Copper is a good conductor because its resistivity is low. We expect to find that this wire is very long and thin.
To solve for the variables, we will use the equation for resistance as a function of resistivity, length, and cross-sectional area.
A = m / (l*p) (*where A=area, m=mass, l=length, p=density)
R = l / (r*A) (*where R=resistance, l=length, r=resistivity, A=area)
Now substitute A into equation for R.
R = l / (r*m / (l*p)) = (r*p*l2) / m
Now solve for L:
l = squareroot(m*R/r*p)
where m = 1.8×10-3 kg
R = 0.52 Ω
r = 1.70 × 10−8 Ω · m
p = 8.92 × 103 kg/m3
Plug in values and solve for l:
l = squareroot((1.8×10-3 kg * 0.52 Ω) / (1.70 × 10−8 Ω · m r * 8.92 × 103 kg/m3 ))
l = 2.48m
Part B, the diameter can be found by noting that:
A = pi*r^2 (*r=radius, and r=d/2 d=diameter)
A=pi(d/2)^2 = m/(l*p)
solving for d:
d = squareroot (4*m/(pi*l*p))
plugging in the values we get that d:
d = squareroot (4*1.8×10-3 kg / (pi*2.48m * 8.92 × 103 kg/m3))
d = 3.219*10^-4 = 3.21 µm