document.write( "Question 308822: I was wondering what the pressure must be a mile deep in the water. \r
\n" ); document.write( "\n" ); document.write( "If water weighs 59.52 lbs/ft^3, what pressure in pounds per square inch would it cause at 5,280 feet below the surface? Hint: Calculate the weight of a column of water with an area of 1 square inch and 5,280 feet high.
\n" ); document.write( "
\n" ); document.write( "

Algebra.Com's Answer #220801 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
Calculate the volume of the 1 inch square x 1 mile column.
\n" ); document.write( "Convert the 1 mile to feet since the density is in feet.
\n" ); document.write( "Convert the 1 inch square to 1 ft square also.
\n" ); document.write( "1 in*1 in*(1 ft/12 in)^2=1/144 ft^2
\n" ); document.write( "1 mile=5280 ft
\n" ); document.write( "\"V=A%2Ah=%281%2F144%29%285280%29\"
\n" ); document.write( "\"V=36.67\" ft^3
\n" ); document.write( "Find the weight of the column by multiplying the volume and the density of water.
\n" ); document.write( "\"W=VA=36.67%2A59.52\"
\n" ); document.write( "\"W=2182.6\" lbs
\n" ); document.write( "That weight acts on the 1 inch square area.
\n" ); document.write( "\"P=2182.6%2F1\"
\n" ); document.write( "\"P=2182.6\" psi or about 148 atmospheres.
\n" ); document.write( "
\n" );