Explanation:
When Michelson-Morley apparatus is turned through 
then position of two mirrors will be changed. The resultant path difference will be as follows.
Formula for change in fringe shift is as follows.
n = 
v = 
According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.
l = 11 m
c = 
m/s
Hence, putting the given values into the above formula as follows.
v =
= 
= 
Thus, we can conclude that velocity deduced is .
<span>At time t1 = 0 since the body is at rest, the body has an angular velocity, v1, of 0. At time t = X, the body has an angular velocity of 1.43rad/s2. Since Angular acceleration is just the difference in angular speed by time. We have 4.44 = v2 -v1/t2 -t1 where V and t are angular velocity and time. So we have 4.44 = 1.43 -0/X - 0. Hence X = 1.43/4.44 = 0.33s.</span>
Answer:
the correct answer is c v₁> 12.5 m / s
Explanation:
This is a one-dimensional kinematics exercise, let's start by finding the link to get up to speed.
v² = v₀² + 2 a₁ x
as part of rest v₀ = 0
a₁ = v² / 2x
a₁ = 25² / (2 120)
a₁ = 2.6 m / s²
now we can find the velocity for the distance x₂ = 60 m
v₁² = 0 + 2 a1 x₂
v₁ = Ra (2 2,6 60)
v₁ = 17.7 m / s
these the speed at 60 m
we see that the correct answer is c v₁> 12.5 m / s
The labeled points which is Letter B in the given Image is the point that the axis of rotation passes through. This problem is an example of rotational dynamics, formerly an object moves in a straight line then the motion is translational but when an object at inactivity lean towards to continue at inactivity and an object in rotation be possible to continue rotating with continuous angular velocity unless bound by a net external torque to act then is rotational. In a rotational motion, the entity is not treated as a constituent part but is treated in translational motion. It points out with the study of torque that outcomes angular accelerations of the object.
