### Example Problem: Pressure and Force at Depth**Problem Statement:**(a) Calculate the absolute pressure at the bottom of a freshwater lake at a depth of 32.6 meters. Assume the density of water is 1.00 × 10³ kg/m³ and the air above is at a pressure of 101.3 kPa.**Answer:**\[ 421106\ \text{Pa} \]**Detailed Explanation and Solution:**The absolute pressure at a certain depth underwater can be calculated using the formula:\[ P = P_0 + \rho g h \]where:- \( P \) is the absolute pressure at depth,- \( P_0 \) is the atmospheric pressure at the surface (101.3 kPa or 101300 Pa),- \( \rho \) is the density of the water (1.00 × 10³ kg/m³),- \( g \) is the acceleration due to gravity (9.81 m/s²),- \( h \) is the depth of the water (32.6 m).Using the given values in the formula:\[ P = 101300\ \text{Pa} + (1.00 \times 10^3\ \text{kg/m}^3 \times 9.81\ \text{m/s}^2 \times 32.6\ \text{m}) \]\[ P = 101300\ \text{Pa} + 319806\ \text{Pa} \]\[ P = 421106\ \text{Pa} \]So, the absolute pressure at the bottom of the lake is 421106 Pa.---(b) What force is exerted by the water on the window of an underwater vehicle at this depth if the window is circular and has a diameter of 33.0 cm?**Original User Response:**\[ 143995.51\ \text{N} \]**Feedback:**Your response differs from the correct answer by more than 100%.**Detailed Explanation and Solution:**To calculate the force exerted on the window, we need to use the formula:\[ F = P \times A \]where:- \( F \) is the force exerted,- \( P \) is the pressure at depth (calculated as 421106 Pa),- \( A \) is the area of the circular window.First,

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Description: ### Example Problem: Pressure and Force at Depth**Problem Statement:**(a) Calculate the absolute pressure at the bottom of a freshwater lake at a depth of 32.6 meters. Assume the density of water is 1.00 × 10³ kg/m³ and the air above is at a pressur

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a) Calculate the absolute pressure at the bottom of a fresh-water lake at a depth of 32.6 m. Assume the density of the water is 1.00 x 10³ kg/m³ and the a 421106 Pa b) What force is exerted by the water on the window of an underwater vehicle at this depth if the window is circular and has a diameter of 33.0 cm? 143995.51 x 'our response differs from the correct answer by more than 100%. N Need Heln? Read It

a) Calculate the absolute pressure at the bottom of a fresh-water lake at a depth of 32.6 m. Assume the density of the water is 1.00 x 10³ kg/m³ and the a 421106 Pa b) What force is exerted by the water on the window of an underwater vehicle at this depth if the window is circular and has a diameter of 33.0 cm? 143995.51 x 'our response differs from the correct answer by more than 100%. N Need Heln? Read It

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter11: Fluid Statics

Section: Chapter Questions

Problem 75PE: Pressure in the spinal fluid is measured as shown in Figure 11.43. If the pressure in the spinal...

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Concept explainers

Fluid Pressure

The term fluid pressure is coined as, the measurement of the force per unit area of a given surface of a closed container. It is a branch of physics that helps to study the properties of fluid under various conditions of force.

Gauge Pressure

Pressure is the physical force acting per unit area on a body; the applied force is perpendicular to the surface of the object per unit area. The air around us at sea level exerts a pressure (atmospheric pressure) of about 14.7 psi but this doesn’t seem to bother anyone as the bodily fluids are constantly pushing outwards with the same force but if one swims down into the ocean a few feet below the surface one can notice the difference, there is increased pressure on the eardrum, this is due to an increase in hydrostatic pressure.

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### Example Problem: Pressure and Force at Depth

**Problem Statement:**

(a) Calculate the absolute pressure at the bottom of a freshwater lake at a depth of 32.6 meters. Assume the density of water is 1.00 × 10³ kg/m³ and the air above is at a pressure of 101.3 kPa.

**Answer:**
\[ 421106\ \text{Pa} \]

**Detailed Explanation and Solution:**

The absolute pressure at a certain depth underwater can be calculated using the formula:
\[ P = P_0 + \rho g h \]
where:
- \( P \) is the absolute pressure at depth,
- \( P_0 \) is the atmospheric pressure at the surface (101.3 kPa or 101300 Pa),
- \( \rho \) is the density of the water (1.00 × 10³ kg/m³),
- \( g \) is the acceleration due to gravity (9.81 m/s²),
- \( h \) is the depth of the water (32.6 m).

Using the given values in the formula:
\[ P = 101300\ \text{Pa} + (1.00 \times 10^3\ \text{kg/m}^3 \times 9.81\ \text{m/s}^2 \times 32.6\ \text{m}) \]
\[ P = 101300\ \text{Pa} + 319806\ \text{Pa} \]
\[ P = 421106\ \text{Pa} \]

So, the absolute pressure at the bottom of the lake is 421106 Pa.

---

(b) What force is exerted by the water on the window of an underwater vehicle at this depth if the window is circular and has a diameter of 33.0 cm?

**Original User Response:**
\[ 143995.51\ \text{N} \]
**Feedback:**
Your response differs from the correct answer by more than 100%.

**Detailed Explanation and Solution:**

To calculate the force exerted on the window, we need to use the formula:
\[ F = P \times A \]
where:
- \( F \) is the force exerted,
- \( P \) is the pressure at depth (calculated as 421106 Pa),
- \( A \) is the area of the circular window.

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