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LenaWriter [7]
3 years ago
10

Two helium-filled balloons are released simultaneously at points A and B on the x axis in an earth-based reference frame. Balloo

n A is to the left of balloon B. Which one of the following statements is true for an observer moving in the +x direction?
A. The observer sees balloon B released before balloon A.
B. The observer always sees the balloons released simultaneously.
C. The observer could see either balloon released first depending on her speed and the distance between A and B.
D. The observer sees balloon A released before balloon B.
Physics
1 answer:
Vedmedyk [2.9K]3 years ago
5 0

Answer:

The correct answer is C

Explanation:

In this exercise you are asked to analyze the following situation: for a fixed observer, the two balloons are launched at the same time.

If the observer is mobile moving towards the positive side of axis ax we have several

possibilities when starting the movement

* the observer is to the left of the two balloons

* the observer is between the two balloons

* the observer is to the right of the two baloomls

The time it takes for the signal to arrive to know which ball goes first is

             v = d / t

             t = d / v

If the signal goes at the speed of light, the speed is a constant and the time will depend only on the distance, so we see that the trigger changes depending on the relative position between a given ball and the observer.

Consequently, it will be seen which comes out first, depending on the relative position with the observer.

           

The correct answer is C

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Answer:

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Here, the sound covered a total of 340 meters, since it bounces off the canyon wall and travels back to Miles in the form of echo.

The time elapsed can be determined by,

v = \frac{2D}{t}

⇒ t = \frac{2D}{v}

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So that,

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t = \frac{340}{343}

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t = 1 s

It takes 1s for Miles to hear his echo.

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Solution:

As per the question:

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D = 1,00,000 ly

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Now,

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The area to each civilization, A' is:

A' = \frac{A}{10000} = \frac{7.8539\times 10^{9}}{10000} = 7.8539\times 10^{5}\ ly^{2}

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r = \sqrt{\frac{A'}{\pi}}

r = \sqrt{\frac{7.8539\times 10^{5}}{\pi}} = 499.968\ ly

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A vehicle of mass 1600kg take off from rest and coveres a distance of 280m in 60s. It continues at this speed for 98seconds and
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<span>2. V = Vo + a*t. </span>
<span>0 = 15 - 5*t, t = 3 s. </span>

<span>3. d1 = Vo*t + 0.5a*t^2. </span>
<span>Vo = 0, t = 5, a = 3 m/s^2, d1 = ?. </span>

<span>d2 = Vo*t. Vo = 15 m/s, t = 15 s, </span>
<span>d2 = ?. </span>

<span>V^2 = Vo^2 + 2a*d3. </span>
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NNADVOKAT [17]

Answer:

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