Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 22

How would the volume of \(1.00 \mathrm{cm}^{3}\) of aluminum on Earth change if it were placed in a vacuum chamber and the pressure changed to that of the Moon (less than \(10^{-9} \mathrm{Pa}\) )?

Short Answer

Expert verified
Answer: The volume of 1.00 cm³ of aluminum remains essentially the same due to its properties as a solid and the negligible effect of the pressure change from Earth to the Moon.

Step by step solution

01

Identify the properties of aluminum

Aluminum is a solid, and its volume remains relatively constant under different pressures, especially if the variationpressure is not too high.
02

Compare the pressures of Earth and the Moon

The atmospheric pressure on the surface of the Earth at sea level is approximately \(1.01 \times 10^5 \mathrm{Pa}\) while the pressure on the surface of the Moon is less than \(10^{-9} \mathrm{Pa}\). However, since aluminum is a solid, this change in pressure shouldn't have a significant effect on its volume.
03

Consider the negligible effect of pressure on the aluminum volume

Since the pressure change of going from Earth to the Moon is not sufficiently large to cause a significant impact on the volume of a solid, like aluminum, we can safely assume that the volume change will be negligible. As a result, the volume of \(1.00 \mathrm{cm}^{3}\) of aluminum on Earth will remain essentially the same when it is placed in a vacuum chamber with the pressure of the moon. #Conclusion# In summary, the volume of \(1.00 \mathrm{cm}^{3}\) of aluminum on Earth will not change significantly when placed in a vacuum chamber with the pressure of the Moon, due to the nature of aluminum as a solid and the pressure change not being large enough to have a substantial effect on the aluminum's volume..

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

A small rowboat has a mass of \(47 \mathrm{kg} .\) When a \(92-\mathrm{kg}\) person gets into the boat, the boat floats \(8.0 \mathrm{cm}\) lower in the water. If the boat is then pushed slightly deeper in the water, it will bob up and down with simple harmonic motion (neglecting any friction). What will be the period of oscillation for the boat as it bobs around its equilibrium position?
(a) Given that the position of an object is \(x(t)=A\) cos \(\omega t\) show that \(v_{x}(t)=-\omega A\) sin \(\omega t .\) [Hint: Draw the velocity vector for point \(P\) in Fig. \(10.17 \mathrm{b}\) and then find its \(x\) component.] (b) Verify that the expressions for \(x(t)\) and \(v_{x}(t)\) are consistent with energy conservation. [Hint: Use the trigonometric identity $\sin ^{2} \omega t+\cos ^{2} \omega t=1.1$.]
Show, using dimensional analysis, that the frequency \(f\) at which a mass- spring system oscillates radical is independent of the amplitude \(A\) and proportional to Whim. IHint: Start by assuming that \(f\) does depend on \(A\) (to some power).]
A grandfather clock is constructed so that it has a simple pendulum that swings from one side to the other, a distance of \(20.0 \mathrm{mm},\) in $1.00 \mathrm{s} .$ What is the maximum speed of the pendulum bob? Use two different methods. First, assume SHM and use the relationship between amplitude and maximum speed. Second, use energy conservation.
The ratio of the tensile (or compressive) strength to the density of a material is a measure of how strong the material is "pound for pound." (a) Compare tendon (tensile strength \(80.0 \mathrm{MPa}\), density $1100 \mathrm{kg} / \mathrm{m}^{3}\( ) with steel (tensile strength \)\left.0.50 \mathrm{GPa}, \text { density } 7700 \mathrm{kg} / \mathrm{m}^{3}\right)$ which is stronger "pound for pound" under tension? (b) Compare bone (compressive strength \(160 \mathrm{MPa}\), density $1600 \mathrm{kg} / \mathrm{m}^{3}$ ) with concrete (compressive strength $\left.0.40 \mathrm{GPa}, \text { density } 2700 \mathrm{kg} / \mathrm{m}^{3}\right):$ which is stronger "pound for pound" under compression?
See all solutions

Recommended explanations on Physics Textbooks

Particle Model of Matter

Read Explanation

Force

Read Explanation

Electrostatics

Read Explanation

Famous Physicists

Read Explanation

Medical Physics

Read Explanation

Physical Quantities And Units

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.

Sign-up for free