document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=617865>Question 617865</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The conductance of a wire varies directly as the square of the wire's diameter and inversely as it's length. Fifty meters of wire with a diameter 2 mm has a conductance of 0.12 ohms. If a wire of the same material has length of 75m and diameter of 2.5mm, what is its conductance?  I need help. I have been on this problem for over an hour. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #388632 by </B><A HREF=/tutors/aboutme.mpl?userid=MathTherapy><B>MathTherapy(10552)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=MathTherapy><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/cgi-bin/show-face.mpl?user=MathTherapy><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=388632\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The conductance of a wire varies directly as the square of the wire's diameter and inversely as it's length. Fifty meters of wire with a diameter 2 mm has a conductance of 0.12 ohms. If a wire of the same material has length of 75m and diameter of 2.5mm, what is its conductance?  I need help. I have been on this problem for over an hour.\r<BR>\n" );
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document.write( "Let conductance of wire = C; D = diameter; and L = length\r<BR>\n" );
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document.write( "The we have: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=C+=+%28kD%5E2%29%2FL ALIGN=MIDDLE  ALT=\"C+=+%28kD%5E2%29%2FL\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0.12+=+%28k%282%29%5E2%29%2F50 ALIGN=MIDDLE  ALT=\"0.12+=+%28k%282%29%5E2%29%2F50\"   DISABLED_style= \"max-width:90%; height:auto;\" > ------- Substituting values for C, D, and L\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0.12+=+%284k%29%2F50 ALIGN=MIDDLE  ALT=\"0.12+=+%284k%29%2F50\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "4k = 50 * 0.12 ---- Cross-multiplying\r<BR>\n" );
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document.write( "4k = 6\r<BR>\n" );
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document.write( "k = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=6%2F4 ALIGN=MIDDLE  ALT=\"6%2F4\"   DISABLED_style= \"max-width:90%; height:auto;\" >, or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3%2F2 ALIGN=MIDDLE  ALT=\"3%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >, or 1.5\r<BR>\n" );
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document.write( "With k = 1.5; D = 2.5; and L = 75, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=C+=+%28kD%5E2%29%2FL ALIGN=MIDDLE  ALT=\"C+=+%28kD%5E2%29%2FL\"   DISABLED_style= \"max-width:90%; height:auto;\" > becomes: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=C+=+%281.5%282.5%29%5E2%29%2F75 ALIGN=MIDDLE  ALT=\"C+=+%281.5%282.5%29%5E2%29%2F75\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=C+=+%281.5%286.25%29%29%2F75 ALIGN=MIDDLE  ALT=\"C+=+%281.5%286.25%29%29%2F75\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=C+=+%289.375%29%2F75 ALIGN=MIDDLE  ALT=\"C+=+%289.375%29%2F75\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "C = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight_green%280.125%29 ALIGN=MIDDLE  ALT=\"highlight_green%280.125%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Send comments and “thank-yous” to “D” at  MathMadEzy@aol.com </SPAN>\n" );
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