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.
Answer:
The weight of the landing craft in the vicinity of Callisto's surface is 3480 N.
Explanation:
The engine of the craft provides an upward thrust of 
so that the space craft descends at a constant speed.
This implies that the net force on the space craft is zero.
The upward thrust will be equal to the downward gravitational pull by Callisto.
So the weight of the craft near the vicinity will be 3480 N.
Aspirate or inhale or respire or
Answer:
H = start height (v = 0)
h = present height
v = present speed
assuming no friction
total energy = PE + KE
mgH = mgh + .5mv^2
if PE = KE then
mgH = mgh + mgh
h = H/2
potential energy = kinetic energy when object is at half its start height.
Explanation:
