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poizon [28]
3 years ago
6

Firecracker A is 300 m from you. Firecracker B is 600 m from you in the same direction. You see both explode at the same time. D

efine event 1 to be "firecracker A explodes" and event 2 to be "firecracker B explodes." Does.event 1 occur before, after, or at the same time as event 2? Explain.
Physics
1 answer:
ss7ja [257]3 years ago
3 0

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

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a)

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c) 3.06 s

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The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards

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0 = 15 - 9.81 * t

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The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.

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0 = 0 + 15 * t - 1/2 * 9.81 t^2

0 = 15 * t - 4.9 * t^2

0 = t * (15 - 4.9 * t)

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