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!! :)
duplicate post, I just did it
Yes, you are on the right track! To solve both problems, you need to use the concept of inverse and direct variation.
For the first problem, you are given that t varies inversely as the cube of z and directly as the square of r. This means that the equation relating t, z, and r is of the form t = k * r² / z³, where k is a constant of variation.
To find the value of k, you can substitute the given values of t, z, and r into the equation. You are given that t = 4 when z = 3 and r = 6, so you can plug these values in to get:
4 = k * (6²) / (3³)
Simplifying the equation further:
4 = k * 36 / 27
To solve for k, you can multiply both sides of the equation by 27:
108 = 36k
Dividing both sides by 36:
k = 108 / 36 = 3
Now that you have the value of k, you can use it to find the value of t when z = 9 and r = 12:
t = 3 * (12²) / (9³)
t = 3 * 144 / 729
t = 432 / 729
So, when z = 9 and r = 12, t is approximately 0.5923.
For the second problem, you are given that the conductance (c) of a wire varies directly as the square of its diameter (d) and inversely as its length (l). This means that the equation relating c, d, and l is of the form c = k * (d² / l), where k is a constant of variation.
To find the value of k, you can substitute the given values of c, d, and l into the equation. You are given that c = 0.54 when d = 3 mm and l = 20 m, so you can plug these values in to get:
0.54 = k * (3²) / 20
Simplifying the equation further:
0.54 = k * 9 / 20
To solve for k, you can multiply both sides of the equation by 20:
0.54 * 20 = 9k
10.8 = 9k
Dividing both sides by 9:
k = 10.8 / 9 = 1.2
Now that you have the value of k, you can use it to find the value of c when d = 5 mm and l = 50 mm:
c = 1.2 * (5²) / 50
c = 1.2 * 25 / 50
c = 30 / 50
So, when d = 5 mm and l = 50 mm, the conductance is 0.6 mho.