document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=193537>Question 193537</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/9780321331793/algebrahomeworkh></A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> A heavy metal sphere with radius 10 cm is dropped into a<BR>\n" );
document.write( "right circular cylinder with base radius of 10 cm. If the<BR>\n" );
document.write( "original cylinder has water in it that is 20 cm high, how<BR>\n" );
document.write( "high is the water after the sphere is placed in it?\r<BR>\n" );
document.write( "\n" );
document.write( "I do not understand how to solve this.  Please help.  Thank you. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #145279 by </B><A HREF=/tutors/aboutme.mpl?userid=RAY100><B>RAY100(1637)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=RAY100><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=145279\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Volume  of  sphere =4/3 *pi*r^3=(4/3)pi(10)^3<BR>\n" );
document.write( "Volume  of  cylinder  = Area  of  Base  *Height=pi(r)^2*h<BR>\n" );
document.write( "the  differential  height  in  the  cylinder  as  the  sphere displaces  the  liquid  is<BR>\n" );
document.write( "found  by  setting  sphere  volume  equal  to  cylinder  volume,  and  solve  for h<BR>\n" );
document.write( "(4/3 )pi (r^3)=pi(r^2)(h) <BR>\n" );
document.write( "4/3*10^3=10^2 *h<BR>\n" );
document.write( "4/3 *10=h<BR>\n" );
document.write( "40/3=13.33=h <BR>\n" );
document.write( "Final  Height  = original  +  displaced=20 +13.33=33.33  ANSWER<BR>\n" );
document.write( ",<BR>\n" );
document.write( ",<BR>\n" );
document.write( ",<BR>\n" );
document.write( "For  FUN<BR>\n" );
document.write( "lets  assume  that  the  water  cannot  pass  around  sphere,but  air  can,   after all  both  radius  are  =10,  and  it  might  be  a  tight  fit,  then  what  happens,<BR>\n" );
document.write( "The  sphere  settles  into  the  water  until the  water  reaches  the  major  diameter.<BR>\n" );
document.write( "how  much  water  goes  into  this  cusp  is  the  difference  of  the  hemisphere  volume  and  the cylinder  volume.,.and  the  corresponding equivalent  height of  the  cylinder  is  how  far  the  sphere  sinks  into  the  water.\r<BR>\n" );
document.write( "\n" );
document.write( "calculating<BR>\n" );
document.write( "pi*10^2(10)-(4/6)pi(10^3)=pi (10^2)h<BR>\n" );
document.write( "10-(4/6)10=h<BR>\n" );
document.write( "3.33=h<BR>\n" );
document.write( "original  less  3.33=  20-3.33=16.67 as  new  height to  bottom  of  sphere<BR>\n" );
document.write( "height  to  major  dia  of  sphere  is  16.67+10=26.67,  also  new  height  of  water<BR>\n" );
document.write( "just  for  fun </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );