/*! 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} 13 PE Conversations with astronauts on... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: 13 PE (page 82)

Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person鈥檚 voice was so loud in the astronaut鈥檚 space helmet that it was picked up by the astronaut鈥檚 microphone and transmitted back to Earth. It is reasonable to assume that the echo time equals the time necessary for the radio wave to travel from the Earth to the Moon and back (that is, neglecting any time delays in the electronic equipment). Calculate the distance from Earth to the Moon given that the echo time was 2.56 s and that radio waves travel at the speed of light ( 3 x 108m/s ).

Short Answer

Expert verified

The distance from the Earth to the Moon for echo time is \({\bf{7}}{\bf{.68 \times 1}}{{\bf{0}}^{\bf{8}}}\;{\bf{m}}\).

Step by step solution

01

Determination of equation for distance from Earth to Moon based on echo time

Given and known data:

The echo time is t = 2.56 sec.

The speed of radio wave travel = \(c = 3 \times {10^8}\;{\rm{m}}/{\rm{s}}\).

The echo time is the time lag between the transmitting and receiving ends. This duration increases with the distance between the sender and the receiver.

The distance between the Earth and the Moon is given by

d = c.t

Here, d is the distance between the Earth and the Moon, c is the speed of light, and t is the echo time.

02

Determination of distance from Earth to Moon

Substitute all the values in the above equation.

\(\begin{array}{l}d = \left( {3 \times {{10}^8}\;{\rm{m}}/{\rm{s}}} \right)\left( {2.56\;{\rm{s}}} \right)\\d = 7.68 \times {10^8}\;{\rm{m}}\end{array}\)

Therefore, the distance from Earth to the Moon for echo time is \(7.68 \times {10^8}\;{\rm{m}}\).

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 91影视!

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

Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s. Calculate his (a) acceleration and (b) deceleration. Express each in multiples of g (\({\bf{9}}{\bf{.80}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\) ) by taking its ratio to the acceleration of gravity.

A soft tennis ball is dropped onto a hard floor from a height of \[{\bf{1}}.{\bf{50}}{\rm{ }}{\bf{m}}\]and rebounds to a height of\[{\bf{1}}.{\bf{10}}{\rm{ }}{\bf{m}}\]. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts\[{\bf{3}}.{\bf{50}}{\rm{ }}{\bf{ms}}{\rm{ }}\left( {{\bf{3}}.{\bf{50}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{3}}}}{\bf{s}}} \right)\]. (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

Find the following for path D in Figure 2.59:

(a) The distance travelled.

(b) The magnitude of the displacement from start to finish.

(c) The displacement from start to finish.

A bicycle racer sprints at the end of a race to clinch a victory. The racer has an initial velocity of 11.5 m/sand accelerates at the rate of 0.500for 7.00 s.

(a) What is his final velocity?

(b) The racer continues at this velocity to the finish line. If he was 300 mfrom the finish line when he started to accelerate, how much time did he save?

(c) One other racer was 5.00 mahead when the winner started to accelerate, but he was unable to accelerate, and travelled at 11.8 m/suntil the finish line. How far ahead of him (in meters and in seconds) did the winner finish?

A powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/ h) in only 3.90 s. (a) What is its average acceleration? (b) How far does it travel in that time?
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Dynamics

Read Explanation

Translational Dynamics

Read Explanation

Modern Physics

Read Explanation

Oscillations

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation
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.