To determine the depth at which the rock is submerged, we can use the hydrostatic pressure equation:

$P = \rho g h$

Where:

• $P$ is the hydrostatic pressure (in lbf/in²),
• $\rho$ is the density of the fluid (in lbm/in³),
• $g$ is the acceleration due to gravity (in ft/s² or in/s²),
• $h$ is the depth (in inches).

Given:

• $P = 18.2 \, \text{lbf/in}^2$,
• $\rho = 0.25 \, \text{lbm/in}^3$.

We need to convert the units and constants to ensure consistency. The acceleration due to gravity in imperial units is $g = 386.4 \, \text{in/s}^2$.

Step 1: Rearrange the hydrostatic pressure formula for $h$

$h = \frac{P}{\rho g}$

Step 2: Plug in the values

$h = \frac{18.2 \, \text{lbf/in}^2}{(0.25 \, \text{lbm/in}^3)(386.4 \, \text{in/s}^2)}$

Now, I will calculate the value of $h$.The depth at which the rock is submerged is approximately $0.188$ inches, rounded to three significant figures.

Would you like any further details or explanations?

Here are some related questions to deepen your understanding:

1. How does changing the fluid density affect the depth calculation?
2. What would happen if the fluid was water instead of the given fluid?
3. How does the depth change if the pressure is doubled?
4. How does atmospheric pressure factor into submerged depth calculations?
5. How does the gravitational constant vary across different units?

Tip: Always ensure unit consistency when working with physical formulas to avoid incorrect results.