A spaceship carrying a light clock moves at a speed of 0.960c relative to an observer on Earth. If the clock on the ship advances by 1.00 s

Question

A spaceship carrying a light clock moves at a speed of 0.960c relative to an observer on Earth. If the clock on the ship advances by 1.00 s as measured by the space travelers aboard the ship, how long did that advance take as measured by the observer on Earth?

A) 0.96 s
B) 1.2 s
C) 2.6 s
D) 3.6 s
E) 5.8 s

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Ngọc Hoa 4 years 2021-07-29T02:36:48+00:00 1 Answers 51 views 0

Answers ( )

  1. diemthu
    0
    2021-07-29T02:37:51+00:00
    Reply

    Answer:

    (A) The time taken as measured by the observer on Earth is 0.96 s.

    Explanation:

    Given;

    relative speed of spaceship to an observer on Earth, v = 0.960c

    (\frac{speed \ of \ spaceship}{speed \ of \ Earth}=\frac{V}{C} = 0.96 )  

    time of the spaceship as it advances, t = 1.00 s

    time of advance observed on Earth, t = ?

    Speed = distance / time

    Distance = speed x time

    Assuming constant distance in the spaceship advancement;

    V_{space}*t_{space} = V_{Earth}*t_{Earth}\\\\\frac{V_{space}}{ V_{Earth}} = \frac{t_{Earth}}{t_{space}}\\\\ 0.96 = \frac{t_{Earth}}{1}\\\\t_{Earth} = 0.96 \ s

    Therefore, the time taken as measured by the observer on Earth is 0.96 s.

    The correct option is “A” 0.96 s

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