Special Relativity: Simultineity question

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The discussion focuses on a problem involving special relativity and simultaneity, specifically calculating the distance between two explosions occurring on a moving rod. The rod, at rest, measures 10 meters but appears contracted to approximately 7.1 meters in the O-frame due to its velocity of 0.70c. Using Lorentz transformations, the correct distance between the explosions in the O-frame is determined to be approximately 14 meters, assuming the explosions occur at opposite ends of the rod.

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A rod whose length at rest is 10 meters is moving to the right along the positive x-axis of the O-frame with velocity V = 0.70c. An observer riding on the rod (i.e. in the O'-frame) sets off two explosions simultaneously - so he says. If the rod is parallel to the x-axis how far apart are the two explosions (in meters) in the O-frame?

I found the length the rod appears in the O-frame (about 7.1m) but am not really sure how to answer the question if I am not given how far in meters apart the explosions happen in the O'-frame. Can someone explain? thanks.
 
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This does seem to be a misprint, you should have been given the locations of the explosions in *some* frame. If the explosions were done at opposite ends of the rod, you are on the right track to use Lorentz transformations to find the new x's. You might seek clarification of the information given in the problem.
 
Ok assuming the explosions happened at opposite ends of the rod..

x' = (x-vt) / sqrt(1-v^2/c^2) so...

x'2 - x'1 = ((x2-vt) - (x1-vt))/sqrt(1-v^2/c^2)

where x2 and x1 are the points at which the explosions occur in the O-frame and x'2 and x'1 are their O'-frame counterparts so picking appropriate x2 and x1..

x2 = 10
x1 = 0

x'2 - x'1 = (10-0) / sqrt(1-(.7c)^2/c^2)
= 10/sqrt(0.51)
= 14m approximately.

Is this correct? Edit I have 10 submissions but I already used 4 so I am asking ahead of time so I Don't waste them unnecessarily.

NEvermind I got it. Silly me it was 14.
 
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