Urgent!
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 t when z=6 and r=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 (1/z^3)(r^2)
for the given:
4 = k(1/27)(36)
k = 4(27)/36 = 3
so your equation is
t = 3r^2/z^3
for the data ...
t = 3(81)/216 = 9/8 or 1.125
#2
c = k d^2/l
sub in the given:
.54 = k (9)/20
k = .54(20)/9 = 1.2
so you have c = 1.2 d^2/l
in your first data, the length was given in metres, but in the second set of data the length is in mm
Is that a typo or do you want that?
I am going to assume that the original 20 m was really 20 mm,
c = 1.2(25)/50 = .6
No
Yes, you are on the right track! To solve both problems, you need to use the given formulas and solve for the constant of variation (k) first, and then use that constant to find the solution in each case.
Let's solve the first problem step-by-step:
1. Given the equation: t = k * (r^2) / (z^3)
2. To find k, substitute the given values: t = 4, z = 3, and r = 6.
4 = k * (6^2) / (3^3)
3. Simplify the equation:
4 = k * 36 / 27
4 = 4/3 * k
4. Solve for k:
k = 3
5. Now that we've found the value of k, we can use it to find the value of t when z = 6 and r = 9.
t = 3 * (9^2) / (6^3)
t = 3 * 81 / 216
t = 243 / 216
t ≈ 1.125
Therefore, when z = 6 and r = 9, t is approximately 1.125.
Now let's solve the second problem:
1. Given the equation: c = (d^2) / l
2. To find c, substitute the given values: c = 0.54, l = 20m, and d = 3mm.
0.54 = (3^2) / 20
3. Simplify the equation:
0.54 = 9 / 20
4. Solve for c:
c = 0.54 * 20 / 9
c ≈ 1.2 mho
5. Now, we can use the value of c to find the conductance when l = 50mm and d = 5mm.
c = (5^2) / 50
c ≈ 0.5 mho
Therefore, when l = 50mm and d = 5mm, the conductance is approximately 0.5 mho.
I hope this helps! Let me know if you have any further questions.
Yes, you are going in the right direction. To solve both problems, you need to use the concept of inverse variation and direct variation.
Let's solve the first problem:
Given:
t varies inversely as the cube of z
t varies directly as the square of r
t = 4 when z = 3 and r = 6
We can write this relationship as:
t = k * (r^2 / z^3)
Plug in the given values:
4 = k * (6^2 / 3^3)
Simplifying this equation, we have:
4 = k * 36 / 27
4 = 4k / 3
Multiply both sides by 3:
12 = 4k
k = 12 / 4
k = 3
Now, substitute the values of z = 6 and r = 9 into the equation:
t = 3 * (9^2 / 6^3)
t = 3 * (81 / 216)
t = 3 * 0.375
t = 1.125
So, when z = 6 and r = 9, the value of t is approximately 1.125.
Now, let's solve the second problem:
Given:
conductance varies directly as the square of the diameter
conductance varies inversely as the length
conductance = c, diameter = d, length = l
c = d^2 / l
Plug in the given values:
0.54 = (3^2) / 20
To find the conductance of the wire with length 50 mm and diameter 5 mm, we use the proportionality property:
(3^2) / 20 = (5^2) / l
Cross-multiply:
3^2 * l = 5^2 * 20
9l = 25 * 20
9l = 500
Divide both sides by 9:
l = 500 / 9
l ≈ 55.56
So, the conductance of the wire with length 50 mm and diameter 5 mm would be approximately 55.56 mho.