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iragen [17]
2 years ago
11

An astronaut travels to a star system 3.9 lyly away at a speed of 0.90 cc . Assume that the time needed to accelerate and decele

rate is negligible.
Physics
1 answer:
kakasveta [241]2 years ago
4 0

Total time elapsed is =8.2y

The starting event is the astronaut leaving Earth. The finishing event is the astronaut arriving at the star system. The time between these events on Earth is:

Δt=3.9ly/0.9c

Δt=4.3y

For the astronaut, two events occur at the same position and can be measured with just one clock. Hence,

Δτ

=  \sqrt{1 -  \frac{v {}^{2} }{c {}^{2} }  }  \times Δt

Δτ

=  \sqrt{1 - ( \frac{0.9c}{c} ) {}^{2} }  \times (4.3y)

Δτ=1.8ly

The total elapsed time is:

T elapsed=Δt+3.9

T elapsed=4.3+3.9

T elapsed=8.2y

learn more about time from here: brainly.com/question/28208983

#SPJ4

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Explanation:

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I hope the answer is helpful.

Thanks for asking.

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