Respuesta :
Answer:
a) d = 7.62 10⁻⁶ m, b) l = 3.25 10⁴ m
Explanation:
Resistance is expressed by the formula
R = ρ l / A (1)
density is defined by
density = m / V
the volume of a wire is the cross section by the length
V = A l
we substitute
density = m / A l
A = m / density l
we substitute in 1
R = ρ l density l / m
R =ρ density l² / m
l = √ (R m /ρ density)
let's calculate the cable length
l = √(11.7 13.5 10⁻³ / (1.68 10⁻⁸ 8.9 10³))
l = √(10.56 10⁸)
l = 3.25 10⁴ m
now we can find the cable diameter with the density equation
A = m / density l
A = 13.5 10⁻³ / (8.9 10³ 3.25 10⁴)
A = 4,557 10⁻¹¹ m²
the area of the circle is
A = π r² = π d² / 4
d = √ (4A /π)
d = √ (4 4,557 10⁻¹¹/π)
d = 7.62 10⁻⁶ m
The diameter of the wire is 0.202m and the length of the wire is 4.47*10^-5m
Data;
- Resistor = 11.7Ω
- mass = 13.5g
- resistivity of copper = 1.68 * 10 ^-8 Ω.m
- density of copper = 8.9*10^3 kg/m^3
Resistivity of Copper
The resistivity of copper is calculated by
[tex]R = \frac{\rho L}{A}\\[/tex]
Let's calculated the volume of the wire first;
[tex]\rho = \frac{mass}{volume} \\volume = \frac{mass}{density} \\volume = \frac{13.5*10^-^3}{8.9*10^3} \\v = 1.52*10^-6m^3[/tex]
The diameter of the wire will be
[tex]R = \frac{\rho L}{A}\\ R = \frac{\rho LA}{A^2}\\ R = \frac{\rho V}{A^2} \\A^2 = \frac{\rho V}{R}\\ A^2 = \frac{8.9*10^3 * 1.516*10^-^6}{11.7} \\A^2 = 0.0011\\A = \sqrt{0.0011} \\A = 0.034m^2[/tex]
Taking the area
[tex]a = \pi r^2\\0.034 = 3.14 * r^2\\r^2 = \frac{0.034}{3.14} \\r^2 = 0.01083\\r = \sqrt{0.01083}\\ r = 0.104m\\d = 2r\\d = 2 * 0.104\\d = 0.208m[/tex]
Length of the wire can be calculated as
[tex]V = AL\\L = V/A\\L = \frac{1.52*10^-^6}{0.034}\\ L = 4.47*10^-^5m[/tex]
The diameter of the wire is 0.202m and the length of the wire is 4.47*10^-5m
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