For the answer to the question above,
we can get the number of fringes by dividing (delta t) by the period of the light (Which is λ/c).
fringe = (delta t) / (λ/c)
We can find (delta t) with the equation:
delta t = [v^2(L1+L2)]/c^3
Derivation of this formula can be found in your physics text book. From here we find (delta t):
600,000^2 x (11+11) / [(3x10^8)^3] = 2.93x10^-13
2.93x10^-13/ (589x10^-9 / 3x10^8) = 149 fringes
This answer is correct but may seem large. That is because of your point of reference with the ether which is usually at rest with respect to the sun, making v = 3km/s.
Displacement can be zero so velocity is zero but the distance won’t be zero so speed won’t be zero
Answer:
C. I think
Explanation:
C. permanent positive charges
I would assume you answer them by doing the math required.
Answer:
58515.9 m/s
Explanation:
We are given that
We have to find the speed (vf).
Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.
Using the formula
Where mass of sun=
