document.write( "Question 268047: a sphere of diameter 25 cm is half full with acid all of which is drained to a tall cyliderical beaker 16 cm in diameter. what is the depth of the acid in the beaker \n" ); document.write( "
Algebra.Com's Answer #196552 by CharlesG2(834)\"\" \"About 
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a sphere of diameter 25 cm is half full with acid all of which is drained to a tall cyliderical beaker 16 cm in diameter. what is the depth of the acid in the beaker\r
\n" ); document.write( "\n" ); document.write( "volume of a sphere is 4/3 * pi * r^3 (diameter is 2 times the radius r)
\n" ); document.write( "volume = 4/3 * pi * (12.5)^3
\n" ); document.write( "volume = 4/3 * 1953.125 * pi
\n" ); document.write( "volume = 2604 1/6 * pi
\n" ); document.write( "volume of sphere is 1/2 full --> 1302 1/12 * pi cm^3 acid
\n" ); document.write( "volume of beaker = area of base * (height H)
\n" ); document.write( "volume = r^2 * H * pi
\n" ); document.write( "volume = 8^2 * H * pi
\n" ); document.write( "volume = 64 * H * pi cm^3 is volume of beaker
\n" ); document.write( "set the volumes equal and solve for H which will be the depth of the acid in the beaker
\n" ); document.write( "1302 1/12 * pi = 64 * H * pi (divide out pi from both sides)
\n" ); document.write( "1302 1/12 = 64 * H
\n" ); document.write( "20.345052 or 20.345 cm is approximately H\r
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