Question
A trash compactor can reduce the volume of its contents to 0.350 their original value. Neglecting the mass of air expelled, by what factor is the density of the rubbish increased?

$2.86$

Solution Video

# OpenStax College Physics Solution, Chapter 11, Problem 7 (Problems & Exercises) (1:09)

View sample solution

## Calculator Screenshots

Video Transcript

This is College Physics Answers with Shaun Dychko. We're told that a trash compactor reduces the volume of its contents to 0.35 times the original volume. So the density in the first case before compaction is the mass divided by the initial volume <i>V one</i>. So the density in the first case before compaction is the mass divided by the initial volume <i>V one</i>. Then after compaction we have density two which equals the same mass divided by a new volume <i>V two</i>. We're told that <i>V two</i> is 0.35 times <i>V one</i>. So we take the ratio of the densities, <i>rho two</i> divided by <i>rho one</i>, that is <i>m</i> over <i>V two</i> which is <i>rho two</i>, divided by <i>m</i> over <i>V one</i> but I didn't write divide by because I don't like dividing a fraction by a fraction. So instead I'm multiplying by the reciprocal of density one, doing that here. So mass over <i>V two</i> time <i>V one</i> over <i>m</i> and that equals <i>V one</i> over <i>V two</i> because the <i>m</i>'s cancel. Now <i>V two</i>, we can make a substitution for it here. It is 0.35 times <i>V one</i> and so this ends up being one over 0.35 which is 2.86. So the density in the second case is greater than the density in the first case by a factor of 2.86.