Question

1. how much copper (in metric tons) is needed to pull an 8 - wire cable between san jose and los angeles (which is a distance of 548 km), if the density of copper is 8.96 g/cm³, and the radius of the wire is 0.83 mm? the area of a circle a = πr², where r is the radius. 1 metric ton = 2200 lb. 2.2 lb = 1 kg.

Explanation:

Step1: Convert units of length

First, convert the length of the cable from km to cm. Since 1 km = 100000 cm, a 548 - km cable has a length $L=548\times100000 = 5.48\times10^{7}$ cm. Also, convert the radius of the wire from mm to cm. Since 1 mm = 0.1 cm, the radius $r = 0.83\times0.1=0.083$ cm.

Step2: Calculate the volume of one wire

The volume of a cylinder (which a wire can be considered as) is $V = A\times L$, where $A=\pi r^{2}$. Substitute $r = 0.083$ cm into the area formula: $A=\pi\times(0.083)^{2}\text{ cm}^2$. Then the volume of one wire $V_1=A\times L=\pi\times(0.083)^{2}\times5.48\times 10^{7}\text{ cm}^3$.

Step3: Calculate the volume of 8 - wire cable

The volume of the 8 - wire cable $V = 8\times V_1=8\times\pi\times(0.083)^{2}\times5.48\times 10^{7}\text{ cm}^3$.
\[

$$\begin{align*} V&=8\times3.14\times(0.083)^{2}\times5.48\times 10^{7}\\ &=8\times3.14\times0.006889\times5.48\times 10^{7}\\ &=8\times3.14\times0.006889\times54800000\\ &=8\times3.14\times377517.2\\ &=9489997.824\text{ cm}^3 \end{align*}$$

\]

Step4: Calculate the mass of the copper

Use the density formula $
ho=\frac{m}{V}$, where $
ho = 8.96$ g/cm³. Rearranging for mass gives $m=
ho\times V$. So $m = 8.96\times9489997.824$ g.
\[m=8.96\times9489997.824 = 85030370.4\] g.
Convert grams to kg: Since 1 kg = 1000 g, $m=\frac{85030370.4}{1000}=85030.3704$ kg.
Convert kg to metric tons: Since 1 metric ton = 1000 kg, $m=\frac{85030.3704}{1000}=85.0303704$ metric tons.

Answer:

85.03 metric tons