Two events are observed in a frame of reference S to occur at the same space point, the second occurring 1.80 s after the first. In a frame

Question

Two events are observed in a frame of reference S to occur at the same space point, the second occurring 1.80 s after the first. In a frame S′ moving relative to S, the second event is observed to occur 2.05 s after the first. What is the difference between the positions of the two events as measured in S^?

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Jezebel 6 months 2021-07-29T02:16:06+00:00 1 Answers 95 views 0

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    2021-07-29T02:18:02+00:00

    Answer:

    The difference between the positions of the two events as measured in = 3.53 *10^8 m/s

    Explanation:

    As we know –

    \Delta x = -\gamma \mu\Delta t

    Here,

    \Delta x is the difference between the positions of the two events as measured in S^

    \gamma = \frac{1}{\sqrt{1-\frac{\mu^2}{c^2} } }

    And

    \mu = 0.547 c

    Substituting the given values in above equation, we get –

    \Delta x = (0.547 c)*\frac{1}{\sqrt{1-\frac{\mu^2}{c^2} } }*2.15\\\Delta x = (0.547 c)*\frac{1}{\sqrt{1-\frac{(0.547 c)^2}{c^2} } }*2.15\\\Delta x = (0.547 *3*10^8)*\frac{1}{\sqrt{(1-\(0.547 )^2 } }*2.15\\\Delta x = 3.53 *10^8meter per second

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