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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 aluminum sphere's radius to the lead sphere's radius is approximately 1.61.

Step by step solution

01

Understand the Problem

We are given two spheres, one made of lead and the other made of aluminum, both having equal mass. We need to find the ratio of their radii.
02

Identify Relevant Formulas

The volume of a sphere is given by the formula \( V = \frac{4}{3}\pi r^3 \). Mass is equal to density times volume, given by \( m = \rho V \). Since the masses are equal, we have \( \rho_{lead} V_{lead} = \rho_{aluminum} V_{aluminum} \).
03

Express Volumes in Terms of Radii

Using the sphere volume formula, express the volumes for lead and aluminum spheres:- \( V_{lead} = \frac{4}{3}\pi r_{lead}^3 \)- \( V_{aluminum} = \frac{4}{3}\pi r_{aluminum}^3 \)
04

Set Up the Mass Equation

Since the masses are the same, set their equations equal:\( \rho_{lead} \cdot \frac{4}{3} \pi r_{lead}^3 = \rho_{aluminum} \cdot \frac{4}{3} \pi r_{aluminum}^3 \).
05

Simplify the Equation

Cancel out the common terms \( \frac{4}{3}\pi \):\( \rho_{lead} r_{lead}^3 = \rho_{aluminum} r_{aluminum}^3 \).
06

Solve for the Radius Ratio

Rearrange the simplified equation to find the radius ratio:\( \left(\frac{r_{aluminum}}{r_{lead}}\right)^3 = \frac{\rho_{lead}}{\rho_{aluminum}} \).
07

Calculate the Ratio Using Densities

Use the known densities of lead \( \rho_{lead} \approx 11340 \, \text{kg/m}^3 \) and aluminum \( \rho_{aluminum} \approx 2700 \, \text{kg/m}^3 \) to calculate:\( \left(\frac{r_{aluminum}}{r_{lead}}\right)^3 = \frac{11340}{2700} \).Simplify to find:\( \left(\frac{r_{aluminum}}{r_{lead}}\right)^3 = 4.2 \).
08

Find the Final Radius Ratio

Take the cube root to find the final ratio:\( \frac{r_{aluminum}}{r_{lead}} = \sqrt[3]{4.2} \). Using approximation, this gives \( \frac{r_{aluminum}}{r_{lead}} \approx 1.61 \).

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Key Concepts

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

Sphere Volume Calculation
To understand the calculation of a sphere's volume, imagine a perfectly round object like a basketball. The volume of this sphere is the space it occupies, which can be mathematically described using a specific formula. The formula to calculate the volume of a sphere is \( V = \frac{4}{3}\pi r^3 \), where \( V \) represents volume and \( r \) is the radius of the sphere. This formula is derived considering that a sphere is a symmetrical three-dimensional shape.
  • \( \pi \) is a constant, approximately equal to 3.14159.
  • \( r^3 \) means we multiply the radius by itself twice, showing how the volume increases rapidly with the growth of radius.
Calculating this volume is crucial, especially when comparing the volumes of spheres made from different materials, such as in physics problems and engineering applications.
Density and Mass Relationship
Density is a measure of how much mass is contained in a given volume. It's like understanding how 'heavy' something feels for its size. The formula to represent this relationship is \( \rho = \frac{m}{V} \), where \( \rho \) is density, \( m \) is mass, and \( V \) is volume.
  • This relationship is pivotal in physics as it helps to identify how compact a material is.
  • A higher density means more mass in a smaller volume, whereas a lower density implies lighter mass in the same volume.
When the mass remains constant, the volume and the density are inversely related. This means, if the volume increases, the density decreases, assuming the mass doesn't change. In our problem, both spheres made of different materials have the same mass, but differing volumes due to their material's differing densities.
Material Density Comparison
When comparing two materials, such as lead and aluminum, their densities serve as a critical point of comparison. Lead is denser than aluminum, meaning for the same mass, lead takes up less space than aluminum. Here, we used the densities of lead \( \rho_{lead} \approx 11340 \text{ kg/m}^3 \) and aluminum \( \rho_{aluminum} \approx 2700 \text{ kg/m}^3 \).
  • The higher the density, the smaller the volume for the same mass.
  • The equation \( \rho_{lead} V_{lead} = \rho_{aluminum} V_{aluminum} \) signifies equal masses but differing volumes.
This comparison enabled us to find the radius ratio of aluminum to lead spheres, resulting in \( \frac{r_{aluminum}}{r_{lead}} \approx 1.61 \). This ratio shows that the aluminum sphere needs a larger radius to reach the same mass as the smaller lead sphere due to its lower density.

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Most popular questions from this chapter

The densities of air, helium, and hydrogen (at \(p\) \(=\) 1.0 atm and \(T\) \(=\) 20\(^\circ\)C) are 1.20 kg/m\(^3\), 0.166 kg/m\(^3\), and 0.0899 kg/m\(^3\), respectively. (a) What is the volume in cubic meters displaced by a hydrogen- filled airship that has a total "lift" of 90.0 kN? (The "lift" is the amount by which the buoyant force exceeds the weight of the gas that fills the airship.) (b) What would be the "lift" if helium were used instead of hydrogen? In view of your answer, why is helium used in modern airships like advertising blimps?

The lower end of a long plastic straw is immersed below the surface of the water in a plastic cup. An average person sucking on the upper end of the straw can pull water into the straw to a vertical height of 1.1 m above the surface of the water in the cup. (a) What is the lowest gauge pressure that the average person can achieve inside his lungs? (b) Explain why your answer in part (a) is negative.

A swimming pool is 5.0 m long, 4.0 m wide, and 3.0 m deep. Compute the force exerted by the water against (a) the bottom and (b) either end. (\(Hint:\) Calculate the force on a thin, horizontal strip at a depth h, and integrate this over the end of the pool.) Do not include the force due to air pressure.

Black smokers are hot volcanic vents that emit smoke deep in the ocean floor. Many of them teem with exotic creatures, and some biologists think that life on earth may have begun around such vents. The vents range in depth from about 1500 m to 3200 m below the surface. What is the gauge pressure at a 3200-m deep vent, assuming that the density of water does not vary? Express your answer in pascals and atmospheres.

(a) As you can tell by watching them in an aquarium, fish are able to remain at any depth in water with no effort. What does this ability tell you about their density? (b) Fish are able to inflate themselves using a sac (called the \(swim\) \(bladder\)) located under their spinal column. These sacs can be filled with an oxygen\(-\)nitrogen mixture that comes from the blood. If a 2.75-kg fish in freshwater inflates itself and increases its volume by 10%, find the \(net\) force that the \(water\) exerts on it. (c) What is the net \(external\) force on it? Does the fish go up or down when it inflates itself?
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