So if I'm using the area 8558.25 in that math I need to have it as 8558.25squared and then sqaure root it?
3 answers
Volume in mm^3 = height in mm which is what the weather person reports times area in mm^2
and about your earlier question:
I've got the total volume/capacity that fell (3.51125KL) and the total area that it fell in (8558.25m).
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this is incorrect
the area must be in length squared, like m^2
By the way, convert the depth from mm into meters and do it in meters^3.
Also by the way
1000 liters = 1 m^3
so 1 kL = 1 m^3
I've got the total volume/capacity that fell (3.51125KL) and the total area that it fell in (8558.25m).
-------------------------------
this is incorrect
the area must be in length squared, like m^2
By the way, convert the depth from mm into meters and do it in meters^3.
Also by the way
1000 liters = 1 m^3
so 1 kL = 1 m^3
rewriting what I guess you mean
I've got the total volume/capacity that fell (3.51125KL) and the total area that it fell in (8558.25m^2).
I say using approximate numbers:
volume = 3.5 kL = 3.5 m^3
area = 8500 m^2
then height = volume/area = 3.5 m^3/8500 m^2
= .0004 m = .4 mm
I've got the total volume/capacity that fell (3.51125KL) and the total area that it fell in (8558.25m^2).
I say using approximate numbers:
volume = 3.5 kL = 3.5 m^3
area = 8500 m^2
then height = volume/area = 3.5 m^3/8500 m^2
= .0004 m = .4 mm