Exercise 37.12 Part A An unstable particle is created in the upper atmosphere fr
ID: 1793365 • Letter: E
Question
Exercise 37.12 Part A An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the earth with a speed of 0.99540 crelative to the earth. A scientist at rest on the earth's surface measures that the particle is created at an altitude of 41.0 km As measured by the scientist, how much time does it take the particle to travel the 41.0 km to the surface of the earth? t= 1.37×10-4 Submit My Answers Give Up Correct Significant Figures Feedback: Your answer 1.37299.104- 1.3730x10-4 s was either rounded differently or used a different number of significant figures than required for this part Part B Use the length-contraction formula to calculate the distance from where the particle is created to the surface of the earth as measured in the particle's frame Submit My Answers Give Up Incorrect, Iry Again, 9 attempts remaining Part C In the particle's frame, how much time does it take the particle to travel from where it is created to the surface of the earth? Submit My Answers Give Up ContinueExplanation / Answer
a)For Scientist, since v=0.9954c d=41km=41000m, v=d/t=>t=d/v
t=41000 x 10^-8 / (3 x 0.9954) = 0.13729 ms
b)Since 41x10^3 m is the contracted by a factor of root(1-v^2/c^2). So the place where the particle orginated is greater than 41 km, since root(1-v^2/c^2) is always <=1. Dividing 41km by root(1-v^2/c^2) would give the place of creation, when the particle was at rest.
d= 41000/(root(1-v^2/c^2))=41000/0.099538= 41.1902 km
c)it agrees with its speed, it just measures a shorter distance travelled
... so "how far it went" / "its speed" = "the time it took"
answer to (b) / (0.99540c) = 0.41380 x 10^-6 s
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