Respuesta :
Explanation:
Given that,
Speed = 0.994c
Distance [tex]d =1.45\times10^{-3}\ m[/tex]
(a). We need to calculate the proper distance
The proper distance is same as measured by the observer at ground.
The proper distance is [tex]1.45\times10^{-3}\ m[/tex]
(b). We need to calculate the distance measured by a hypothetical person traveling with the particle
Using formula of distance
[tex]d'=d\sqrt{1-\dfrac{v^2}{c^2}}[/tex]
Put the value into the formula
[tex]d'=1.45\times10^{-3}\sqrt{1-\dfrac{(0.994c)^2}{c^2}}[/tex]
[tex]d'=1.45\times10^{-3}\sqrt{1-(0.994)^2}[/tex]
[tex]d'=0.0001586 = 1.586\times10^{-4}\ m[/tex]
The distance measured by a hypothetical person traveling with the particle is [tex]1.586\times10^{-4}\ m[/tex].
(c). We need to calculate the proper lifetime
Using formula of time
[tex]v = \dfrac{d}{t}[/tex]
[tex]0.994\times3\times10^{8}=\dfrac{1.586\times10^{-4}}{t}[/tex]
[tex]t=\dfrac{1.45\times10^{-3}}{0.994\times3\times10^{8}}[/tex]
[tex]t=4.86\times10^{-12}\ sec[/tex]
The proper lifetime is [tex]4.86\times10^{-12}\ sec[/tex]
(d). We need to calculate the dilated lifetime
Using formula of dilated time
[tex]t'=t\sqrt{1-\dfrac{v^2}{c^2}}[/tex]
[tex]t'=4.86\times10^{-12}\times\sqrt{1-\dfrac{(0.994c)^2}{c^2}}[/tex]
[tex]t'=4.86\times10^{-12}\times\sqrt{1-(0.994)^2}[/tex]
[tex]t'=5.32\times10^{-13}\ sec[/tex]
The dilated lifetime is [tex]5.32\times10^{-13}\ sec[/tex].
Hence, This is the required solution.
