## University Physics with Modern Physics (14th Edition)

We can find the volume of the rod. $V = \pi~r^2~h$ $V = (\pi)(0.01425~m)^2(0.858~m)$ $V = 5.47\times 10^{-4}~m^3$ We can find the mass of the rod. $M = \rho~V$ $M = (7800~kg/m^3)(5.47\times 10^{-4}~m^3)$ $M = 4.27~kg$ We can find the weight of the rod. $weight = (4.27~kg)(9.80~m/s^2) = 41.9~N$ Since the weight is only 41.9 N, we should be able to deliver the iron rod without a cart.