/*! 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 20 An astronaut's pack weighs 17.5 ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 20

An astronaut's pack weighs 17.5 \(\mathrm{N}\) when she is on earth but only 3.24 \(\mathrm{N}\) when she is at the surface of an asteroid. (a) What is the acceleration due to gravity on this asteroid? (b) What is the mass of the pack on the asteroid?

Short Answer

Expert verified
(a) \( g = 1.815 \, \text{m/s}^2 \), (b) Mass = 1.7857 kg.

Step by step solution

01

Understand the Problem

We are given the weight of an astronaut's pack on Earth and on the surface of an asteroid. We need to find the acceleration due to gravity on the asteroid and the mass of the pack on the asteroid.
02

Recall the Formula for Weight

Weight is the force of gravity acting on an object and is given by the formula: \( W = m \cdot g \), where \( W \) is weight, \( m \) is mass, and \( g \) is the acceleration due to gravity.
03

Calculate Mass of the Pack

Use the weight of the pack on Earth to find its mass. On Earth, \( g = 9.8 \, \text{m/s}^2 \), and the weight \( W = 17.5 \, \text{N} \). Therefore, \( m = \frac{W}{g} = \frac{17.5}{9.8} = 1.7857 \, \text{kg}\).
04

Calculate Gravitational Acceleration on the Asteroid

Using the mass from Step 3 and the weight on the asteroid (3.24 N), use the formula \( W = m \cdot g \), where \( m = 1.7857 \, \text{kg} \) to find \( g \). Therefore, \( g = \frac{W}{m} = \frac{3.24}{1.7857} = 1.815 \text{m/s}^2 \).
05

Confirm Mass Consistency

The mass of the pack remains the same regardless of location in the universe: \( 1.7857 \, \text{kg} \). This is a constant property of matter.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Acceleration Due to Gravity
Acceleration due to gravity is essentially the rate at which an object accelerates when it's falling down due to the gravitational pull of a celestial body. This acceleration is different for every planet, moon, or asteroid based on their mass and size. On Earth, the acceleration due to gravity is approximately 9.8 m/s虏.
  • This value is calculated based on Earth's mass and radius.
  • It represents how strong Earth's gravitational pull is.
For our asteroid, we found the gravitational acceleration to be 1.815 m/s虏. This is much smaller than Earth's, which is expected since asteroids generally have much less mass than planets like Earth.
When you know the weight of an object and its mass, you can use the relation between weight and gravitational force to solve any such problem involving different celestial bodies.
Understanding Mass Determination
Mass is a measure of how much matter is in an object. It remains the same regardless of where the object is located in the universe. Unlike weight, which can vary if you鈥檙e on Earth or an asteroid, mass is constant.
  • Mass is measured in kilograms (kg).
  • It can be calculated if you know the weight and gravitational acceleration.
In our scenario, the pack has a mass of 1.7857 kg, found using its weight on Earth (17.5 N) and Earth's gravitational acceleration (9.8 m/s虏).
This step emphasizes the importance of distinguishing between mass and weight. While they are related, these properties are not the same. Knowing the mass allows us to find the gravitational acceleration on the asteroid by using the weight measured there.
Using the Weight Formula
Weight is defined as the force exerted on an object due to gravity. It is calculated using the weight formula: \[ W = m \cdot g \]Where:- \( W \) is the weight of the object,- \( m \) is its mass,- \( g \) is the acceleration due to gravity.
Using this formula, you can determine how much an object will weigh under different gravitational conditions. For example:
  • On Earth, the astronaut's pack weighs 17.5 N.
  • On the asteroid, it weighs only 3.24 N because the gravitational pull is weaker.
This formula helps you convert between mass and weight whenever you know one of these properties and the local gravitational acceleration. When dealing with gravity calculations, always remember to distinguish clearly between mass and weight to accurately solve problems.

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

An oil tanker's engines have broken down, and the wind is blowing the tanker straight toward a recf at a constant spocd of 1.5 \(\mathrm{m} / \mathrm{s}(\text { Fig. } 4.37) .\) When the tanker is 500 \(\mathrm{m}\) m from the reef, the wind dies down just as the engineer gets the engines going again. The rudder is stuck, so the only choice is to try to accelerate straight backward away from the reef. The mass of the tanker and cargo is \(3.6 \times 10^{7} \mathrm{kg}\) , and the engines produce a net horizontal force of \(8.0 \times 10^{4} \mathrm{N}\) on the tanker. Will the ship hit the reef? If it does, will the oil be safe? The hull can withstand an impact at a speed of 0.2 \(\mathrm{m} / \mathrm{s}\) or less. You can ignore the retarding force of the water on the tanker's hull.

A hockey puck with mass 0.160 \(\mathrm{kg}\) is at rest at the origin \((x=0)\) on the horizontal, frictionless surface of the rink. At time \(t=0\) a player applies a force of 0.250 \(\mathrm{N}\) to the puck, parallel to the \(x\) -axis; he continues to apply this force until \(t=2.00 \mathrm{s}\) . (a) What are the position and speed of the puck at \(t=2.00 \mathrm{s?}\) (b) If the same force is again applied at \(t=5.00 \mathrm{s}\) , what are position and speed of the puck at \(t=7.00 \mathrm{s} ?\)

Superman throws a \(2400-N\) boulder at an adversary. What horizontal force must Superman apply to the boulder to give it a horizontal acceleration of 12.0 \(\mathrm{m} / \mathrm{s}^{2}\) ?

An object with mass \(m\) initially at rest is acted on by a force \(\vec{F}=k_{1} \hat{z}+k_{2} t^{3}\) , where \(k_{1}\) and \(k_{2}\) are constants. Calculate the velocity \(\vec{v}(t)\) of the object as a function of time.

An athlete whose mass is 90.0 \(\mathrm{kg}\) is performing weight-lifting exercises. Starting from the rest position, he lifts, with constant acceleration, a barbell that weighs 490 \(\mathrm{N}\) . He lifts the barbell a distance of 0.60 \(\mathrm{m}\) in 1.6 \(\mathrm{s}\) (a) Draw a clearly labeled free-body force diagram for the barbell and for the athlete. (b) Use the diagrams in part (a) and Newton's laws to find the total force that his feet exert on the ground as he lifts the barbell.
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Electricity and Magnetism

Read Explanation

Translational Dynamics

Read Explanation

Mechanics and Materials

Read Explanation

Famous Physicists

Read Explanation

Torque and Rotational Motion

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.