A waste treatment pond is 50 meters long, 15 meters wide, and has an average depth of 2 meters. The density of the waste is 85.3 lbm/ft³. Calculate the weight of the pond contents in $1{0}^{4}10^4$ kilograms.

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Calculate the Volume of the Pond:The dimensions of the pond are:

The volume $VV$ of the pond is:

$V=length\times width\times depthV\; =\; \backslash text\{length\}\; \backslash times\; \backslash text\{width\}\; \backslash times\; \backslash text\{depth\}$

$V=50\hspace{0.17em}m\times 15\hspace{0.17em}m\times 2\hspace{0.17em}mV\; =\; 50\; \backslash ,\; \backslash text\{m\}\; \backslash times\; 15\; \backslash ,\; \backslash text\{m\}\; \backslash times\; 2\; \backslash ,\; \backslash text\{m\}$

$V=1500\hspace{0.17em}{3}^{}V\; =\; 1500\; \backslash ,\; \backslash text\{m\}^3$

Convert Density to kg/m³:The given density of the waste is $85.3\hspace{0.17em}{3}^{}85.3\; \backslash ,\; \backslash text\{lbm/ft\}^3$85.3. We need to convert this to kg/m³.

${3}^{}=85.3\hspace{0.17em}{3}^{}\times 0.45359237\hspace{0.17em}kg/lbm\xf70.0283168\hspace{0.17em}{3}^{}\mathrm{/}{3}^{}\backslash text\{Density\; in\; kg/m\}^3\; =\; 85.3\; \backslash ,\; \backslash text\{lbm/ft\}^3\; \backslash times\; 0.45359237\; \backslash ,\; \backslash text\{kg/lbm\}\; \backslash div\; 0.0283168\; \backslash ,\; \backslash text\{m\}^3/\backslash text\{ft\}^3$

${3}^{}=85.3\times 0.45359237\xf70.0283168\backslash text\{Density\; in\; kg/m\}^3\; =\; 85.3\; \backslash times\; 0.45359237\; \backslash div\; 0.0283168$

${3}^{}\approx 1367.89\hspace{0.17em}{3}^{}\backslash text\{Density\; in\; kg/m\}^3\; \backslash approx\; 1367.89\; \backslash ,\; \backslash text\{kg/m\}^3$

Calculate the Mass of the Pond Contents:$Mass=Volume\times Density\backslash text\{Mass\}\; =\; \backslash text\{Volume\}\; \backslash times\; \backslash text\{Density\}$

$Mass=1500\hspace{0.17em}{3}^{}\times 1367.89\hspace{0.17em}{3}^{}\backslash text\{Mass\}\; =\; 1500\; \backslash ,\; \backslash text\{m\}^3\; \backslash times\; 1367.89\; \backslash ,\; \backslash text\{kg/m\}^3$

$Mass\approx 2051835\hspace{0.17em}kg\backslash text\{Mass\}\; \backslash approx\; 2051835\; \backslash ,\; \backslash text\{kg\}$

Convert Mass to $1{0}^{4}10^4$104 kg:$Massin1{0}^{4}\hspace{0.17em}kg=\frac{2051835\hspace{0.17em}kg}{1{0}^{4}}\backslash text\{Mass\; in\; \}\; 10^4\; \backslash ,\; \backslash text\{kg\}\; =\; \backslash frac\{2051835\; \backslash ,\; \backslash text\{kg\}\}\{10^4\}$ $Massin1{0}^{4}\hspace{0.17em}kg\approx 205.18\backslash text\{Mass\; in\; \}\; 10^4\; \backslash ,\; \backslash text\{kg\}\; \backslash approx\; 205.18$

Answer:The weight of the pond contents is approximately $205.18\times 1{0}^{4}205.18\; \backslash times\; 10^4$205.18×104 kilograms.