Question

Please explain and help me get the correct answer. Thank you. Practice work that is not graded.

Answer 1

First, from the formula given we solve the equation for D:

`\begin{gathered} 4PD=D^2LN\pi, \\ (4PD)/(LN\pi)=D^2, \\ \sqrt{(4PD)/(LN\pi)}=D^{}. \end{gathered}`

Now, substituting the given data:

(a) PD=405 cu in, L=4.7 in, and N=7,

in the above equation we get:

`D=\sqrt[]{(4*405)/(4.7*7*\pi)}in=\sqrt[]{15.6736175}in\approx3.96\text{ in}`

(b) PD=399.4 cu in, L=2 in, and N=6,

in the above equation we get:

`D=\sqrt[]{(4*399.4)/(2*6*\pi)}in=\sqrt[]{42.37765618}in\approx6.51\text{ in}`

Answer:

(a) 3.96 in.

(b) 6.51 in.