a steel pipe 100 cm long has an outside diameter of 2 cm and an inside diameter of 1.8 cm. if the density of…

a steel pipe 100 cm long has an outside diameter of 2 cm and an inside diameter of 1.8 cm. if the density of the steel is 7.8 grams per cm³, what is the mass of the pipe, to the nearest gram?

a steel pipe 100 cm long has an outside diameter of 2 cm and an inside diameter of 1.8 cm. if the density of the steel is 7.8 grams per cm³, what is the mass of the pipe, to the nearest gram?

Answer

Explanation:

Step1: Calculate outer - radius and inner - radius

The outer - radius $R=\frac{2}{2}=1$ cm and the inner - radius $r = \frac{1.8}{2}=0.9$ cm.

Step2: Calculate the volume of the steel in the pipe

The volume of a cylinder is $V=\pi h(R^{2}-r^{2})$. Here, $h = 100$ cm. So $V=\pi\times100\times(1^{2}-0.9^{2})=\pi\times100\times(1 - 0.81)=\pi\times100\times0.19\approx3.14\times100\times0.19 = 59.66$ $cm^{3}$.

Step3: Calculate the mass of the pipe

We know that density $\rho=\frac{m}{V}$, so $m=\rho V$. Given $\rho = 7.8$ g/$cm^{3}$ and $V\approx59.66$ $cm^{3}$. Then $m=7.8\times59.66 = 465.348\approx465$ g.

Answer:

465