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,

You have to use the specific heat equation.
Q = cmΔT where Q is the energy, c is specific heat, m is mass, and ΔT is change in temp.
So we can substitute our variables into the equation.
30000J = (390g)(3.9J*g/C)ΔT
Solving for ΔT, we get:
30000J/[(390g)*(3.9J*g/C) = ΔT
ΔT = 19.72386588C
I'm assuming the temperature is C, since it was not specified.
Hope this helps!
Answer:
yes it flows through flow paths.
Explanation:
Answer:
The ball will have a kinetic energy of 0.615 Joules.
Explanation:
Use the kinetic energy formula

The kinetic energy at the moment of leaving the hand will be 0.615 Joules. (From there on, as it ball is traveling upwards, this energy will be gradually traded off with potential energy until the ball's velocity becomes zero at the apex of the flight)
I think it’s C. Stroke if not then D