A spacecraft is moving past the earth at a constant speed of 0.60 times the speed of light. The astronaut measures the time interval between

Question

A spacecraft is moving past the earth at a constant speed of 0.60 times the speed of light. The astronaut measures the time interval between ticks of the spacecraft clock to be 3.2 s. What is the time interval (in seconds) that an earth observer measures

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Ngọc Hoa 4 years 2021-08-01T04:49:11+00:00 1 Answers 9 views 0

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    2021-08-01T04:50:30+00:00

    Answer:

    the time interval that an earth observer measures is 4 seconds

    Explanation:

    Given the data in the question;

    speed of the spacecraft as it moves past the is 0.6 times the speed of light

    we know that speed of light c = 3 × 10⁸ m/s

    so speed of spacecraft v = 0.6 × c = 0.6c

    time interval between ticks of the spacecraft clock Δt₀ = 3.2 seconds

    Now, from time dilation;

    t = Δt₀ / √( 1 – ( v² / c² ) )

    t = Δt₀ / √( 1 – ( v/c )² )

    we substitute

    t = 3.2 / √( 1 – ( 0.6c / c )² )

    t = 3.2 / √( 1 – ( 0.6 )² )

    t = 3.2 / √( 1 – 0.36 )

    t = 3.2 / √0.64

    t = 3.2 / 0.8

    t = 4 seconds

    Therefore, the time interval that an earth observer measures is 4 seconds

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