/*! 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} Q49P Question: After landing on an un... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Q49P (page 462)

Question: After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length \({\bf{50}}.{\bf{0}}\,{\bf{cm}}\) . She finds that the pendulum makes \(100\) complete swings in \({\bf{136}}{\rm{ }}{\bf{s}}\). What is the value of \(g\) on this planet?

Short Answer

Expert verified

Acceleration due to gravity on an unfamiliar planet is \(10.67\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}\)

Step by step solution

01

Identification of given data

Time period of pendulum \(T = \frac{{136\,{\rm{s}}}}{{100}}\,\,{\rm{or}}\,\,{\rm{1}}{\rm{.36}}\,{\rm{s}}\)

Length of pendulum \(L = 50\,{\rm{cm}}\,\,{\rm{or}}\,\,{\rm{0}}{\rm{.5}}\,{\rm{m}}\)

02

Significance of time period of simple pendulum

The letter "\(T\)" stands for the period of time needed for the pendulum to complete one complete oscillation.

\(T = 2\pi \sqrt {\frac{L}{g}} \) 鈥�(i)

Where, \(L\) is the length of pendulum and \(g\) is the acceleration due to gravity

03

Determining the acceleration due to gravity on an unfamiliar planet

Rearranging the equation (i)

\(g = \frac{{4{\pi ^2}L}}{{{T^2}}}\)

Substitute all the values in above equation

\(\begin{array}{c}g = \frac{{4{\pi ^2} \times 0.5\,{\rm{m}}}}{{{{\left( {1.36\,{\rm{s}}} \right)}^2}}}\\ = 10.67\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}\end{array}\)

Hence the acceleration due to gravity on an unfamiliar planet is \(10.67\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}\)

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

Can you find two vectors with different lengths that have a vector sum of zero? What length restrictions are required for three vectors to have a vector sum of zero? Explain.

(a) The recommended daily allowance (RDA) of the trace metal magnesium is 410 mg/day for males. Express this quantity in 碌g/day. (b) For adults, the RDA of the amino acid lysine is 12 mg per kg of body weight. How many grams per day should a 75-kg adult receive? (c) A typical multivitamin tablet can contain 2.0 mg of vitamin B2 (riboflavin), and the RDA is 0.0030 g/day. How many such tablets should a person take each day to get the proper amount of this vitamin, if he gets none from other sources? (d) The RDA for the trace element selenium is 0.000070 g/day. Express this dose in mg/day.

A lunar lander is makingits descent to Moon Base I (Fig. E2.40). The lander descendsslowly under the retro-thrust of its descent engine. The engine iscut off when the lander is 5.0 m above the surface and has a downwardspeed of 0.8m/s . With the engine off, the lander is in freefall. What is the speed of the lander just before it touches the surface?The acceleration due to gravity on the moon is 1.6m/s2.

In hot-air ballooning, a large balloon is filled with air heated by a gas burner at the bottom. Why must the air be heated? How does the balloonist control ascent and descent?

At what distance above the surface of the earth is the acceleration due to the earth鈥檚 gravity 0.980 m/s2 if the acceleration due to gravity at the surface has magnitude 9.8 m/s2?
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Solid State Physics

Read Explanation

Turning Points in Physics

Read Explanation

Electromagnetism

Read Explanation

Mechanics and Materials

Read Explanation

Space Physics

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.