Let's solve the problem step by step: --- ## **Step 1: Write Down Given Data and the Resistance Formula** **Given:** - Length of each wire, \( L = 40. \) cm \( = .400 \) m - Diameter of each wire, \( d = 2.00 \) mm \( = 2.00 \times 10^{-3} \) m - Resistivity of gold, \( \rho_{Au} = 2.44 \times 10^{-8} \ \Omega \cdot m \) - Resistivity of silver, \( \rho_{Ag} = 1.59 \times 10^{-8} \ \Omega \cdot m \) **Resistance Formula:** \[ R = \frac{\rho L}{A} \] Where \( A \) is the cross-sectional area of the wire. **Explanation:** We need to find the equivalent resistance of the two wires connected in series. Both wires have the same length and diameter, so we first calculate the cross-sectional area. \[ A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{2.00 \times 10^{-3}}{2}\right)^2 = \pi (1.00 \times 10^{-3})^2 = \pi \times 10^{-6} \ \text{m}^2 \] --- ## **Step 2: Calculate Individual Resistances and Total Resistance** ### **Gold:** \[ R_{Au} = \frac{\rho_{Au} L}{A} = \frac{2.44 \times 10^{-8} \times .400}{\pi \times 10^{-6}} \] \[ R_{Au} = \frac{9.76 \times 10^{-9}}{3.1416 \times 10^{-6}} = 3.11 \times 10^{-3} \ \Omega \] ### **Silver:** \[ R_{Ag} = \frac{\rho_{Ag} L}{A} = \frac{1.59 \times 10^{-8} \times .400}{\pi \times 10^{-6}} \] \[ R_{Ag} = \frac{6.36 \times 10^{-9}}{3.1416 \times 10^{-6}} = 2.02 \times 10^{-3} \ \Omega \] ### **Total Series Resistance:** \[ R_{total} = R_{Au} + R_{Ag} = (3.11 \times 10^{-3}) + (2.02 \times 10^{-3}) = 5.13 \times 10^{-3} \ \Omega \] **Explanation:** We calculated the resistance of each wire and added them because they are in series. --- ## **Final Answer** \[ \boxed{5.13 \times 10^{-3} \ \Omega} \] is the equivalent resistance between points A and B.