Convert foot-pound • pound force to cubic foot of atmosphere
Learn how to convert
1
foot-pound • pound force to
cubic foot of atmosphere
step by step.
Calculation Breakdown
Set up the equation
\(1.0\left(foot-pound \times pound \text{ } force\right)={\color{rgb(20,165,174)} x}\left(cubic \text{ } foot \text{ } of \text{ } atmosphere\right)\)
Define the base values of the selected units in relation to the SI unit \(\left(joule\right)\)
\(\text{Left side: 1.0 } \left(foot-pound \times pound \text{ } force\right) = {\color{rgb(89,182,91)} 1.3558179483314\left(joule\right)} = {\color{rgb(89,182,91)} 1.3558179483314\left(J\right)}\)
\(\text{Right side: 1.0 } \left(cubic \text{ } foot \text{ } of \text{ } atmosphere\right) = {\color{rgb(125,164,120)} 2869.2044809344\left(joule\right)} = {\color{rgb(125,164,120)} 2869.2044809344\left(J\right)}\)
Insert known values into the conversion equation to determine \({\color{rgb(20,165,174)} x}\)
\(1.0\left(foot-pound \times pound \text{ } force\right)={\color{rgb(20,165,174)} x}\left(cubic \text{ } foot \text{ } of \text{ } atmosphere\right)\)
\(\text{Insert known values } =>\)
\(1.0 \times {\color{rgb(89,182,91)} 1.3558179483314} \times {\color{rgb(89,182,91)} \left(joule\right)} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} {\color{rgb(125,164,120)} 2869.2044809344}} \times {\color{rgb(125,164,120)} \left(joule\right)}\)
\(\text{Or}\)
\(1.0 \cdot {\color{rgb(89,182,91)} 1.3558179483314} \cdot {\color{rgb(89,182,91)} \left(J\right)} = {\color{rgb(20,165,174)} x} \cdot {\color{rgb(125,164,120)} 2869.2044809344} \cdot {\color{rgb(125,164,120)} \left(J\right)}\)
\(\text{Cancel SI units}\)
\(1.0 \times {\color{rgb(89,182,91)} 1.3558179483314} \cdot {\color{rgb(89,182,91)} \cancel{\left(J\right)}} = {\color{rgb(20,165,174)} x} \times {\color{rgb(125,164,120)} 2869.2044809344} \times {\color{rgb(125,164,120)} \cancel{\left(J\right)}}\)
\(\text{Conversion Equation}\)
\(1.3558179483314 = {\color{rgb(20,165,174)} x} \times 2869.2044809344\)
Switch sides
\({\color{rgb(20,165,174)} x} \times 2869.2044809344 = 1.3558179483314\)
Isolate \({\color{rgb(20,165,174)} x}\)
Multiply both sides by \(\left(\dfrac{1.0}{2869.2044809344}\right)\)
\({\color{rgb(20,165,174)} x} \times 2869.2044809344 \times \dfrac{1.0}{2869.2044809344} = 1.3558179483314 \times \dfrac{1.0}{2869.2044809344}\)
\(\text{Cancel}\)
\({\color{rgb(20,165,174)} x} \times {\color{rgb(255,204,153)} \cancel{2869.2044809344}} \times \dfrac{1.0}{{\color{rgb(255,204,153)} \cancel{2869.2044809344}}} = 1.3558179483314 \times \dfrac{1.0}{2869.2044809344}\)
\(\text{Simplify}\)
\({\color{rgb(20,165,174)} x} = \dfrac{1.3558179483314}{2869.2044809344}\)
Solve \({\color{rgb(20,165,174)} x}\)
\({\color{rgb(20,165,174)} x}\approx0.0004725414\approx4.7254 \times 10^{-4}\)
\(\text{Conversion Equation}\)
\(1.0\left(foot-pound \times pound \text{ } force\right)\approx{\color{rgb(20,165,174)} 4.7254 \times 10^{-4}}\left(cubic \text{ } foot \text{ } of \text{ } atmosphere\right)\)
