The rate of dust deposition is
[tex]Rate= \frac{10 \,tons}{mi^{2}-30\,days} = \frac{1}{3} \, \frac{tons}{mi^{2}-day} [/tex]
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
[tex]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} [/tex]
Answer: [tex]4.865\, \frac{mg}{m^{2}-h} [/tex]
