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A cube of compressible material (such as Styrofoam or balsa wood) has a density \(\rho\) and sides of length \(L .\) (a) If you keep its mass the same, but compress each side to half its length, what will be its new density, in terms of \(\rho ?\) (b) If you keep the mass and shape the same, what would the length of each side have to be (in terms of \(L )\) so that the density of the cube was three times its original value?

Short Answer

Expert verified
(a) New density: \(8\rho\); (b) New side length: \(\frac{L}{\sqrt[3]{3}}\).

Step by step solution

01

Understanding the Problem

We have a cube with initial side length \( L \) and density \( \rho \). We need to find its new density when all its sides are compressed to half the length, keeping mass constant.
02

Calculate Initial Mass and Volume

The initial volume of the cube is \( L^3 \). The initial mass is \( \rho L^3 \), since density \( \rho = \frac{mass}{volume} \).
03

Determine New Volume After Compression

Each side of the cube is compressed to half: \( L/2 \). Therefore, the new volume is \( (L/2)^3 = \frac{L^3}{8} \).
04

Find New Density for Compressed Cube

Because the mass remains constant, the new density \( \rho_{new} \) is given by: \( \rho_{new} = \frac{mass}{new \, volume} = \frac{\rho L^3}{\frac{L^3}{8}} = 8\rho \).
05

Understanding the Second Problem

For the second part, the density needs to be three times the original density (\( 3\rho \)), keeping mass constant.
06

Establish Equation for Desired Density

We want \( 3\rho = \frac{mass}{(side \, length)^3} = \frac{\rho L^3}{x^3} \). Simplifying, \( x^3 = \frac{L^3}{3} \).
07

Solve for New Side Length

Taking the cube root, \( x = \left( \frac{L^3}{3} \right)^{1/3} = \frac{L}{\sqrt[3]{3}} \). This is the length of each side for the density to be three times the original.

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Key Concepts

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

Compressible Materials
Compressible materials, such as Styrofoam and balsa wood, have the unique property of changing their volume when subjected to pressure, while maintaining their mass. This physical characteristic is essential in scenarios where materials need to be reshaped or fitted into different spaces without altering their foundational weight.
Compressibility refers to the ability of a material to experience a change in volume under pressure. This can be observed when squeezing a sponge or foam rubber, which reduce their size but do not lose any of their material substance.
In the context of our exercise, compressing a cube means decreasing its side lengths while the amount of material (its mass) remains the same. If we keep the mass of the material constant, any change in volume due to compression will result in a change in density. This is because density is calculated as mass divided by volume. So, when the volume decreases due to compression, the density increases, assuming mass conservation.
Volume Calculation
Calculating volume, especially for simple shapes like cubes, is straightforward. For a cube with side length, the volume is given by the formula:
  • Volume = side length × side length × side length
  • Or, mathematically: Volume = \( L^3 \)
When compressing a cube's side length to half, each side becomes \( \frac{L}{2} \).
  • Thus, the new volume is \( \left(\frac{L}{2}\right)^3 = \frac{L^3}{8} \)
The decrease in volume demonstrates how shape modification (without mass loss) affects the density. Understanding volume calculation is critical, as it directly influences how we perceive and measure density in physical science and engineering.
In the example above, compressing the object led to its volume being reduced by a factor of eight. This calculation helps us derive the new density values when compressible materials alter their shape.
Mass Conservation
Mass conservation is a foundational principle in physics, stating that mass cannot be created or destroyed within an isolated system, only rearranged. In our exercise, the mass of the compressible material remains unchanged, even when its volume changes.
Maintaining constant mass is essential when calculating changes in other properties, such as density. Density is defined as mass divided by volume (\( \rho = \frac{mass}{volume} \)). Therefore, if the volume of an object decreases while its mass remains constant, the density increases proportionally.
For instance:
  • Before compression, the cube’s mass is \( \rho L^3 \)
  • Post-compression the mass remains the same, but the volume changes to \( \frac{L^3}{8} \).
  • Therefore, the new density is calculated as: \( 8\rho \), showing that density has increased eight-fold.
This principle ensures that changes in volume due to external forces will adjust the object's density accordingly while preserving the total mass, a critical understanding in many scientific applications.

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Most popular questions from this chapter

Exploring Europa's oceans. Europa, a satellite of Jupiter, appears to have an ocean beneath its icy surface. Proposals have been made to send a robotic submarine to Europa to see if there might be life there. There is no atmosphere on Europa, and we shall assume that the surface ice is thin enough that we can neglect its weight and that the oceans are fresh water having the same density as on the earth. The mass and diameter of Europa have been measured to be \(4.78 \times 10^{22}\) kg and 3130 \(\mathrm{km}\) , respectively. (a) If the submarine intends to submerge to a depth of \(100 \mathrm{m},\) what pressure must it be designed to withstand? (b) If you wanted to test this submarine before sending it to Europa, how deep would it have to go in our oceans to experience the same pressure as the pressure at a depth of 100 \(\mathrm{m}\) on Europa?

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