The proper time interval of the rocket ship adventure.

Question

Want to see the full answer?

Answer 1

Question

Chapter 26, Problem 3MCQ

To determine

The proper time interval of the rocket ship adventure.

To determine

Calculation:

To the captain’s frame:

The flashlight flash as measured by the captain can be found

$Δ t_{o} = 1 \times \sqrt{1 - \frac{{(0.5 c)}^{2}}{c^{2}}}$

$Δ t_{o} = 0.87 s$

To the space dock frame:

Suppose if the flashlight is flashes in the space dock and the flashes are recorded by the space dock personal is the proper value and is 1 s

The flashlight flash as measured by the captain is the time dilation value and its value is

$Δ t = \frac{1}{\sqrt{1 - \frac{{(0.5 c)}^{2}}{c^{2}}}}$

$Δ t = 1.154$

Analysis of options:

Option a: 1.15 s:

This is not the right answer.

This gives the time dilation value relative to the space dock. So, it is wrong.

Option b: Every 1.00 s:

Hence, it is the wrong answer.

This is measured by the space dock personal and it would be the proper time if the event was held in the space dock. But in our problem statement, it is being held in the rocket. So, it is wrong.

Option c: 0.87 s.:

It is the correct answer.

This is the proper time interval, as calculated above. It is measured by the captain of the rocket the flash light has been flashed from the rocket only. Hence, it is the correct answer.

Option d: 0.13 s:

The proper time and the time interval measured from another reference differ by a factor $\sqrt{1 - \frac{v^{2}}{c^{2}}}$ . The above value does not hold this. Hence, it is the wrong answer.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Four capacitors are connected as shown in the figure below. (Let C = 12.0 μF.) a C 3.00 με Hh. 6.00 με 20.0 με HE (a) Find the equivalent capacitance between points a and b. 5.92 HF (b) Calculate the charge on each capacitor, taking AV ab = 16.0 V. 20.0 uF capacitor 94.7 6.00 uF capacitor 67.6 32.14 3.00 µF capacitor capacitor C ☑ με με The 3 µF and 12.0 uF capacitors are in series and that combination is in parallel with the 6 μF capacitor. What quantity is the same for capacitors in parallel? μC 32.14 ☑ You are correct that the charge on this capacitor will be the same as the charge on the 3 μF capacitor. μC

In the pivot assignment, we observed waves moving on a string stretched by hanging weights. We noticed that certain frequencies produced standing waves. One such situation is shown below: 0 ст Direct Measurement ©2015 Peter Bohacek I. 20 0 cm 10 20 30 40 50 60 70 80 90 100 Which Harmonic is this? Do NOT include units! What is the wavelength of this wave in cm with only no decimal places? If the speed of this wave is 2500 cm/s, what is the frequency of this harmonic (in Hz, with NO decimal places)?

Four capacitors are connected as shown in the figure below. (Let C = 12.0 µF.) A circuit consists of four capacitors. It begins at point a before the wire splits in two directions. On the upper split, there is a capacitor C followed by a 3.00 µF capacitor. On the lower split, there is a 6.00 µF capacitor. The two splits reconnect and are followed by a 20.0 µF capacitor, which is then followed by point b. (a) Find the equivalent capacitance between points a and b. µF(b) Calculate the charge on each capacitor, taking ΔVab = 16.0 V. 20.0 µF capacitor µC 6.00 µF capacitor µC 3.00 µF capacitor µC capacitor C µC