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!! :)

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)

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.

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

Yes, you are on the right track! Let's solve the first problem first.

1. The relationship between t, z, and r can be expressed as follows: t = k * (r^2) / (z^3), where k is the constant of proportionality.

2. We are given that t = 4 when z = 3 and r = 6. We can plug these values into the equation to find the value of k: 4 = k * (6^2) / (3^3).

3. Simplifying, we have: 4 = k * 36 / 27. Multiplying both sides by 27, we get: 108 = 36k. Dividing both sides by 36, we find: k = 3.

4. Now let's solve for t when z = 9 and r = 10 using the equation: t = k * (r^2) / (z^3).

t = 3 * (10^2) / (9^3).

Simplifying further, we have: t = 300 / 729 = 100 / 243.

Therefore, when z = 9 and r = 10, t is equal to 100 / 243.

Now let's move on to the second problem:

1. The relationship between conductance (c), diameter (d), and length (l) can be expressed as: c = k * (d^2) / l, where k is the constant of proportionality.

2. We are given that when l = 20 m and d = 3 mm, c = 0.54 mho. Plugging these values into the equation, we get: 0.54 = k * (3^2) / 20.

3. Simplifying, we have: 0.54 = k * 9 / 20. Multiplying both sides by 20, we get: 10.8 = 9k. Dividing both sides by 9, we find: k = 1.2.

4. Now let's solve for c when l = 50 mm and d = 5 mm using the equation: c = k * (d^2) / l.

c = 1.2 * (5^2) / 50.

Simplifying further, we have: c = 1.2 * 25 / 50 = 0.6 mho.

Therefore, when l = 50 mm and d = 5 mm, the conductance is 0.6 mho.