Solution Exercise 13
[latex]r_{0} = 2,500\ m[/latex]
[latex]\mathrm{\Delta}s\ = \ 21.5\ \ m[/latex]
[latex]T\ =[/latex] [latex]\frac{2.3\ \ Q}{4\ \pi\ \Delta s} = \frac{2.3\ \ \left( 0.00231\frac{m^{3}}{s} \right)}{4\ \pi\ 21.5\ m}\ = 2.0\ x\ 10^{- 5}\frac{m^{2}}{s}[/latex]
[latex]S\ = \frac{2.25\ \ T\ \ t}{{r_{0}}^{2}} = \frac{2.25\ \ \left( 2.0\ x\ 10^{- 5}\ \frac{m^{2}}{s} \right)\ \ 86,400\ s}{{(2,500\ m)}^{2}} = \ 6.1x10^{- 7}[/latex]