To find the depth below the surface of the sea where the pressure intensity is $91.58 \, \text{kPa}$, we can use the formula for pressure in a fluid:

$P = \rho g h$

Where:

• $P$ is the pressure intensity (in Pascals, Pa),
• $\rho$ is the density of the fluid (in $\text{kg/m}^3$),
• $g$ is the acceleration due to gravity (approximately $9.81 \, \text{m/s}^2$),
• $h$ is the depth below the surface (in meters).

Step 1: Convert the pressure to Pascals

Since $1 \, \text{kPa} = 1000 \, \text{Pa}$, the given pressure of $91.58 \, \text{kPa}$ becomes:

$P = 91.58 \times 1000 = 91580 \, \text{Pa}$

Step 2: Calculate the density of sea water

The density of sea water $\rho$ is related to the density of pure water $\rho_{\text{water}}$ (which is $1000 \, \text{kg/m}^3$) by the specific gravity $S$. The specific gravity of sea water is given as $0.989$, so the density of sea water is:

$\rho = S \times \rho_{\text{water}} = 0.989 \times 1000 = 989 \, \text{kg/m}^3$

Step 3: Solve for depth $h$

Rearranging the pressure formula to solve for depth:

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

Substituting the known values:

$h = \frac{91580}{989 \times 9.81}$

Now, let me calculate the depth.The depth of the point below the water surface where the pressure intensity is $91.58 \, \text{kPa}$ is approximately $9.44 \, \text{meters}$.

Do you want further details or have any questions?

Here are 5 related questions:

1. How would the depth change if the specific gravity of seawater increased?
2. What is the pressure at a depth of 15 meters in seawater with the same specific gravity?
3. How does the pressure vary with depth in a fluid with variable density?
4. What is the significance of specific gravity in determining fluid properties?
5. How would the calculation differ in freshwater, where specific gravity is 1?

Tip: Always ensure that units are consistent when applying formulas in fluid mechanics.