Answer
The pipe weighs 61.6 N.
Work Step by Step
We can find the volume of copper used for the pipe. Let $R_1$ be the inside radius and let $R_2$ be the outside radius.
$V = \pi~R_2^2~L - \pi~R_1^2~L$
$V = \pi~L~(R_2^2 -R_1^2)$
$V = (\pi)(1.50~m)~[(0.0175~m)^2 -(0.0125~m)^2]$
$V = 7.07\times 10^{-4}~m^3$
We can find the mass of the pipe.
$M = \rho~V$
$M = (8.90\times 10^3~kg/m^3)(7.07\times 10^{-4}~m^3)$
$M = 6.29~kg$
We can find the weight of the pipe.
$weight = Mg$
$weight = (6.29~kg)(9.80~m/s^2)$
$weight = 61.6~N$
The pipe weighs 61.6 N.
