Problem 89 - Fundamentals of Physics

Question

For seawater of density \(1.03 \mathrm{~g} / \mathrm{cm}^{3}\), find the weight of water on top of a submarine at a depth of \(255 \mathrm{~m}\) if the horizontal crosssectional hull area is \(2200.0 \mathrm{~m}^{2}\). (b) In atmospheres, what water pressure would a diver experience at this depth?

Step-by-Step Solution

Answer

The weight of water is 5662740000 N, and the pressure is approximately 25.38 atm.

Step 1: Convert Units for Density

Convert density from grams per cubic centimeter to kilograms per cubic meter for consistency in calculations. Given density of seawater is \(1.03 \text{ g/cm}^3\), convert it to \(\text{kg/m}^3\): \[1.03 \text{ g/cm}^3 = 1.03 \times 1000 \text{ kg/m}^3 = 1030 \text{ kg/m}^3\]

Step 2: Calculate Volume of Water

Determine the volume of water exerting pressure on the submarine. The volume is the product of the cross-sectional area and the depth: \[ V = A \times h = 2200 \text{ m}^2 \times 255 \text{ m} = 561000 \text{ m}^3\]

Step 3: Calculate Mass of Water

The mass of the water is given by multiplying the volume by the density: \[ m = \rho \times V = 1030 \text{ kg/m}^3 \times 561000 \text{ m}^3 = 577830000 \text{ kg} \]

Step 4: Calculate Weight of Water

Weight is mass times gravitational acceleration \(g = 9.8 \text{ m/s}^2\): \[ W = m \times g = 577830000 \text{ kg} \times 9.8 \text{ m/s}^2 = 5662740000 \text{ N} \]

Step 5: Calculate Pressure in Pascals

Pressure exerted by the water is calculated as \( P = \rho \times g \times h \): \[ P = 1030 \text{ kg/m}^3 \times 9.8 \text{ m/s}^2 \times 255 \text{ m} = 2571660 \text{ N/m}^2 \text{ or } 2571660 \text{ Pa} \]

Step 6: Convert Pressure to Atmospheres

Convert the pressure from Pascals to atmospheres, knowing that 1 atmosphere is approximately 101325 Pa: \[ \frac{2571660 \text{ Pa}}{101325 \text{ Pa/atm}} \approx 25.38 \text{ atm} \]

Key Concepts

Density of Seawater

The density of seawater is a critical factor when discussing fluid pressure and buoyancy principles. It is typically measured in grams per cubic centimeter (g/cm³) or kilograms per cubic meter (kg/m³). The standard density of seawater, due to dissolved salts, is about 1.03 g/cm³. This translates to 1030 kg/m³ when converted. Such conversions are fundamental in physics to maintain consistency across units.

• To convert from g/cm³ to kg/m³, multiply by 1000, as 1 g/cm³ equals 1000 kg/m³.
• Seawater's density affects submergence depth and buoyant forces on objects like submarines.

The density remains relatively constant but can vary slightly due to salinity and temperature changes.

Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at equilibrium at a given point within the fluid, due to gravity. This concept is pivotal in understanding the forces applied by water on submerged objects.
The formula for hydrostatic pressure is based on depth (h), density of the fluid (\(\rho\)), and gravitational acceleration (g):

1. \( P = \rho \cdot g \cdot h \)

Where:
- \( P \) is the pressure,
- \( \rho \) is the density of the fluid (e.g., seawater),
- \( g \) is the acceleration due to gravity (≈ 9.8 m/s²), and
- \( h \) is the depth below the surface.The deeper an object is, the greater the pressure exerted upon it because more fluid is above it, increasing weight and, hence, pressure. This relationship explains why divers and submarines experience intense pressure changes with depth. It’s also why understanding and calculating hydrostatic pressure accurately is crucial for designing underwater structures and vehicles.

Unit Conversion in Physics

Unit conversion in physics ensures that all measurements are consistent and comparable. In computations, especially involving different physical quantities, converting units is often required to prevent errors and misinterpretation.
Here are some pointers for unit conversions:

• Always convert to the correct standard unit before performing calculations, such as converting densities from g/cm³ to kg/m³, or pressure from Pascals to atmospheres.
• Know that 1 atmosphere (atm) equals 101325 Pascals (Pa); this is vital when calculating pressure effects in different mediums.
• Standard unit conversion in physics often employs multiplication with powers of ten, as seen in the conversion from cm to m where 1 cm = 0.01 m.

Using consistent units across calculations not only simplifies the process but also ensures accuracy in the outcomes. For problems involving fluid pressure and densities, mastering these conversions is essential to derive meaningful and correct results.