The light-second is a unit of distance; 1 light-second is the distance that light travels in 1 second. (a) Find the conversion between light-seconds and meters: 1 lightsecond \(=? \mathrm{m} .\) (b) What is the speed of light in units of light-seconds per second?

Short Answer

Expert verified
Answer: The speed of light is 1 light-second/second.

Step by step solution

01

(a) Conversion between light-seconds and meters

To find the distance light travels in one second, we can simply multiply the speed of light in meters per second by the time interval of 1 second. In this case, the speed of light is approximately 299,792,458 meters per second. So, the conversion between light-seconds and meters is: 1 light-second = 299,792,458 meters
02

(b) Speed of light in light-seconds per second

Now that we have the conversion between light-seconds and meters, we can determine the speed of light in light-seconds per second. Since 1 light-second is the distance light travels in 1 second, the speed of light in light-seconds per second is simply 1 light-second per second. So, the speed of light is 1 light-second/second.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A neutron (mass 1.00866 u) disintegrates into a proton (mass $1.00728 \mathrm{u}\( ), an electron (mass \)0.00055 \mathrm{u}$ ), and an antineutrino (mass 0 ). What is the sum of the kinetic energies of the particles produced, if the neutron is at rest? $\left(1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^{2} .\right)$
Derivation of the Doppler formula for light. A source and receiver of EM waves move relative to one another at velocity \(v .\) Let \(v\) be positive if the receiver and source are moving apart from one another. The source emits an EM wave at frequency \(f_{\mathrm{s}}\) (in the source frame). The time between wavefronts as measured by the source is \(T_{\mathrm{s}}=1 / f_{\mathrm{s}}\) (a) In the receiver's frame, how much time elapses between the emission of wavefronts by the source? Call this \(T_{\mathrm{T}}^{\prime} .\) (b) \(T_{\mathrm{T}}^{\prime}\) is not the time that the receiver measures between the arrival of successive wavefronts because the wavefronts travel different distances. Say that, according to the receiver, one wavefront is emitted at \(t=0\) and the next at \(t=T_{\mathrm{o}}^{\prime} .\) When the first wavefront is emitted, the distance between source and receiver is \(d_{\mathrm{r}}\) When the second wavefront is emitted, the distance between source and receiver is \(d_{\mathrm{r}}+v T_{\mathrm{r}}^{\prime} .\) Each wavefront travels at speed \(c .\) Calculate the time \(T_{\mathrm{r}}\) between the arrival of these two wavefronts as measured by the receiver. (c) The frequency detected by the receiver is \(f_{\mathrm{r}}=1 / T_{\mathrm{r}} .\) Show that \(f_{\mathrm{r}}\) is given by $$f_{\mathrm{r}}=f_{\mathrm{s}} \sqrt{\frac{1-v / \mathrm{c}}{1+v / \mathrm{c}}}$$
Consider the following decay process: \(\pi^{+} \rightarrow \mu^{+}+v .\) The mass of the pion \(\left(\pi^{+}\right)\) is \(139.6 \mathrm{MeV} / c^{2},\) the mass of the muon \(\left(\mu^{+}\right)\) is \(105.7 \mathrm{MeV} / c^{2},\) and the mass of the neutrino \((v)\) is negligible. If the pion is initially at rest, what is the total kinetic energy of the decay products?
Kurt is measuring the speed of light in an evacuated chamber aboard a spaceship traveling with a constant velocity of \(0.60 c\) with respect to Earth. The light is moving in the direction of motion of the spaceship. Siu- Ling is on Earth watching the experiment. With what speed does the light in the vacuum chamber travel, according to Siu-Ling's observations?
In a beam of electrons used in a diffraction experiment, each electron is accelerated to a kinetic energy of \(150 \mathrm{keV} .\) (a) Are the electrons relativistic? Explain. (b) How fast are the electrons moving?
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free