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A plastic cube of density [tex] \rho [/tex] is compressed so that each of its sides becomes one-half of its original length. What is the new density of the material of the cube?

A. [tex]0.125 \rho[/tex]
B. [tex]0.50 \rho[/tex]
C. [tex]2.0 \rho[/tex]
D. [tex]8.0 \rho[/tex]

Sagot :

Sure, let's work through the problem step-by-step.

1. Initial State of the Cube:
- Let's denote the initial side length of the cube as [tex]\( s \)[/tex].
- The initial volume [tex]\( V \)[/tex] of the cube is given by [tex]\( V = s^3 \)[/tex].
- The initial density of the cube is [tex]\( p \)[/tex].

2. Volume Compression:
- Each side of the cube becomes one-half of its original length after compression. So, the new side length [tex]\( s' \)[/tex] is [tex]\( s' = \frac{s}{2} \)[/tex].
- The new volume [tex]\( V' \)[/tex] of the cube after compression is given by:
[tex]\[ V' = (s')^3 = \left(\frac{s}{2}\right)^3 = \frac{s^3}{8} \][/tex]

3. Density Calculation:
- Density is defined as mass per unit volume. The mass [tex]\( m \)[/tex] of the cube remains constant during compression.
- The initial mass of the cube can be expressed as:
[tex]\[ m = \text{Initial Density} \times \text{Initial Volume} = p \times s^3 \][/tex]
- The new density [tex]\( p' \)[/tex] is the mass divided by the new volume:
[tex]\[ p' = \frac{m}{V'} = \frac{p \times s^3}{\frac{s^3}{8}} = p \times \frac{8s^3}{s^3} = 8p \][/tex]

So, the new density of the material of the cube, after compression, is [tex]\( 8.0p \)[/tex].

Therefore, the correct answer is D 8.0p.