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Spacecraft A and B pass in opposite directions at a relative speed of 4.00×10^7m/s. An internal clock in spacecraft A causes it to emit a radio signal for 1.00 s. The computer in spacecraft B corrects for the beginning and end of the signal having traveled different distances, to calculate the time interval during which ship A was emitting the signal. What is the time interval that the computer in spacecraft B calculates?​

Sagot :

TNQRidn

Answer:

The analysis of simultaneity shows that Einstein’s postulates imply an important effect: Time intervals have different values when measured in different inertial frames. Suppose, for example, an astronaut measures the time it takes for a pulse of light to travel a distance perpendicular to the direction of his ship’s motion (relative to an earthbound observer), bounce off a mirror, and return (Figure  5.4.1a5.4.1a ). How does the elapsed time that the astronaut measures in the spacecraft compare with the elapsed time that an earthbound observer measures by observing what is happening in the spacecraft?

Examining this question leads to a profound result. The elapsed time for a process depends on which observer is measuring it. In this case, the time measured by the astronaut (within the spaceship where the astronaut is at rest) is smaller than the time measured by the earthbound observer (to whom the astronaut is moving). The time elapsed for the same process is different for the observers, because the distance the light pulse travels in the astronaut’s frame is smaller than in the earthbound frame, as seen in Figure  5.4.1b5.4.1b . Light travels at the same speed in each frame, so it takes more time to travel the greater distance in the earthbound frame.

Explanation:

NOT SURE

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