The answer will depend on the density of the compost. For this question, I am going to use 800 pounds/cubic yard (see: http://www.dot.ca.gov/hq/LandArch/policy/compost_specs.htm)
40 tons=80000 pounds
Volume=Mass/Density
80000 lbs/(800lbs/cu yd)=100 cubic yards
So we have 100 cubic yards of compost spread over one acre.
One acre is 43560 square feet = 43560/9=4840 square yards.
Volume = Surface area*depth
100=4840*depth --> depth=(100/4840) yards
Now convert depth to inches by multiplying by 36 (since there are 36 inches in a yard).
(100/4840) yards=(100/4840)*36 inches=90/121 inches, or about 3/4"
There are 27,154 gallons of water in one acre of land that is one inch deep.
1 acre = 43560 sq.feet. 1 foot deep water over an acre is 43560 cubic feet, which is 325851.4 gallons in 1 foot deep. If it is 1 inch deep (1/12 foot) then it is 27154.3 gallons in 1 inch deep.
2
1 acre = 43,560 square feet1 acre 5 feet deep = 43,560 x 5 = 217,800 cubic feet = 1,629,257 gallons (rounded)
27,154
The answer will depend on how deep the lake is!
36
A 1000 acre lake that is 7 feet deep (uniformly) will have 7000 acre feet of water in it. An acre foot is an acre of water one foot deep, and the unit is used to measure reservoir water capacity. The conversion factor is that one acre foot equals about 325,851.5 gallons. The lake in question is holding 7000 times 325,851.5 gallons of water.
Three acres three inches deep is 403.33 cubic yards.
10 acre = 435600 sq ft So, if it is 165 ft wide, it must be 435600/165 = 2640 ft deep.
my butt iches
11.22 inches