document.write( "Question 885774: A stream of water in steady flow from a kitchen faucet. At the faucet the diameter of stream is 0.96 cm. The stream fills a 125 cm^3 in 16.3 seconds. Find the diameter of the stream 13 cm below the opening of the faucet. \n" ); document.write( "
Algebra.Com's Answer #535417 by Alan3354(69443)\"\" \"About 
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A stream of water in steady flow from a kitchen faucet. At the faucet the diameter of stream is 0.96 cm. The stream fills a 125 cm^3 in 16.3 seconds. Find the diameter of the stream 13 cm below the opening of the faucet.
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\n" ); document.write( "Find the speed of the stream at the faucet:
\n" ); document.write( "Flow rate = Vol/sec = area*speed = 125/16.3 cc/sec
\n" ); document.write( "area*speed = 125/16.3
\n" ); document.write( "\"area+=+pi%2Ar%5E2+=+0.48%5E2%2Api+=+0.2304pi\" (area at the faucet)
\n" ); document.write( "\"speed+=+%28125%2F16.3%29%2F%280.2304pi%29\"
\n" ); document.write( "speed =~ 10.595 cm/sec (speed at the faucet)
\n" ); document.write( "------
\n" ); document.write( "Use \"h%28t%29+=+-4.9t%5E2+-+10.595t\" to find the time it takes to fall 13 cm
\n" ); document.write( "13 cm = 0.13 meters
\n" ); document.write( "\"h%28t%29+=+-4.9t%5E2+-+10.595t+=+-0.13\"
\n" ); document.write( "\"-4.9t%5E2+-+10.595t+%2B+0.13+=+0\"
\n" ); document.write( "\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation \"ax%5E2%2Bbx%2Bc=0\" (in our case \"-4.9x%5E2%2B-10.595x%2B0.13+=+0\") has the following solutons:
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\n" ); document.write( " \"x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca\"
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\n" ); document.write( " For these solutions to exist, the discriminant \"b%5E2-4ac\" should not be a negative number.
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\n" ); document.write( " First, we need to compute the discriminant \"b%5E2-4ac\": \"b%5E2-4ac=%28-10.595%29%5E2-4%2A-4.9%2A0.13=114.802025\".
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\n" ); document.write( " Discriminant d=114.802025 is greater than zero. That means that there are two solutions: \"+x%5B12%5D+=+%28--10.595%2B-sqrt%28+114.802025+%29%29%2F2%5Ca\".
\n" ); document.write( "
\n" ); document.write( " \"x%5B1%5D+=+%28-%28-10.595%29%2Bsqrt%28+114.802025+%29%29%2F2%5C-4.9+=+-2.17444598843739\"
\n" ); document.write( " \"x%5B2%5D+=+%28-%28-10.595%29-sqrt%28+114.802025+%29%29%2F2%5C-4.9+=+0.0122010904782066\"
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\n" ); document.write( " Quadratic expression \"-4.9x%5E2%2B-10.595x%2B0.13\" can be factored:
\n" ); document.write( " \"-4.9x%5E2%2B-10.595x%2B0.13+=+%28x--2.17444598843739%29%2A%28x-0.0122010904782066%29\"
\n" ); document.write( " Again, the answer is: -2.17444598843739, 0.0122010904782066.\n" ); document.write( "Here's your graph:
\n" ); document.write( "\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-4.9%2Ax%5E2%2B-10.595%2Ax%2B0.13+%29\"

\n" ); document.write( "\n" ); document.write( "====================
\n" ); document.write( "t = 0.0122 seconds
\n" ); document.write( "The water is accelerated at 9.8m/sec/sec
\n" ); document.write( "Its speed after 0.0122 seconds is 0.10595 m/sec + 9.8*0.0122
\n" ); document.write( "= 0.22551 m/sec
\n" ); document.write( "= 22.551 cm/sec
\n" ); document.write( "-------------
\n" ); document.write( "Area at that speed = (125/16.3)/22.551 sq cm
\n" ); document.write( "Area =~ 0.3400608 sq cm = pi*r^2
\n" ); document.write( "r = 0.329 cm
\n" ); document.write( "d = 0.658 cm
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\n" ); document.write( "So you have 2 answers that don't agree.
\n" ); document.write( "You can see how it's done.
\n" ); document.write( "Check the work, pick the one you like.\r
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