi/4
</span>
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
D because choppy water is already a little bit of wave action and then the wind adds a lot of speed to the waves and height to the waves to meek the best waves. Hope this helped.