t = k r^2/z^3
4 = k 36/27
k = 3
so t = 3 r^2/z^3
============================
c = k d^2/L
.54 mho = k 3^2/50
k = 50 * .54 /9 = 3 again LOL
(note - in your units which are not the ones a physicist would tend to use)
If t varies inversely as the cube of z and directly as the square of r, and t=4 when z=3 and r=6, find 9. T= k (square) r/(cubed) z 4=k (square) 6/ (cubed) 3
The electrical conductance of a wire varies directly as the square of its a diameter and inversely as its length. The conductance of a wire 20 m long and 3 mm in diameter is 0.54 mho. If a wire of the same material has length 50 mm and diameter 5 mm, what is its conductance? Let conductance = c, diameter = d, length = l c= (square) d/l
Am I going in the right direction??
Please help and give solutions!! :)
4 answers
t = kr^2/z^3
4 = 36k/27
k = 4*27/36 = 3
c = kd^2/l
so, cl/d^2 = k, a constant
That means you want c such that
c*50/25 = .54*20/9 = 0.6
I'm assuming you meant a 50m wire, not 50mm.
Note that you can find the new c without actually having to find the value of k.
4 = 36k/27
k = 4*27/36 = 3
c = kd^2/l
so, cl/d^2 = k, a constant
That means you want c such that
c*50/25 = .54*20/9 = 0.6
I'm assuming you meant a 50m wire, not 50mm.
Note that you can find the new c without actually having to find the value of k.
I see that I have mixed up the units. Multiply c by 1000, since the length is in meters. Makes more sense, since you know that a useful wire has very high conductance (low resistance).
Why did you not respond to my original concern here?
http://www.jiskha.com/display.cgi?id=1455973924
http://www.jiskha.com/display.cgi?id=1455973924
1 answer