A hydraulic jack has an input cylinder diameter of 3.2 in and an output cylinder diameter of 9.2 in. an input force of 485 lb is observed to produce an output force of 2800 lb. Calculate the efficiency

$\begin{gathered} F1=485\text{ lb} \\ r1=(3.2in)/(2)=1.6in \\ F2=2800lb \\ r2=(9.2in)/(2)=4.6\text{ in} \\ efficiency=? \\ First\text{ Pascal to find the output force} \\ (F1)/(A1)=(F_(out))/(A2) \\ F_(out)=(A2F1)/(A1) \\ \\ F_(out)=(\pi(r2)^2F1)/(\pi(r1)^2) \\ \\ F_(out)=((r2)^2F1)/((r1)^2) \\ \\ F_(out)=((4.6in)^2(485lb))/((1.6in)^2) \\ \\ f_(out)=4008,8\text{ lb} \\ \\ efficiency=(2800lb)/(4008,8lb)*100 \\ \\ efficiency=69.84\text{ \%} \\ The\text{ efficiency is 69.84 \%} \\ \end{gathered}$

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