Respuesta :
Answer:
Q1) Time taking by particle to travel the 40 km wrt. earth = [tex]1.34\times10^{-6}[/tex] sec.
Q2) The distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.
Q3) The time taking by particle to travel from where it is created to the surface of the earth = [tex]1.285\times10^{-5}[/tex] sec.
Explanation:
Given :
Speed of particle wrt. earth [tex]v=0.99537c[/tex]
Distance between where particle is created and earth surface = 40 km
we know that,
⇒ [tex]v = \frac{x}{t}[/tex]
Where [tex]x = 40\times10^{3} m[/tex], [tex]v = 0.99537c[/tex], we know speed of light [tex]c = 3 \times10^{8}[/tex]
∴ [tex]t = \frac{x}{v}[/tex]
[tex]= \frac{40 \times10^{3} }{0.99537\times3\times10^{8} }[/tex]
[tex]t = 1.34\times10^{-6} sec[/tex]
∴ Thus, time taking by particle to travel the 40 km wrt. earth [tex]t = 1.34\times10^{-6} sec[/tex]
According to the lorentz transformation,
⇒ [tex]l = l_{o} \sqrt{1-\frac{v^2}{c^2} }[/tex]
Where [tex]l =[/tex] improper length, [tex]l_{o} =[/tex]proper length (distance measured wrt. rest frame) = 40 km
[tex]l = 40 \sqrt{1-\frac{v^2}{c^2 }[/tex]
[tex]l = 40 \times 0.096[/tex]
[tex]l = 3.84[/tex] km
∴ Thus, the distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.
According to the time dilation,
[tex]\Delta t = \frac{\Delta t_{o} }{\sqrt{1-\frac{v^2}{c^2} } }[/tex]
Where [tex]\Delta t =[/tex] improper time (wrt. earth frame time) [tex]=1.34\times10^{-6} sec[/tex] , [tex]\Delta t _{o} =[/tex] proper time (wrt. particle frame).
[tex]1.34\times10^{-6} = \frac{ \Delta t_{o}}{0.096}[/tex]
[tex]\Delta t_{o} = 1.285 \times10^{-5} sec[/tex]
Thus, the time taking by particle to travel from where it is created to the surface of the earth [tex]= 1.285 \times10^{-5} sec[/tex].
