Physics 2: A spaceship approaches earth with speed of 0.55c. A passenger in the ship measures his heartbeat..
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Physics 2: A spaceship approaches earth with speed of 0.55c. A passenger in the ship measures his heartbeat..

[From: ] [author: ] [Date: 12-04-19] [Hit: ]
v = 0.This is the length of time for one beat (in seconds).γ =1.=> ∆t = 1.H = heart rate as observed on earth.=> H = 70/1.......
A spaceship approaches earth with speed of 0.55c. A passenger in the ship measures his heartbeat as 70 beats per minute. What is his heartbeat rate according to an observer that is at rest relative to earth?

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Time dilation formula:

∆t= γ∆t_0

γ=(1 - (v^2/c^2))^-(1/2)

∆t_0 = time interval between two events in the frame where they occur at the same spatial coordinates.
∆t = time interval between the same two events in a frame travelling at a speed v relative to the first.
v = 0.55c

Using this information:

∆t_0 = 60/70

This is the length of time for one beat (in seconds).

γ =1.197

=> ∆t = 1.197(60/70) = 60/H

H = heart rate as observed on earth.

=> H = 70/1.197 = 58.48 beats per minute

i.e. slower. Hence time DIALATION.
Hope this is clear.
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