Respuesta :
Answer:
he difference between the positions of the two events as measured in S = 4.42 x 10 8 m
Explanation:
The concept of time dilation is applied to solve the question as shown in the attachment.
Two variables are observed at the very same space point in a frame of reference [tex]S'[/tex] , with the second incident happening first. Time compression among two occurrences in frames [tex]S'[/tex] is, [tex]\Delta t= 1.70\ s[/tex] Let the second frame [tex]S'[/tex] move relative to [tex]S'[/tex] with speed, v Time compression across two events for the frame [tex]S'[/tex] is, [tex]\Delta t'= 2.25\ s[/tex] the velocity of the object, c
Solution:
From the special theory of relativity, we have a formula between the time dilations as
[tex]\to \Delta t' = \frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}} \\\\\to \sqrt{1-\frac{v^2}{c^2}} =\frac{\Delta t}{\Delta t'} =\frac{1.7\ s}{2.25\ s}\\\\ V = 0.655\ c\\\\[/tex]
Hence velocity of the frame [tex]S'[/tex] is, v = 0.655 c but, from the Lorentz
transformations for positions between the frames as
[tex]\to \Delta x'= \frac{\Delta x- v \Delta t}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
but, [tex]\Delta x = 0[/tex] (from frame S')
so,
[tex]\to \Delta x'=\frac{0-(0.655c)(1.7 s)}{\aqrt{1- \frac{(0.655c)^2}{c^2}}}\\\\[/tex]
[tex]= (-1.4736\ c)\ m\\\\ =(-1.4736)(3 \times 10^8 \ \frac{m}{s}) \\\\= 4.42 \times 10^8 \ m\\\\[/tex]
In this case, the - ve sign denotes the motion of the frame [tex]S'[/tex] in the opposite direction, resulting in a difference of [tex](\Delta x')[/tex] between the positions of the two events as measured in [tex]S'[/tex] is [tex]4.42 \times 10^8 \ m\\\\[/tex].
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