/*! 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 18 A spaceship is moving at a const... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A spaceship is moving at a constant velocity of \(0.70 c\) relative to an Earth observer. The Earth observer measures the length of the spaceship to be $40.0 \mathrm{m}$. How long is the spaceship as measured by its pilot?

Short Answer

Expert verified
Answer: The length of the spaceship as measured by the pilot is approximately 56.19 meters.

Step by step solution

01

Understand the length contraction formula

The length contraction formula is given by: \(L = L_{0} \sqrt{1 - \frac{v^2}{c^2}}\) where \(L\) is the contracted length measured by the observer, \(L_0\) is the proper length measured in the rest frame of the object (in this case, the pilot), \(v\) is the relative velocity of the object, and \(c\) is the speed of light.
02

Rearrange the formula to solve for the proper length

We need to find \(L_0\), the proper length as measured by the pilot. Rearrange the formula to solve for \(L_0\): \(L_0 = \frac{L}{\sqrt{1 - \frac{v^2}{c^2}}}\)
03

Substitute the given values and solve for the proper length

Plug in the given values: \(L = 40.0\,\text{m}\) and \(v = 0.70c\): \(L_0 = \frac{40.0\,\text{m}}{\sqrt{1 - \frac{(0.70c)^2}{c^2}}}\) Simplify the expression in the square root: \(L_0 = \frac{40.0\,\text{m}}{\sqrt{1 - (0.70^2)}}\)
04

Calculate the proper length

Now, calculate the proper length: \(L_0 = \frac{40.0\,\text{m}}{\sqrt{1 - 0.49}}\) \(L_0 = \frac{40.0\,\text{m}}{\sqrt{0.51}}\) \(L_0 \approx 56.19\,\text{m}\) So, the length of the spaceship as measured by the pilot is approximately 56.19 meters.

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

Refer to Example \(26.2 .\) One million muons are moving toward the ground at speed \(0.9950 c\) from an altitude of \(4500 \mathrm{m} .\) In the frame of reference of an observer on the ground, what are (a) the distance traveled by the muons; (b) the time of flight of the muons; (c) the time interval during which half of the muons decay; and (d) the number of muons that survive to reach sea level. [Hint: The answers to (a) to (c) are not the same as the corresponding quantities in the muons' reference frame. Is the answer to (d) the same?]
A starship takes 3.0 days to travel between two distant space stations according to its own clocks. Instruments on one of the space stations indicate that the trip took 4.0 days. How fast did the starship travel, relative to that space station?
A futuristic train moving in a straight line with a uniform speed of \(0.80 c\) passes a series of communications towers. The spacing between the towers, according to an observer on the ground, is \(3.0 \mathrm{km} .\) A passenger on the train uses an accurate stopwatch to see how often a tower passes him. (a) What is the time interval the passenger measures between the passing of one tower and the next? (b) What is the time interval an observer on the ground measures for the train to pass from one tower to the next?
A spaceship is traveling away from Earth at \(0.87 c .\) The astronauts report home by radio every 12 h (by their own clocks). (a) At what interval are the reports sent to Earth, according to Earth clocks? (b) At what interval are the reports received by Earth observers, according to their own clocks?
The Tevatron is a particle accelerator at Fermilab that accelerates protons and antiprotons to high energies in an underground ring. Scientists observe the results of collisions between the particles. The protons are accelerated until they have speeds only \(100 \mathrm{m} / \mathrm{s}\) slower than the speed of light. The circumference of the ring is \(6.3 \mathrm{km} .\) What is the circumference according to an observer moving with the protons? [Hint: Let \(v=c-u\) where $v \text { is the proton speed and } u=100 \mathrm{m} / \mathrm{s} .$]
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.