We now consider the motion of a flash of light that is emitted from the ship des

Question

We now consider the motion of a flash of light that is emitted from the ship described in Part A. Imagine that there is a flashlight on the pool table that emits a flash of light directed toward the bow of the ship, which is still traveling northward at 25.0 m/s . The light encounters two photo diodes on the table that are spaced exactly 1.00 m apart along a north-south axis. What is the time table that elapses between when the flash of light encounters the first photo diode and when it strikes the second?

Express your answer in seconds to nine significant figures.Remember that an answer such as 1/300000025 will be accepted at itscalculated value (if you do not put commas in your answer). Also,the speed of light in vacuum is c = 299,792,458 m/s

Explanation / Answer

In a particular frame of reference,suppose two events,as measured by an observer at rest in this same frame (which we call the rest frame of this observer),is ?to.Then an observer in a second rame moving with constant speed u relative to the rest frame will measure the time interval to be ?t,where ?t = (?to/(1 - u2/c2)^1/2) (time dilation) -----------(1) We recall that no inertial observer can travel at u = c and note that (1 - u2/c2)^1/2 is imaginary for u > c. The flash of light moves a total distance 2d,so the time interval is ?to = (2d/c) where d = 1.00 m and c = 299,792,458 m/s is the speed of light in vacuum The bow of the ship is travelling northward at a speed u = 25 m/s Substituting the above values in equation (1),we get the the time table that elapses between when the flash of light encounters the first photo diode and when it strikes the second.

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