document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=56706>Question 56706</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If a rope is wound around a wooden pole, the number of pounds of frictional force, F, between the pole and the rope is a function of the number of turns, N, according to the equation f=14*10^0.70n.  What is the frictional force when the number of turns is: 0.5, 1, 3? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #38565 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=38565\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> f(n)=14*10^0.70n. <BR>\n" );
document.write( "What is the frictional force when the number of turns is: 0.5, 1, 3?<BR>\n" );
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document.write( "f(0.5)=14*10^(35)=1.4*10^36 lbs. of frictional force<BR>\n" );
document.write( "f(1)=14*10^0.7=1.4*10^1.7 = 1.4(10^1.7)=70.166... lbs. of frictional force<BR>\n" );
document.write( "f(3)=14*10^(2.1)= 14*125.8925...=1762.4955... lbs. of frictional forch.<BR>\n" );
document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
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