A muon formed high in the Earth's atmosphere is measured by an observer on the Earth's surface to travel at speed V - 0.983c for a distance of 3.53 km before it decays into an electron, a neutrino, and an antineutrino ( -e" + v + v). (a) For what time interval does the muon live as measured in its referencerame? 2.32 X Your response is within 10% of the correct value. This may be due to roundoff ett you could have a mistake th your calculation Carry out all intermediate results to at least four-digit accuracy to minimize roundoff errotus (b) How far does the Earth travel as measured in the frame of the muon?

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klefenz klefenz · Answer 1 · 2019-09-15T17:55:08+02:00

Answer:

The moun lives 2.198*10^-6 s as measured by its own frame of reference

The Earth moved 648 m as measured by the moun's frame of reference

Explanation:

From the point of view of the observer on Earth the muon traveled 3.53 km at 0.983c

0.983 * 3*10^8 = 2.949*10^8 m/s

Δt = d/v = 3530 / 2.949*10^8 = 1.197*10^-5 s

The muon lived 1.197*10^-5 s from the point of view of the observer.

The equation for time dilation is:

[tex]\Delta t' = \Delta t * \sqrt{1 - \frac{v^2}{c^2}}[/tex]

Then:

[tex]\Delta t' = 1.197*10^-5 * \sqrt{1 - \frac{(0.983c)^2}{c^2}} = 2.198*10^-6 s[/tex]

From the point of view of the moun Earth moved at 0.983c (2.949*10^8 m/s) during a time of 2.198*10^-6, so it moved

d = v*t = 2.949*10^8 * 2.198*10^-6 = 648 m