Answer:
Approximately [tex]45.2\; {\rm kg}[/tex] (assuming that the density of the copper is [tex]8.96\; {\rm g \cdot cm^{-3}}[/tex].)
Explanation:
Convert length [tex]h[/tex] of the pipe to centimeters:
[tex]h = 1.70\; {\rm m} = 170\; {\rm cm}[/tex].
If a cylinder is of radius [tex]r[/tex] and height [tex]h[/tex], the volume of that cylinder will be [tex]\pi\, r^{2}\, h[/tex].
Let [tex]r[/tex] denote the inner diameter of this pipe, and let [tex]R[/tex] denote the outer diameter of the pipe.
If the pipe is filled with copper, volume of the entire pipe cylinder will be [tex]V = \pi\, R^{2}\, h[/tex]. In reality, the inside of the pipe is a hollow cylinder of radius [tex]r[/tex] and volume [tex]\pi\, r^{2}\, h[/tex].
To find the volume of copper in this pipe, subtract the volume of the hollow cylinder [tex]\pi\, r^{2}\, h[/tex] from the volume of the entire pipe cylinder [tex]\pi\, R^{2}\, h[/tex]:
[tex]V = \pi\, R^{2} \, h - \pi\, r^{2}\, h = (R^{2} - r^{2})\, \pi\, h[/tex].
Substitute in [tex]R = 3.90\; {\rm cm}[/tex], [tex]r = 2.40\; {\rm cm}[/tex], and [tex]h = 170\; {\rm cm}[/tex] to find the volume of this pipe cylinder:
[tex]\begin{aligned}V &= (R^{2} - r^{2})\, \pi\, h \\ &= ((3.90\; {\rm cm})^{2} - (2.40\; {\rm cm})^{2})\, \pi\, (170\; {\rm m}) \\ &\approx 5047.0\; {\rm cm^{3}}\end{aligned}[/tex].
Multiply the volume [tex]V[/tex] of copper in this pipe by density [tex]\rho[/tex] to find mass [tex]m[/tex]:
[tex]\begin{aligned}m &= \rho\, V \\ &\approx (8.96\; {\rm g \cdot cm^{-3}})\, (5047.0\; {\rm cm^{3}) \\ &\approx 45220\; {\rm g} \\ &= 45.2\; {\rm kg}\end{aligned}[/tex].
