Question

determine how much longer a 1.00 s proper time interval appears to a stationary observer when a clock is moving with a speed of 0.60c.​

Answer 1

The time dilation formula,
`\( \Delta t' = \frac{\Delta t}{\sqrt{1 - (v^2)/(c^2)}} \)`, can be used to determine how much longer a 1.00 s proper time interval appears to a stationary observer when a clock is moving with a speed of 0.60c. Substituting the given values, the observed time interval
`(\( \Delta t' \))` is approximately 1.25 s.

Step-by-step explanation:

The time dilation phenomenon is a consequence of Einstein's theory of special relativity, stating that time appears to pass more slowly for an observer in motion relative to a stationary observer. The time dilation formula is given by
`\( \Delta t' = \frac{\Delta t}{\sqrt{1 - (v^2)/(c^2)}} \)`, where
`\( \Delta t' \)` is the observed time interval,
`\( \Delta t \)` is the proper time interval,
`\( v \)` is the speed of the moving clock, and
`\( c \)` is the speed of light.

In this scenario, the proper time interval
`(\( \Delta t \))` is given as 1.00 s, and the speed of the moving clock
`(\( v \))` is 0.60c. Substituting these values into the time dilation formula, we get
`\( \Delta t' = \frac{1.00 \, \text{s}}{√(1 - (0.60)^2)} \)`, which simplifies to
`\( \Delta t' \approx 1.25 \, \text{s} \)`. This implies that, according to the stationary observer, the 1.00 s proper time interval on the moving clock would appear to be 1.25 s, demonstrating the time dilation effect at relativistic speeds.