/*! 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 7 A cube of compressible material ... [FREE SOLUTION] | 91Ó°ÊÓ

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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
New density is \(8\rho\) when sides are halved; sides must be \(\frac{L}{\sqrt[3]{3}}\) for the density to triple.

Step by step solution

01

Understanding the problem

We are given a cube with an initial side length of \(L\) and density \(\rho\). We need to find out the new density if the side length is compressed to half while keeping the mass constant and also determine the side length if the density needs to triple, still keeping the mass constant.
02

Relate density to volume and mass

The density of a cube \(\rho\) is given by \(\rho = \frac{m}{V}\), where \(m\) is the mass and \(V\) is the volume. Since volume \(V\) of a cube is \(L^3\), we have \(\rho = \frac{m}{L^3}\).
03

Calculate new volume

When each side is compressed to half its original length \(L\), the new side length is \(\frac{L}{2}\). The new volume \(V'\) is \((\frac{L}{2})^3 = \frac{L^3}{8}\).
04

New density in compressed cube

The mass \(m\) remains constant, so the new density \(\rho'\) is \(\rho' = \frac{m}{V'} = \frac{m}{\frac{L^3}{8}} = 8\frac{m}{L^3}\). Since \(\rho = \frac{m}{L^3}\), we have \(\rho' = 8\rho\).
05

Condition for density being three times

We want the new density \(\rho' = 3\rho\), keeping the mass constant. Thus, \(3 \rho = \frac{m}{L'^3}\), where \(L'\) is the new side length.
06

Solve for new side length

We equate \(3 \rho = \frac{m}{L'^3}\) and solve for \(L'\):\(3 \rho = \frac{m}{L'^3} = \frac{\rho L^3}{L'^3}\) \(L'^3 = \frac{L^3}{3}\)\(L' = \left(\frac{L^3}{3}\right)^{1/3} = \frac{L}{\sqrt[3]{3}}\).

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Key Concepts

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

Volume
Understanding the concept of volume is crucial when dealing with solid objects like a cube. Volume determines how much space an object occupies. For a cube, volume is calculated by multiplying the length of its sides. If all sides are of equal length (which is the case for a standard cube), then the volume is given by:
  • Volume = side length × side length × side length = \( L^3 \).
When it's about compressible materials, knowing how volume changes with alteration in dimensions helps us analyze how properties like density will be affected. If you compress a cube by changing its side length, you're effectively altering its volume, which in turn changes the density if the mass stays constant.
In our specific exercise, the change in volume happens when each dimension is reduced to half, diminishing the original volume to one-eighth of its initial value.
Compressible materials
Compressible materials like Styrofoam or balsa wood can significantly change in volume when pressure is applied. This is due to their ability to contract and take up less space without losing mass. In scientific terms, when we talk about compressible materials, we consider the effect on their density and structural properties when dimensions are altered.
For such materials, compressing to half the side length of a cube means decreasing each side from \( L \) to \( \frac{L}{2} \). This in turn reduces the cube's volume massively, while maintaining the same mass. Materials that are not compressible would not shrink in such a manner. For this reason, they are ideal for studying concepts like density changes due to compression, allowing us to relate theoretical calculations with actual manipulation of materials.
Cube
A cube is a three-dimensional geometric shape with six matching square faces. When working with cube-related problems, it's useful to recognize that each of these surfaces adds up to define properties like volume and surface area.
The uniformity in shape simplifies calculations, as seen with volume \( V = L^3 \) and surface area \( A = 6L^2 \) for a cube. Cubes are useful in physics and engineering problems because of their symmetry and the simplicity they introduce in mathematical treatments.
  • In exercises like the one provided, adjusting a cube's side lengths impacts not just its volume, but its density too.
Changing these dimensions amidst real-world constraints, like keeping mass constant, brings focus to conserving and understanding material properties.
Mass conservation
Mass conservation is a principle stating that the mass of a closed system remains constant, regardless of changes inside the system. This concept is critical when analyzing real-world situations involving compressible materials.
In our exercise, while the volume of the cube changes with compression, the mass remains the same. This consistency helps us derive the new density \( \rho' \). By ensuring mass doesn't change, we see that to keep density as precise, adjustments in the volume must be compensated accordingly.
  • The calculation involves recognizing that density \( \rho = \frac{m}{V} \) adapts based on volume changes while conserving mass.
This principle simplifies the derivation of the new side lengths necessary to adjust to any required density, like needing the density to be triple.

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