/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Find the speed of light in feet ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the speed of light in feet per nanosecond, to three significant figures.

Short Answer

Expert verified
Answer: The speed of light is approximately 0.981 ft/ns.

Step by step solution

01

Know the speed of light in meters per second

The speed of light in a vacuum is a constant value, approximately 2.99 x 10^8 meters per second.
02

Convert meters to feet

To convert meters to feet, we can use the conversion factor 1 meter = 3.281 feet. So, multiplying the speed of light in meters per second by the conversion factor will give us the speed of light in feet per second: (2.99 x 10^8 m/s) x (3.281 ft/m) = 9.81 x 10^8 ft/s.
03

Convert seconds to nanoseconds

We know that 1 second = 10^9 nanoseconds. So to convert the speed of light from feet per second to feet per nanosecond, we will divide the speed of light in feet per second by 10^9: (9.81 x 10^8 ft/s) / (10^9 ns/s) = 0.981 ft/ns
04

Round to three significant figures

Finally, rounding 0.981 feet per nanosecond to three significant figures, we get the result: Speed of light ≈ 0.981 ft/ns

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 91Ó°ÊÓ!

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 square of area \(100 \mathrm{~m}^{2}\) that is at rest in the reference frame is moving with a speed \((\sqrt{3} / 2) c\). Which of the following statements is incorrect? a) \(\beta=\sqrt{3} / 2\) b) \(\gamma=2\) c) To an observer at rest, it looks like another square with an area less than \(100 \mathrm{~m}^{2}\) d) The length along the moving direction is contracted by a factor of \(\frac{1}{2}\)

Using relativistic expressions, compare the momentum of two electrons, one moving at \(2.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}\) and the other moving at \(2.00 \cdot 10^{3} \mathrm{~m} / \mathrm{s}\). What is the percentage difference between classical momentum values and these values?

A wedge-shaped spaceship has a width of \(20.0 \mathrm{~m}\) a length of \(50.0 \mathrm{~m},\) and is shaped like an isosceles triangle. What is the angle between the base of the ship and the side of the ship as measured by a stationary observer if the ship is traveling by at a speed of \(0.400 c\) ? Plot this angle as a function of the speed of the ship.

Consider a positively charged particle moving at constant speed parallel to a current-carrying wire, in the direction of the current. As you know (after studying Chapters 27 and 28), the particle is attracted to the wire by the magnetic force due to the current. Now suppose another observer moves along with the particle, so according to him the particle is at rest. Of course, a particle at rest feels no magnetic force. Does that observer see the particle attracted to the wire or not? How can that be? (Either answer seems to lead to a contradiction: If the particle is attracted, it must be by an electric force because there is no magnetic force, but there is no electric field from a neutral wire; if the particle is not attracted, you see that the particle is, in fact, moving toward the wire.)

A rod at rest on Earth makes an angle of \(10^{\circ}\) with the \(x\) -axis. If the rod is moved along the \(x\) -axis, what happens to this angle, as viewed by an observer on the ground?

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.