challenge the inner cylinder of the bushing shown is hollow. what is the weight of the bushing if it is made…

Explanation:

Step1: Calculate outer - cylinder volume

The formula for the volume of a cylinder is $V=\pi r^{2}h$. The outer - radius $r_{1}=\frac{5}{2}$ inches and height $h = 8\frac{3}{4}=\frac{8\times4 + 3}{4}=\frac{35}{4}$ inches. So, $V_{1}=\pi r_{1}^{2}h=3.14\times(\frac{5}{2})^{2}\times\frac{35}{4}=3.14\times\frac{25}{4}\times\frac{35}{4}=\frac{3.14\times25\times35}{16}=\frac{2747.5}{16}$ cubic inches.

Step2: Calculate inner - cylinder volume

The inner - radius $r_{2}=\frac{2\frac{3}{4}}{2}=\frac{\frac{2\times4 + 3}{4}}{2}=\frac{11}{8}$ inches and height $h=\frac{35}{4}$ inches. So, $V_{2}=\pi r_{2}^{2}h=3.14\times(\frac{11}{8})^{2}\times\frac{35}{4}=3.14\times\frac{121}{64}\times\frac{35}{4}=\frac{3.14\times121\times35}{256}=\frac{13297.9}{256}$ cubic inches.

Step3: Calculate the volume of the bushing

The volume of the bushing $V = V_{1}-V_{2}=\frac{2747.5}{16}-\frac{13297.9}{256}$. First, make a common denominator of 256: $\frac{2747.5\times16}{16\times16}-\frac{13297.9}{256}=\frac{43960}{256}-\frac{13297.9}{256}=\frac{43960 - 13297.9}{256}=\frac{30662.1}{256}\approx119.77$ cubic inches.

Step4: Calculate the weight of the bushing

The weight $W$ of the bushing is given by the density formula $W=\text{density}\times V$. The density is $0.2835$ pounds per cubic inch. So, $W = 0.2835\times119.77\approx33.9$ pounds.

Answer:

$33.9$