Answer:
2.14 ft
Explanation:
We will use the following equation:
[tex]L = \frac{R_{T} * A}{p_{20C}*(1+\alpha _{20C}*dt ) }[/tex]
Data obtained:
[tex]R_{15.3 F , -9.27 C} = 15.3 ohms\\A_{28-gauge} = 8.2*10^(-8) m^2\\p_{20 C} = 1.723*10^(-6)\\\alpha _{20C} = 0.00393\\ dt = 20 - (-9.27) = 29.28[/tex]
Using the above equation:
[tex]L = \frac{15.3 * 8.2*10^(-8)}{1.723*10^(-6)*(1+0.00393*29.28 ) }\\\\L =0.6530 m = 2.14 ft[/tex]
