Question

10 multiple choice 10 points arresting cables on aircraft carriers are made of a high grade steel with a density of 0.284 lb/in³. what volume, in l, of steel would be required to make a 228 kg cable? 4.50 l 0.0116 l 29.0 l 0.00179 l

Explanation:

Step1: Convert mass to pounds

We know that 1 kg = 2.20462 lb. So for a 228 - kg cable, the mass in pounds is $m=228\times2.20462$ lb.
$m = 228\times2.20462=502.65336$ lb.

Step2: Use the density - volume formula

The density formula is $
ho=\frac{m}{V}$, where $
ho$ is density, $m$ is mass, and $V$ is volume. We can re - arrange it to solve for volume $V=\frac{m}{
ho}$. Given $
ho = 0.284$ lb/in³ and $m = 502.65336$ lb, then $V=\frac{502.65336}{0.284}$ in³.
$V=\frac{502.65336}{0.284}=1770$ in³.

Step3: Convert volume from in³ to L

We know that 1 in³ = 0.0163871 L. So the volume in liters is $V_{L}=1770\times0.0163871$ L.
$V_{L}=1770\times0.0163871 = 29.0$ L.

Answer:

29.0 L