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Chapter 14: Problem 19

Normal atmospheric pressure is \(1.013 \times 10^{5} \mathrm{Pa}\). The approach of a storm causes the height of a mercury barometer to drop by \(20.0 \mathrm{mm}\) from the normal height. What is the atmospheric pressure? (The density of mercury is \(\left.13.59 \mathrm{g} / \mathrm{cm}^{3} .\right)\)

Short Answer

Expert verified
The new atmospheric pressure after the storm is \(98636.982 \, Pa\).

Step by step solution

01

Convert the given parameters

First things, first. The parameters must be within the same unit system. It requires to convert the height drop from mm to m, which will be \(20.0 \, mm = 20.0 \times 10^{-3} \, m\). And the density should be converted from g/cm鲁 to kg/m鲁 resulting in \(13.59 \, g/cm鲁 = 13.59 \times 10^{3} \, kg/m鲁 \).
02

Calculate the pressure decrease

Now we can use the formula for pressure in a fluid column to calculate the pressure decrease: \(Pressure = density * g * height \rightarrow Pressure_{decrease} = 13.59 \times 10^{3} \, kg/m鲁 * 9.81 \, m/s虏 * 20.0 \times 10^{-3} \, m = 2663.018 \, Pa\)
03

Calculate the new pressure

With the pressure decrease calculated, we can find the new pressure by subtracting this from the normal atmospheric pressure: \(Pressure_{new} = 1.013 \times 10^{5} \, Pa - 2663.018 \, Pa = 98636.982 \, Pa\)

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Key Concepts

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

Fluid Mechanics
Fluid mechanics is a branch of physics that deals with the study of fluids (liquids and gases) and the forces on them. It encompasses various concepts such as fluid statics, fluid dynamics, and fluid kinetics, which describe the behavior of fluids at rest, in motion, and the energy changes, respectively.

One critical aspect of fluid mechanics is understanding how pressure within a fluid operates. Pressure is defined as the force exerted per unit area. In fluids, all points at the same depth experience equal pressure, known as hydrostatic equilibrium. This principle allows us to calculate forces on submerged surfaces and is foundational in understanding real-world applications like hydraulic systems, atmospheric phenomena, and undersea exploration.

For students grappling with these concepts, it's essential to visualize fluids as particles that are free to move and collide, imparting forces on their container walls or any object submerged within them. This particle interaction results in the uniform distribution of pressure at a given depth.
Pressure in a Fluid Column
The concept of pressure in a fluid column is vital in the study of fluid mechanics. It refers to the pressure exerted by a static fluid within a container due to the fluid's weight. The formula for calculating pressure at a point within a fluid is given by \( P = \rho g h \), where \( \rho \) is the fluid's density, \( g \) is the acceleration due to gravity, and \( h \) is the height of the fluid column above the point.

For a better understanding, consider a column of fluid as a stack of layers. Each layer's weight adds to the pressure exerted on the layers below. This concept explains why pressure increases with depth in fluids and is crucial when working with devices like barometers and manometers. It also plays a role in engineering applications, such as calculating the pressure at various points within dams, tanks, or the bloodstream in medical applications.

Students should note that the pressure calculation assumes the fluid is incompressible and at a constant gravity, conditions that are typically met for liquids such as water or mercury in everyday situations.
Mercury Barometer
A mercury barometer is an instrument designed to measure atmospheric pressure. It consists of a glass tube, closed at one end, filled with mercury and inverted in a mercury container. The weight of the mercury in the tube exerts a pressure that balances the atmospheric pressure, and the height of the mercury column acts as an indicator of this pressure.

The device operates on the principle that atmospheric pressure can support a column of mercury to a certain height, typically measured in millimeters of mercury (mmHg) or torr. The standard atmospheric pressure at sea level can support a mercury column of about 760 mmHg, equivalent to 1013.25 hPa (hectopascals).

When the mercury level in the barometer drops, it signifies a decrease in atmospheric pressure, often associated with changes in weather, such as the approach of a storm. Students studying meteorology or involved in weather prediction need to understand how to interpret the changes in barometer readings and relate them to upcoming weather conditions.

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Most popular questions from this chapter

How many cubic meters of helium are required to lift a balloon with a \(400-\mathrm{kg}\) payload to a height of \(8000 \mathrm{m} ?\) (Take \(\rho_{\mathrm{He}}=0.180 \mathrm{kg} / \mathrm{m}^{3} .\) ) Assume that the balloon maintains a constant volume and that the density of air decreases with the altitude \(z\) according to the expression \(\rho_{\text {air }}=\rho_{0} e^{-z / 8,000}\) where \(z\) is in meters and \(\rho_{0}=1.25 \mathrm{kg} / \mathrm{m}^{3}\) is the density of air at sca level.

\- A \(10.0-\mathrm{kg}\) block of metal measuring \(12.0 \mathrm{cm} \times 10.0 \mathrm{cm} \times\) \(10.0 \mathrm{cm}\) is suspended from a scale and immersed in water as in Figure P14.25b. The 12.0-cm dimension is vertical, and the top of the block is \(5.00 \mathrm{cm}\) below the surface of the water. (a) What are the forces acting on the top and on the bottom of the block? (Take \(P_{0}=1.0130 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}\) ) (b) What is the reading of the spring scale? (c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block.

The Bernoulli effect can have important consequences for the design of buildings. For example, wind can blow around a skyscraper at remarkably high speed, creating low pressure. The higher atmospheric pressure in the still air inside the buildings can cause windows to pop out. As originally constructed, the John Hancock building in Boston popped window panes, which fell many stories to the sidewalk below. (a) Suppose that a horizontal wind blows in streamline flow with a speed of \(11.2 \mathrm{m} / \mathrm{s}\) outside a large pane of plate glass with dimensions \(4.00 \mathrm{m} \times 1.50 \mathrm{m}\) Assume the density of the air to be uniform at \(1.30 \mathrm{kg} / \mathrm{m}^{3} .\) The air inside the building is at atmospheric pressure. What is the total force exerted by air on the window pane? (b) What If? If a second skyscraper is built nearby, the air speed can be especially high where wind passes through the narrow separation between the buildings. Solve part (a) again if the wind speed is \(22.4 \mathrm{m} / \mathrm{s}\), twice as high.

A spherical aluminum ball of mass \(1.26 \mathrm{kg}\) contains an empty spherical cavity that is concentric with the ball. The ball just barely floats in water. Calculate (a) the outer \(\mathrm{ra}\) dius of the ball and (b) the radius of the cavity.

Tor the cellar of a new house, a hole is dug in the ground, -with vertical sides going down \(2.40 \mathrm{m}\). A concrete foundamall is built all the way across the \(9.60-\mathrm{m}\) width of the excavation. This foundation wall is \(0.183 \mathrm{m}\) away from the front of the cellar hole. During a rainstorm, drainage from the strect fills up the spacc in front of the concrete wall, but not the cellar behind the wall. The water does not soak into the clay soil. Find the force the water causes on the foundation wall. For comparison, the weight of the water is given by \(2.40 \mathrm{m} \times 9.60 \mathrm{m} \times 0.183 \mathrm{m} \times 1000 \mathrm{kg} / \mathrm{m}^{3} \times\) \(9.80 \mathrm{m} / \mathrm{s}^{2}=41.3 \mathrm{kN}\)
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