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# 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?

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Here a trash compactor reduces the volume of its contents to 0.350 their original value.

Let the original volume of the rubbish is $V_{\text{i}}$, then its volume after compression is

\begin{align*} V_{\text{f}} = 0.350\,V_{\text{i}} \end{align*}

Now as we know the density is the mass per unit volume of a substance or object. In equation form, density is defined as

\begin{align*} \rho = \frac{m}{V}\tag{1} \end{align*}

Where $m$ is the mass and $V$ is the volume of the substance or the object.

Hence for a constant mass

\begin{align*} \rho \propto \frac{1}{V} \end{align*}

Therefore the original density of the rubbish is

\begin{align*} \rho_{\text{i}} \propto \frac{1}{V_{\text{i}}}\tag{2} \end{align*}

And the final density of the rubbish is

\begin{align*} \rho_{\text{f}} \propto \frac{1}{V_{\text{f}}}\tag{3} \end{align*}

So from equations (2) and (3), we get

\begin{align*} \frac{\rho_{\text{f}}}{\rho_{\text{i}}} & = \frac{V_{\text{i}}}{V_{\text{f}}}\\ & = \frac{V_{\text{i}}}{0.350\,V_{\text{i}}}\\ & = 2.86 \end{align*}

Therefore, the density of the rubbish is

\begin{align*} \rho_{\text{f}} = 2.86\,\rho_{\text{i}} \end{align*}

Therefore the factor by which the density is increased

\begin{align*} F = 2.86 \end{align*}

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