Answer with Explanation:
We are given that
Angle of incidence,
Angle of refraction,
a.Refractive index of air,
We know that


b.Wavelength of red light in vacuum,

Wavelength in the solution,

c.Frequency does not change .It remains same in vacuum and solution.
Frequency,
Where 
Frequency,
d.Speed in the solution,

The magnitude and direction of the electric field in the wire are mathematically given as
![L &=[(v / L) v / m] \hat{i}](https://tex.z-dn.net/?f=L%20%26%3D%5B%28v%20%2F%20L%29%20v%20%2F%20m%5D%20%5Chat%7Bi%7D)
<h3>What is the magnitude and direction of the electric field in the wire?</h3>
Generally, the equation for is mathematically given as
A cylindrical wire that is straight and parallel to the x-axis has the following dimensions: length L, diameter d, resistivity p, diameter d, potential v, and z length. combining elements from both sides
E d 
![\begin{aligned}&-E \int_0^L d x=\int_v^0 d v \\\therefore E \cdot L &=v \\L &=[(v / L) v / m] \hat{i}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26-E%20%5Cint_0%5EL%20d%20x%3D%5Cint_v%5E0%20d%20v%20%5C%5C%5Ctherefore%20E%20%5Ccdot%20L%20%26%3Dv%20%5C%5CL%20%26%3D%5B%28v%20%2F%20L%29%20v%20%2F%20m%5D%20%5Chat%7Bi%7D%5Cend%7Baligned%7D)
In conclusion, the magnitude and direction of the electric field in the wire are given as
![L &=[(v / L) v / m]](https://tex.z-dn.net/?f=L%20%26%3D%5B%28v%20%2F%20L%29%20v%20%2F%20m%5D)
Read more about electric fields
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When an elevator is accelerating downward, the normal force is equal to mg-ma (hence you feel a little lighter when accelerating downwards)
Therefore, the upward force of the elevator floor on the person must be less than 750N
Answer: I would say the answer is B.
Explanation: It looks and sounds the most correct.
Answer:Atomic model keep changing because the electrons around the nucleus are not fixed and they keeps rotating or changing their position in valence orbits around the nucleus.
Explanation: