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omer10000
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Q1. Time Dilation Question
Homework Statement
Three identical triplets leave Earth when they reach the age of 21, in the year 2121. Each triplet goes on a spaceship journey that takes T years, as measured by a clock in each spaceship. During the journey they travel at a constant speed v, as measured on earth, except during the relatively short acceleration phases of their journey. (Triplets occur when three babies are born together.)
Part A)
The triplets return to Earth in different years according to people who stay on Earth. Rank the triplets on the basis of the year on Earth when they return from their journey. That is, work out the year when each triplet returns, and then rank the triplets according to the largest return year to the smallest return year.
A- T= 20 years, v=0.4
B- T= 10 years, v=0.8
C- T= 10 years, v=0.4
Part B)
Calculate the age of each triplet when they return from their journey. Rank the triplets on the basis of their return age.
A- T= 10 years, v=0.4
B- T= 20 years, v=0.4
C- T= 10 years, v=0.8
Δt' = x*b where b = √(1-(v/c)^2) → x = (Δt')/√(1-(v/c)^2) => PART A
Part A - Solving for x: A = 22 yrs...B = 17 yrs...C = 11
Is this correct? I don't think so; answer might be: B A C. How is this possible?
Part B
Shouldn't the answer be same as Part A? Apparently it may be: C A B. Why?
Q2. Relativity
1. Homework Statement
A spaceship passes Earth moving at 0.80c. The spaceship then passes the Moon. How much time, according to the ship's clock, elapses between the time a person looking out the ship's window sees Earth pass by and when the Moon passes by. The Earth-Moon distance is approximately 3.84*10^8m
Δt = Δx/v
Δt'=Δt*√(1-(v/c)^2)
Δt = Δx/v = (3.84*10^8m)/((3.0*10^8)*0.80c) = 1.6s
Δt'=Δt*√(1-(v/c)^2)=1.6*√(1-0.8^2)=0.96s
|^| Is this in reference to person in spaceship?
Thanks
Homework Statement
Three identical triplets leave Earth when they reach the age of 21, in the year 2121. Each triplet goes on a spaceship journey that takes T years, as measured by a clock in each spaceship. During the journey they travel at a constant speed v, as measured on earth, except during the relatively short acceleration phases of their journey. (Triplets occur when three babies are born together.)
Part A)
The triplets return to Earth in different years according to people who stay on Earth. Rank the triplets on the basis of the year on Earth when they return from their journey. That is, work out the year when each triplet returns, and then rank the triplets according to the largest return year to the smallest return year.
A- T= 20 years, v=0.4
B- T= 10 years, v=0.8
C- T= 10 years, v=0.4
Part B)
Calculate the age of each triplet when they return from their journey. Rank the triplets on the basis of their return age.
A- T= 10 years, v=0.4
B- T= 20 years, v=0.4
C- T= 10 years, v=0.8
Homework Equations
Δt' = x*b where b = √(1-(v/c)^2) → x = (Δt')/√(1-(v/c)^2) => PART A
The Attempt at a Solution
Part A - Solving for x: A = 22 yrs...B = 17 yrs...C = 11
Is this correct? I don't think so; answer might be: B A C. How is this possible?
Part B
Shouldn't the answer be same as Part A? Apparently it may be: C A B. Why?
Q2. Relativity
1. Homework Statement
A spaceship passes Earth moving at 0.80c. The spaceship then passes the Moon. How much time, according to the ship's clock, elapses between the time a person looking out the ship's window sees Earth pass by and when the Moon passes by. The Earth-Moon distance is approximately 3.84*10^8m
Homework Equations
Δt = Δx/v
Δt'=Δt*√(1-(v/c)^2)
The Attempt at a Solution
Δt = Δx/v = (3.84*10^8m)/((3.0*10^8)*0.80c) = 1.6s
Δt'=Δt*√(1-(v/c)^2)=1.6*√(1-0.8^2)=0.96s
|^| Is this in reference to person in spaceship?
Thanks
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