/*! 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 The weight of an object on an or... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 18

The weight of an object on an orbiting space vehicle is measured to be \(42 \mathrm{~N}\) based on an artificial gravitational acceleration of \(6 \mathrm{~m} / \mathrm{s}^{2}\). What is the weight of the object, in \(\mathrm{N}\), on earth, where \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\) ?

Short Answer

Expert verified
Approximately 68.67 N.

Step by step solution

01

- Understanding the Given Values

Identify the given values from the problem.
02

- Calculate the Mass of the Object

Use the formula for weight, which is given by: \[ W = m \times g \] Rearrange to solve for mass: \[ m = \frac{W}{g} \] Substitute the given values of weight (42 N) and artificial gravitational acceleration (6 \mathrm{~m} / \mathrm{s}^{2}): \[ m = \frac{42}{6} = 7 \mathrm{~kg} \]
03

- Calculate the Weight on Earth

Use the calculated mass and Earth's gravitational acceleration to find the weight on Earth. \[ W_{earth} = m \times g_{earth} \] Substitute the mass (7 \mathrm{~kg}) and Earth's gravity (9.81 \mathrm{~m} / \mathrm{s}^{2}): \[ W_{earth} = 7 \times 9.81 \approx 68.67 \mathrm{~N} \]

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Key Concepts

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

artificial gravitational acceleration
Artificial gravitational acceleration is a concept used in space to create a force similar to gravity, allowing astronauts to stay healthy under long periods in space. This is achieved by rotating the spacecraft or a part of it to simulate gravity. The artificial gravitational accelation in this problem is given as 6 m/s虏. This force can be adjusted based on the needs of the mission and the health of the astronauts. It's important to understand that artificial gravity is not real gravity; it's a result of rotational forces acting on the mass of objects. It helps in keeping muscles and bones strong, preventing them from becoming weaker over time.
mass calculation
Mass is a fundamental property of an object that does not change, regardless of location. To determine mass when a gravitational force (weight) and gravitational acceleration are known, you can use the equation: \[ W = m \times g \] Rearranging gives: \[ m = \frac{W}{g} \] In this exercise, the weight measured under artificial gravity is 42 N, and the artificial gravitational acceleration is 6 m/s虏. Using the formula: \[ m = \frac{42}{6} = 7 \text{ kg} \] The calculated mass of the object is 7 kg. Once we know the mass, we can find the weight under different gravitational forces by using the same mass with the new gravitational acceleration.
gravitational force
Gravitational force, often referred to as weight, is the force with which an object is pulled towards the center of a planet, such as Earth, due to gravity. On Earth, gravity accelerates objects downward at 9.81 m/s虏. To calculate the weight of an object on Earth, you use the object's mass and Earth's gravity: \[ W_{earth} = m \times g_{earth} \] Given the mass calculated earlier as 7 kg, we substitute this value into the equation: \[ W_{earth} = 7 \times 9.81 \approx 68.67 \text{ N} \] The weight of the object on Earth is approximately 68.67 N, demonstrating how the gravitational force changes based on the gravitational acceleration at different locations.

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Most popular questions from this chapter

A dish of liquid water is placed on a table in a room. After a while, all of the water evaporates. Taking the water and the air in the room to be a closed system, can the system be regarded as a pure substance during the process? After the process is completed? Discuss.

What are possible boundaries for studying each of the following? (a) a bicycle tire inflating. (b) a cup of water being heated in a microwave oven. (c) a household refrigerator in operation. (d) a jet engine in flight. (e) cooling a desktop computer. (f) a residential gas furnace in operation. (g) a rocket launching.

A simple instrument for measuring the acceleration of gravity employs a linear spring from which a mass is suspended. At a location on earth where the acceleration of gravity is \(9.81 \mathrm{~m} / \mathrm{s}^{2}\), the spring extends \(0.739 \mathrm{~cm}\). If the spring extends \(0.116\) in. when the instrument is on Mars, what is the Martian acceleration of gravity? How much would the spring extend on the moon, where \(g=1.67 \mathrm{~m} / \mathrm{s}^{2}\) ?

Write a report reviewing the principles and objectives of statistical thermodynamics. How does the macroscopic approach to thermodynamics of the present text differ from this? Explain.

Over a limited temperature range, the relation between electrical resistance \(R\) and temperature \(T\) for a resistance temperature detector is $$ R=R_{0}\left[1+\alpha\left(T-T_{0}\right)\right] $$ where \(R_{0}\) is the resistance, in ohms \((\Omega)\), measured at reference temperature \(T_{0}\) (in \(\left.{ }^{\circ} \mathrm{C}\right)\) and \(\alpha\) is a material constant with units of \(\left({ }^{\circ} \mathrm{C}\right)^{-1}\). The following data are obtained for a particular resistance thermometer: \begin{tabular}{lcrr} \hline & \(T\left({ }^{\circ} \mathrm{C}\right)\) & \multicolumn{2}{c}{\(R(\Omega)\)} \\ \hline Test \(1\left(\mathrm{~T}_{0}\right)\) & 0 & \(\left(\mathrm{R}_{0}\right)\) & \(51.39\) \\ Test 2 & 91 & \(51.72\) \\ \hline \end{tabular} What temperature would correspond to a resistance of \(51.47 \Omega\) on this thermometer?
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