Answer:
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Explanation:
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Answer:
e see that the distances are different, the only way that the two beams of light approach simultaneously is that event 2 (farthest) occurs first than event 1
Explanation:
This is an ejercise in special relativity, where the speed of light is constant.
Let's carefully analyze the approach, we see the two events at the same time.
The closest event time is
c = (x₁-300) / t
t = (x₁-300) / c
The time for the other event is
t = (x₂- 600) / c
since they tell us that we see the events simultaneously, we can equalize
(x₁ -300) / c = (x₂ -600) / c
x₁ = x₂ - 300
We see that the distances are different, the only way that the two beams of light approach simultaneously is that event 2 (farthest) occurs first than event 1
Answer: B) 2.5 m/s
Explanation: Find the average of the time and distance, and see how far they go in only 1 second.
1 + 2 + 3 + 4 + 5 = 15
15 divided by 5 = 3
3 seconds
2 + 5 + 7 + 10 + 12 = 36
36 divided by 5 = 7.2
7.2m per 3 seconds.
7.2 divided by 3 = 2.4
Therefore, the answer is technically 2.4m/s
