Question

A rate of deposit of dust from unpolluted air is 10.0 tons per square mile per month (30 days). what is this dust fall, in milligrams per square meter per hour?

Answer 1

The rate of dust deposition is
$Rate= \frac{10 \,tons}{mi^{2}-30\,days} = \frac{1}{3} \, \frac{tons}{mi^{2}-day}$

Note that
1 ton = 9.07185 x 10⁵ g = 9.07185 x 10⁸ mg
1 mi = 1609.34 m = 1.60934 x 10³ m
1 day = 24 h

Therefore
$Rate=( \frac{1}{3}\, \frac{tons}{mi^{2}-day} )*( \frac{1}{24} \, \frac{day}{h} )*( \frac{1}{1.60934\times10^{3}}\, \frac{mi}{m} )^{2}*(9.07185\times10^{8}\, \frac{mg}{ton})\\= 4.865\, \frac{mg}{m^{2}-h}$

Answer: $4.865\, \frac{mg}{m^{2}-h}$