Solution Exercise 12
[latex]t_{0} = 1x10^{- 4}\ days;8.64\ seconds[/latex]
[latex]\mathrm{\Delta}s\ = \ 14\ m[/latex]
[latex]T\ =[/latex] [latex]\frac{2.3\ \ Q}{4\ \pi\ \Delta s} = \frac{2.3\ \ \left( 0.00347\frac{m^{3}}{s} \right)}{4\ \pi\ 14}\ = \ 4.54\ x\ 10^{- 5}\ \frac{m^{2}}{s}[/latex]
[latex]S\ = \frac{2.25\ \ T\ \ t_{0}}{r^{2}} = \frac{2.25\ \ \left( 4.54\ x\ 10^{- 5}\ \frac{m^{2}}{s} \right)\ \ 8.64\ seconds}{{(2\ m)}^{2}} = \ 2.2x10^{- 4}[/latex]