A spacecraft flies away from the earth with a speed of 6.90 Times 10^6 m/s relat

Question

A spacecraft flies away from the earth with a speed of 6.90 Times 10^6 m/s relative to the earth and then returns at the same speed. The spacecraft carries an atomic clock that has been carefully sychronized with identical clock that remains at rest on earth. The spacecraft returns to its starting point 1095 days (3 years) later, as measured by the clock that remained on earth. What is the difference of the elapsed times on the two clocks, measured in hours? Which clock, the one in the spacecraft or the one on the earth, shows the smallest elapsed time? both clocks show the same elapsed time the clock in the spacecraft the clock on the earth Two particles in a High energy accelerator experiment are approaching each other head-an each with a speed 0.9520c as measured in the.What is the magnitude of the velocity of one particle return to the other?

Explanation / Answer

7) v= 6.9*10^6 m/s

v/c = 6.9*10^6/(3*10^8) =0.023

v = 0.023 c

Dt = 1095 days

Dt = Dto/(1-v^2/c^2)^1/2

1095 = Dto/(1-0.023^2)^0.5

Dto = 1094.71033 days

Dt - Dto = 1095 -1094.71033 =0.28967 days = 6.95 h

8) u = v1 -v2 = (1-v1v2/c^2

u = 2fc/(1+f^2)

f = 0.9520 c

u =2*0.9520c/(1+0.9520^2)

u = 0.9988 c

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