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Chapter 9: Problem 21

Calculate the absolute pressure at the bottom of a freshwater lake at a depth of \(27.5 \mathrm{~m}\). Assume the density of the water is \(1.00 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) and the air above is at a pressure of \(101.3 \mathrm{kPa}\). (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 \(35.0 \mathrm{~cm}\) ?

Short Answer

Expert verified
The absolute pressure at the bottom of the lake is \( \mathrm{Pa}\). The force exerted by the water on the window is \( \mathrm{N}\).

Step by step solution

01

Calculate the pressure at the bottom of the lake

First calculate the pressure due to the water column above the point in the lake. This pressure, often referred to as the hydrostatic pressure \(P_{w}\), can be calculated using the formula: \(P_{w} = 蟻gh\), where 蟻 is the density of the fluid (water in this case), \(g\) is the acceleration due to gravity, and \(h\) is the height (or depth) of the water column. Substituting the given values we get, \(P_{w} = (1.00 \times 10^{3} \mathrm{~kg/m^{3}}) \times (9.8 \mathrm{~m/s^{2}}) \times (27.5 \mathrm{~m})\).
02

Calculate the absolute pressure at the bottom of the lake

The absolute pressure \(P\) at the bottom of the lake is the sum of the atmospheric pressure \(P_{atm}\) and the hydrostatic pressure \(P_{w}\). So we have, \(P = P_{atm} + P_{w}\). Substituting the known values we get, \(P = (101.3 \times 10^{3} \mathrm{~Pa}) + P_{w}\).
03

Calculate the area of the window

To compute the force exerted by the water on the window, we need to know the area of the window. The window is circular and its area \(A\) is given by the formula \(A = 蟺d^{2}/4\), where \(d\) is the diameter of the circle. So we have, \(A = 蟺(0.35 \mathrm{~m})^{2}/4\).
04

Calculate the force exerted by water on the window

The force \(F\) exerted by the water on the window can be calculated using the formula \(F = PA\), where \(P\) is the pressure and \(A\) is the area. So we have, \(F = P \times A\). Substituting the known values and solving we get the force exerted by the water on the window.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydrostatic Pressure
Hydrostatic pressure is the pressure exerted by a fluid at equilibrium due to the gravitational pull acting on it. In the context of a body of water, this pressure is induced by the weight of the water column above a given point. To find hydrostatic pressure, we use the formula: \[ P_{w} = \rho gh \] where:
  • \(\rho\) is the density of the fluid.
  • \(g\) is the acceleration due to gravity.
  • \(h\) is the depth or height of the fluid column.
By inserting the values given in the problem, such as the water density of \(1.00 \times 10^{3} \text{ kg/m}^{3}\), gravity \(9.8 \text{ m/s}^{2}\), and the depth \(27.5 \text{ m}\), we can compute the pressure exerted solely by the water above. This understanding of hydrostatic pressure is crucial when working with fluids, and especially in applications like dam constructions and evaluating underwater pressures.
Absolute Pressure
Absolute pressure represents the total pressure at a given point in a fluid, comprised of both atmospheric pressure and the hydrostatic pressure. It is essential in understanding scenarios where the fluid is not just acting under its own weight but is also influenced by the surrounding atmospheric conditions. The absolute pressure \(P\) at a particular depth in a lake, is computed as follows:\[ P = P_{\text{atm}} + P_{w} \] Here, \(P_{\text{atm}}\) is the atmospheric pressure above the fluid surface (given as \(101.3 \text{ kPa}\) or \(101.3 \times 10^{3} \text{ Pa}\) in this case), and \(P_{w}\) is the hydrostatic pressure obtained from the earlier calculation. Summing these pressures gives the absolute pressure, illustrating how pressures add up in fluid systems, factoring in both local fluid weights and external pressure influences.
Force Calculation
When immersed in a fluid, objects experience pressure exerted over their surface area, creating a force. To determine the force on an object, such as a window, we first calculate the area of the object's surface where the force is applied. For a circular window, the area \(A\) can be obtained using the formula: \[ A = \frac{\pi d^2}{4} \] With a diameter \(d\) of \(0.35 \text{ m}\), one substitutes this value into the formula to ascertain the window's area. Once the area is known, the force \(F\) exerted by the water on this area is found through: \[ F = PA \] where \(P\) is the already calculated absolute pressure. This calculation highlights how pressures translate physically into forces, important for designing structures that encounter fluid forces, like submarines and underwater observatories.

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Most popular questions from this chapter

(a) Calculate the mass flow rate (in grams per second) of blood \(\left(\rho=1.0 \mathrm{~g} / \mathrm{cm}^{\mathrm{l}}\right)\) in an aorta with a cross-sectional area of \(2.0 \mathrm{~cm}^{2}\) if the flow speed is \(40 \mathrm{~cm} / \mathrm{s}\). (b) Assume that the aorta branches to form a large number of capillaries with a combined cross-sectional area of \(3.0 \times\) \(10^{3} \mathrm{~cm}^{2} .\) What is the flow speed in the capillaries?

Determining the density of a fluid has many imporLant applications. A car battery contains sulfuric acid, and the battery will not function properly if the acid density is too low. Similarly, the effectiveness of antifreeze in your car's engine coolant depends on the density of the mixture (usually ethylene glycol and water). When you donate blood to a blood bank, its screcning includes a determination of the density of the blood because higher density correlates with higher hemoglobin content. A hydrometer is an instrument used to determine the density of a liquid. \(A\) simple one is sketched in Figure \(\mathrm{P} 9.84\). The bulb of a syringe is squeezed and released to lift a sample of the liquid of interest into a tube containing a calibrated rod of known density. (Assume the rod is cylindrical.) The rod, of length \(L\) and average density \(\rho_{0}\). floats partially immersed in the liquid of density \(p .\) A length \(h\) of the tod protrudes above the surface of the liquid. Show that the density of the liquid is given by $$ \rho=\frac{\rho_{11} L}{I_{A}-h} $$

A rubber ball filled with air has a diameter of \(25.0 \mathrm{~cm}\) and a mass of \(0.540 \mathrm{~kg}\). What force is required to hold the ball in equilibrium immediately below the surface of water in a swimming pool?

Take the density of blood to be \(\rho\) and the distance between the feet and the heart to be \(h_{H}\). Ignore the flow of blood, (a) Show that the difference in blood pressure between the feet and the heart is given by \(P_{F}-P_{H}=\rho g h_{j F}\) (b) Take the density of blood to be \(1.05 \times 10^{4} \mathrm{~kg} / \mathrm{m}^{3}\) and the distance between the heart and the feet to be \(1.20 \mathrm{~m}\). Find the difference in blood pressure between these two points. This problem indicates that pumping blood from the extremities is very difficult for the heart. The veins in the legs have valves in them that open when blood is pumped toward the heart and close when blood flows away from the heart. Also, pumping action produced by physical activities such as walking and breathing assists the heart.

The block of ice (temperature \(0^{\circ} \mathrm{C}\) ) shown in Figure \(\mathrm{P} 9.63\) is drawn over a level surface lubricated by a layer of water \(0.10 \mathrm{~mm}\) thick. Determine the magnitude of the force \(\vec{F}\) needed to pull the block with a constant speed of \(0.50 \mathrm{~m} / \mathrm{s}\). At \(0^{\circ} \mathrm{C}\), the viscosity of water has the value \(\eta=1.79 \times 10^{-3} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\)
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