/*! 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 5 A uniform lead sphere and a unif... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 5

A uniform lead sphere and a uniform aluminum sphere have the same mass. What is the ratio of the radius of the aluminum sphere to the radius of the lead sphere?

Short Answer

Expert verified
The ratio of the radius of the aluminum sphere to the lead sphere is approximately 1.61.

Step by step solution

01

Understand the relationship between mass, density, and volume

The mass of a sphere is given by \( m = \rho V \) where \( \rho \) is the density and \( V \) is the volume. The volume of a sphere is \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius. Since both spheres have the same mass, their masses can be expressed as:\[ m = \rho_{\text{lead}} \cdot V_{\text{lead}} = \rho_{\text{aluminum}} \cdot V_{\text{aluminum}} \]
02

Express each volume in terms of radius

Since the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \), substitute for each material:- For lead: \( V_{\text{lead}} = \frac{4}{3} \pi r_{\text{lead}}^3 \)- For aluminum: \( V_{\text{aluminum}} = \frac{4}{3} \pi r_{\text{aluminum}}^3 \)
03

Set up the equation for equal mass and simplify

Since both spheres have the same mass, \[ \rho_{\text{lead}} \cdot \frac{4}{3} \pi r_{\text{lead}}^3 = \rho_{\text{aluminum}} \cdot \frac{4}{3} \pi r_{\text{aluminum}}^3 \] Simplify the equation by cancelling \( \frac{4}{3} \pi \) from both sides: \[ \rho_{\text{lead}} \cdot r_{\text{lead}}^3 = \rho_{\text{aluminum}} \cdot r_{\text{aluminum}}^3 \]
04

Solve for the radius ratio

Re-arrange the equation to solve for the ratio \( \frac{r_{\text{aluminum}}}{r_{\text{lead}}} \): \[ \left(\frac{r_{\text{aluminum}}}{r_{\text{lead}}}\right)^3 = \frac{\rho_{\text{lead}}}{\rho_{\text{aluminum}}} \] To find the ratio of the radii, take the cube root: \[ \frac{r_{\text{aluminum}}}{r_{\text{lead}}} = \sqrt[3]{\frac{\rho_{\text{lead}}}{\rho_{\text{aluminum}}}} \]
05

Substitute densities and calculate the cube root

The density of lead is approximately \( 11.34 \, \text{g/cm}^3 \), and the density of aluminum is approximately \( 2.70 \, \text{g/cm}^3 \). Substitute these values into the equation: \[ \frac{r_{\text{aluminum}}}{r_{\text{lead}}} = \sqrt[3]{\frac{11.34}{2.70}} \] Calculate the ratio: \[ \frac{r_{\text{aluminum}}}{r_{\text{lead}}} \approx \sqrt[3]{4.2} \approx 1.61 \]
06

Conclusion

Therefore, the ratio of the radius of the aluminum sphere to the radius of the lead sphere is approximately \( 1.61 \).

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 Mass
Mass is a fundamental concept in physics that describes how much matter is present in an object. It is commonly measured in grams (g) or kilograms (kg), and it reflects the amount of substance in an object.
  • Mass is a scalar quantity, meaning it has magnitude but no direction.
  • It remains constant regardless of location, unlike weight, which can change with gravitational pull.
  • In our problem, the mass of the two spheres, one made of aluminum and the other of lead, is the same.
This means that both spheres contain the same total amount of matter, irrespective of their size or density differences.
Exploring Volume
Volume is a measure of the space an object occupies. For three-dimensional objects like spheres, it is typically measured in cubic centimeters (cm鲁) or liters. The volume of a sphere can be calculated using the formula:\[ V = \frac{4}{3} \pi r^3 \]where \( r \) is the radius of the sphere. The formula shows that volume depends strongly on the radius.
  • The larger the radius, the larger the volume.
  • A small increase in the radius can lead to a significant increase in volume because the radius is cubed in the formula.
In the exercise, understanding volume is crucial for calculating how different densities affect the size of the spheres needed to have the same mass.
Significance of Sphere Radius
The radius is a key dimension of a sphere. It is the distance from the center of the sphere to any point on its surface. The radius has a profound impact on other properties of a sphere, such as its volume.In our exercise, both spheres have the same mass but are made of different materials with different densities:
  • Higher density materials like lead result in smaller spheres for the same mass.
  • Lower density materials like aluminum require larger radii to achieve the same mass.
The exercise tasked us with finding the ratio between these radii. Using the density values provided, and knowing the mass remains constant, we derived the ratio\[ \frac{r_{\text{aluminum}}}{r_{\text{lead}}} = \sqrt[3]{\frac{\rho_{\text{lead}}}{\rho_{\text{aluminum}}}} \]This calculation underscores how the density differences influence the sphere's form to maintain equivalent mass.

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 slab of ice floats on a freshwater lake. What minimum volume must the slab have for a 45.0 -kg woman to be able to stand on it without getting her feet wet?

Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water completely fills the pipe. (a) At one point in the pipe the radius is 0.150 \(\mathrm{m}\) . What is the speed of the water at this point if water is flowing into this pipe at a steady rate of 1.20 \(\mathrm{m}^{3} / \mathrm{s} ?\) (b) At a second point in the pipe the water speed is 3.80 \(\mathrm{m} / \mathrm{s} .\) What is the radius of the pipe at this point?

A sealed tank containing seawater to a height of 11.0 m also contains air above the water at a gauge pressure of 3.00 atm. Water flows out from the bottom through a small hole. How fast is this water moving?

An incompressible fluid with density \(\rho\) is in a horizontal test tube of inner cross-sectional area \(A .\) The test tube spins in a horizontal circle in an ultracentrifuge at an angular speed \omega. Gravitational forces are negligible. Consider a volume element of the fluid of area \(A\) and thickness \(d r^{\prime}\) a distance \(r^{\prime}\) from the rotation axis. The pressure on its inner surface is \(p\) and on its outer surface is \(p+d p .\) (a) Apply Newton's second law to the volume element to show that \(d p=\rho \omega^{2} r^{\prime} d r^{\prime}\) . (b) If the surface of the fluid is at a radius \(r_{0}\) where the pressure is \(p_{0},\) show that the pressure \(p\) at a distance \(r \geq r_{0}\) is \(p=p_{0}+\rho \omega^{2}\left(r^{2}-r_{0}^{2}\right) / 2 .\) (c) An object of volume \(V\) and density \(\rho_{\mathrm{ob}}\) has its center of mass at a distance \(R_{\mathrm{cmob}}\) from the axis. Show that the net horizontal force on the object is \(\rho V \omega^{2} R_{\mathrm{cm}},\) where \(R_{\mathrm{cm}}\) is the distance from the axis to the center of mass of the displaced fluid. (d) Explain why the object will move inward if \(\rho R_{\mathrm{cm}}>\rho_{\mathrm{ob}} R_{\mathrm{cmob}}\) and outward if \(\rho R_{\mathrm{cm}}<\rho_{\mathrm{ob}} R_{\mathrm{cmob}} .\) (e) For small objects of uniform density, \(R_{\mathrm{cm}}=R_{\mathrm{cmob}}\) . What happens to a mixture of small objects of this kind with different densities in an ultracentrifuge?

Untethered helium balloons, floating in a car that has all the windows rolled up and outside air vents closed, move in the direction of the car's acceleration, but loose balloons filled with air move in the opposite direction. To show why, consider only the horizontal forces acting on the balloons. Let \(a\) be the magnitude of the car's forward acceleration. Consider a horizontal tube of air with a cross-sectional area \(A\) that extends from the windshield, where \(x=0\) and \(p=p_{0},\) back along the \(x\) -axis. Now consider a volume element of thickness \(d x\) in this tube. The pressure on its front surface is \(p\) and the pressure on its rear surface is \(p+d p .\) Assume the air has a constant density \(\rho .\) (a) Apply Newton's second law to the volume element to show that \(d p=\rho a d x .\) (b) Integrate the result of part (a) to find the pressure at the front surface in terms of \(a\) and \(x\) . (c) To show that considering \(\rho\) constant is reasonable, calculate the pressure difference in atm for a distance as long as 2.5 \(\mathrm{m}\) and a large acceleration of 5.0 \(\mathrm{m} / \mathrm{s}^{2}\) . (d) Show that the net horizontal force on a balloon of volume \(V\) is oVa. (e) For negligible friction forces, show that the acceleration of the balloon (average density \(\rho_{\text { bal }}\) is \(\left(\rho / \rho_{\text { bal }}\right) a,\) so that the acceleration relative to the car is \(a_{\mathrm{rcl}}=\left[\left(\rho / \rho_{\mathrm{bal}}\right)-1\right] a\) (f) Use the expression for \(a_{\mathrm{rcl}}\) in part (e) to explain the movement of the balloons.
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Force

Read Explanation

Electromagnetism

Read Explanation

Thermodynamics

Read Explanation

Kinematics Physics

Read Explanation

Wave Optics

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.