Answer
a) $R_{total}=14.1$
b) $I=0.00674A$
c) $V_{Al}=0.0579V$
$V_{Cu}=0.0367V$
Work Step by Step
a) We can find the total resistance by adding the resistance of the copper wire and the resistance of the aluminum wire. We can use the formula $R=\frac{pl}{A}$, where $A=\pi(\frac{1.4\times10^-4}{2})^2=1.54\times10^{-8}m^{2}$
$R_{Cu}=\frac{1.68\times10^{-8}\times5.0m}{1.54\times10^{-8}}=5.45$
$R_{Al}=\frac{2.65\times10^{-8}\times5.0m}{1.54\times10^{-8}}=8.60$
$R_{total}=5.45+8.60=14.1$
b) We can use Ohm's Law $V=IR$
$I=\frac{V}{R}=\frac{0.095V}{14.1}=0.00674A$
c) $V_{Al}=IR_{Al}=0.00674A\times8.60=0.0579V$
$V_{Cu}=IR_{Cu}=0.00674A\times5.45=0.0367V$