Chapter 12.1, Problem 38SR

(a)

To determine

To simplify: The complex fraction.

Answer to Problem 38SR

The simplified fraction is:

  $P=\frac{814.45}{r^3}$

Explanation of Solution

Given information:

The following formula:

  $P=\frac{3412.94}{\frac{4\pi r^3}{3}}$

where $P$ represents the pressure in pound per square inch (psi) and $r$ is the radius of the ball In inches.

Calculation:

The given fraction can be simplified as follows:

  $\begin{array}{l}P=\frac{3412.94}{\frac{4\pi r^3}{3}} [\text{Original equation]}\\ =3412.94\times \frac{3}{4\pi r^3} [\text{Multiply by reciprocal of denominator]}\\ =3412.94\times \frac{3}{4\left(\frac{22}{7}\right)r^3} [\text{Substitute value of }\pi ]\\ =3412.94\times \frac{3}{\frac{88}{7}{r}^3} [\text{Simplify denominator]}\\ =3412.94\times 3\times \frac{7}{88r^3} [\text{Multiply by reciprocal of denominator]}\\ =\frac{71671.74}{88r^3} [\text{Multiply]}\\ =\frac{814.45}{r^3} [\text{Divide]}\end{array}$

(b)

To determine

To calculate: Radius of the ball.

Answer to Problem 38SR

Radius of the ballis $4.67\text{ inches}$

Explanation of Solution

Given information:

Pressure = $\text{8 psi}$

Calculation:

The radius of the ball is calculated as follows:

  $\begin{array}{l}P=\frac{814.45}{r^3} [\text{Calculated in part a]}\\ 8=\frac{814.45}{r^3} [\text{Substitute]}\\ 8\times r^3=\frac{814.45}{r^3}\times r^3 [\text{Multiply both sides by }{r}^3]\\ 8r^3=814.45 [\text{Simplify]}\\ \frac{8r^3}{8}=\frac{814.45}{8} [\text{Divide both sides by 8]}\\ r^3=101.81 [\text{Simplify]}\\ r=4.6694 [\text{Find cube root of both sides]}\\ r\approx 4.67 [\text{Round off]}\end{array}$