Answer
$b=1100Kg/s$
Work Step by Step
We know that
$A=A_{\circ}e^{\frac{-bt}{2m}}$
$\implies \frac{A}{A_{\circ}}=e^{\frac{-bt}{2m}}$
Taking the natural log of both sides, we obtain:
$ln(\frac{A}{A_{\circ}})=-\frac{bt}{2m}$
This can be rearranged as:
$b=(\frac{2m}{t})ln(\frac{A_{\circ}}{A})$
We plug in the known values to obtain:
$b=\frac{2(2.44\times 10^5)}{3.0\times 10^2}ln2$
$b=1100Kg/s$
