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Sagot :
hello
the relationship between the time and the area of the pipe is an inversely proportional one
[tex]\begin{gathered} t\propto\frac{1}{a} \\ t=\frac{k}{a} \\ k=ta \\ t_1a_1=t_2a_2 \end{gathered}[/tex][tex]\begin{gathered} a_1=113in^2 \\ t_1=6.4h \\ a_2=50.25in^2 \\ t_2=\text{ ?} \end{gathered}[/tex]above are the value of data given in the question
we can now proceed to solve this problem using the formula we've already established.
[tex]\begin{gathered} t_1a_1=t_2a_2 \\ 113\times6.4=t_2\times50.25 \\ 723.2=50.25t_2 \\ \text{divide both sides by the coefficient of t\_2} \\ \frac{723.2}{50.25}=\frac{50.25t_2}{50.25} \\ t_2=14.39h \end{gathered}[/tex]from the calculations above, decreasing the cross-sectional area to 50.25sq in, would increase the time it takes for the tank to empty to 14.39 hours
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