In special relativity, a moving clock observed from a stationary frame is measured to:
Run slow
Run fast
Keep the same time
Stop entirely
Tap an answer to check it.
WhyTime dilation means $\Delta t=\gamma\Delta t_{0}$, so a moving clock is observed to run slow compared with the observer's own clock.
4Fill in the blank · Medium
Under a Galilean transformation, the velocity of an object in a frame moving at speed $v$ is $u'=u-$ .
Answer:
v
WhyThe Galilean velocity transformation subtracts the frame velocity: $u'=u-v$. (It fails at speeds near $c$.)
5Multiple choice · Hard
A pendulum has a period of 3.0 s in its own rest frame. An observer moves at 0.80c relative to it. What period does the observer measure?
5.0 s
1.8 s
8.3 s
1.1 s
Tap an answer to check it.
WhyThe Lorentz factor at 0.80c is 1/sqrt(1 - 0.64) = 1/0.6 = 1.67, so the measured period is 3.0 x 1.67 = 5.0 s.
6Multiple choice · Medium
The proper time interval between two events is the time measured
in the rest frame of the events
by any moving observer
in the fastest moving frame
only on Earth
Tap an answer to check it.
Whyin the frame where the two events occur at the same place (the rest frame).
7Fill in the blank · Medium
A moving clock runs compared with one at rest, an effect called time dilation.
Answer:
slow / slower / slowly
WhyIt runs slow (slower).
8Fill in the blank · Medium
The Lorentz factor gamma is always greater than or equal to .
Answer:
1 / one
WhyGamma is at least 1.
9Multiple choice · Hard
Observer X sees a spacecraft moving at 0.50c in the positive x-direction. Observer Y sees the same spacecraft moving at 0.50c in the negative x-direction. What is the speed of Y in the frame of X?