/*! 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} Q16E An astronaut’s pack weighs 17.... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An astronaut’s pack weighs 17.5 N when she is on the earth but only 3.24 N when she is at the surface of a moon.

(a) What is the acceleration due to gravity on this moon?

(b) What is the mass of the pack on this moon?

Short Answer

Expert verified

(a) The acceleration due to gravity on the surface of the moon where the astronaut's pack weighs 3.24 N is 1.81m/s2.

(b) The mass of the astronaut's pack on the moon is 1.79 kg.

Step by step solution

01

Given data

The weight of the astronaut's pack on Earth is

WE=17.5N

The weight of the astronaut's pack on the moon is

Wm=3.24N

The acceleration due to gravity on Earth is

g=9.8m/s2

02

Gravitational force

The downward gravitational force on a body of mass mis

Fg=mg.....(1)

03

The acceleration due to gravity on the moon

From equation (1), the ratio of weights of the astronaut's pack on the moon and on Earth is

WmWE=mgmmggm=WmWE×ggm=3.24N17.5N×9.8m/s2gm=1.81m/s2

Thus, the acceleration due to gravity on the moon is1.81m/s2

04

Mass of the astronaut's pack

From equation (1), the mass of the astronaut's pack is

m=Wmgm=3.24N1.81m/s2=1.79·1m/s2·1N×1kg·m/s21N=1.79kg

Thus, the mass of the pack is 1.79 kg.

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

Planet Vulcan.Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury. What would be the orbital period of such a planet? (Such a planet was once postulated, in part to explain the precession of Mercury’s orbit. It was even given the name Vulcan, although we now have no evidence that it actually exists. Mercury’s precession has been explained by general relativity.)

Question: A car’s velocity as a function of time is given byvxt=α+βt2, whereα=3.00m/sand β=0.100m/s3.(a) Calculate the average acceleration for the time interval t=0tot=5.00s. (b) Calculate the instantaneous acceleration forrole="math" t=0tot=5.00s.

(c) Draw vx-tandax-tgraphs for the car’s motion betweent=0tot=5.00s.

Question: The purity of gold can be tested by weighing it in air and in water. How? Do you think you could get away with making a fake gold brick by gold-plating some cheaper material?

Water flows steadily from an open tank as in Fig. P12.81. The elevation of point 1 is 10.0 m, and the elevation of points 2 and 3 is 2.00 m. The cross-sectional area at point 2 is 0.0480 m2; at point 3 it is 0.0160 m2. The area of the tank is very large compared with the cross-sectional area of the pipe. Assuming that Bernoulli’s equation applies, compute (a) the discharge rate in cubic meters per second and (b) the gauge pressure at point 2.

Starting from the front door of a ranch house, you walk 60.0 m due east to a windmill, turn around, and then slowly walk 40.0 m west to a bench, where you sit and watch the sunrise. It takes you 28.0 s to walk from the house to the windmill and then 36.0 s to walk from the windmill to the bench. For the entire trip from the front door to the bench, what are your (a) average velocity and (b) average speed?

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.