The diameter of the rod is given as 5/8 inches.
\[ d = \frac{5}{8} \text{ in} = 0.625 \text{ in} \]
The cross-sectional area \( A \) of the rod is:
\[ A = \pi \left( \frac{d}{2} \right)^2 = \pi \left( \frac{0.625}{2} \right)^2 = \pi \left( 0.3125 \right)^2 \approx
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