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 metric unit to measure work which equals one newton meter is called One Joule.
Answer:
See the answers below.
Explanation:
to solve this problem we must make a free body diagram, with the forces acting on the metal rod.
i)
The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.
We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.
For the summation of forces we will take the forces upwards as positive and the negative forces downwards.
ΣF = 0

Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.
ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.
ΣM = 0
![(15*9) - (18*W) = 0\\135 = 18*W\\W = 7.5 [N]](https://tex.z-dn.net/?f=%2815%2A9%29%20-%20%2818%2AW%29%20%3D%200%5C%5C135%20%3D%2018%2AW%5C%5CW%20%3D%207.5%20%5BN%5D)
Answer:
Explanation:
The charges will repel each other and go away with increasing velocity , their kinetic energy coming from their potential energy .
Their potential energy at distance d
= kq₁q₂ / d
= 9 x 10⁹ x 36 x 10⁻¹² / 2 x 10⁻² J
= 16.2 J
Their total kinetic energy will be equal to this potential energy.
2 x 1/2 x mv² = 16.2
= 3 x 10⁻⁶ v² = 16.2
v = 5.4 x 10⁶
v = 2.32 x 10³ m/s
When masses are different , total P.E, will be divided between them as follows
K E of 3 μ = (16.2 / 30+3) x 30
= 14.73 J
1/2 X 3 X 10⁻⁶ v₁² = 14.73
v₁ = 3.13 x 10³
K E of 30 μ = (16.2 / 30+3) x 3
= 1.47 J
1/2 x 30 x 10⁻⁶ x v₂² = 1.47
v₂ = .313 x 10³ m/s