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Chapter 11: Problem 28

A mercury barometer reads \(747.0 \mathrm{mm}\) on the roof of a building and \(760.0 \mathrm{mm}\) on the ground. Assuming a constant value of \(1.29 \mathrm{kg} / \mathrm{m}^{3}\) for the density of air, determine the height of the building.

Short Answer

Expert verified
The height of the building is approximately 137 meters.

Step by step solution

01

Understand the Problem

We have two barometer readings: one at the roof of the building and one at the ground. The difference in these readings is due to the atmospheric pressure, which can be used to calculate the height of the building. The air density is given as a constant at 1.29 kg/m鲁.
02

Identify Relevant Formula

The difference in barometer readings can be used to find pressure difference: \[ \Delta P = P_2 - P_1 = \rho \cdot g \cdot h \] Where \(P_1\) is the pressure at the roof and \(P_2\) is the pressure at the ground, \(\rho\) is the density of air, \(g\) is the acceleration due to gravity, and \(h\) is the height of the building.
03

Convert Pressure Readings into Difference

The pressure difference \(\Delta P\) based on the mercury barometer readings is \[ \Delta P = 760.0 \text{ mmHg} - 747.0 \text{ mmHg} = 13.0 \text{ mmHg} \] To use the formula, convert this difference to Pascals (N/m虏). Recall that 1 mmHg = 133.322 Pa.
04

Calculate Pressure Difference in Pascals

Convert \(13.0 \text{ mmHg}\) to Pascals:\[ \Delta P = 13.0 \times 133.322 = 1733.186 \text{ Pa} \]
05

Use Formula to Solve for Height

Rearrange the formula to solve for height \(h\):\[ h = \frac{\Delta P}{\rho \cdot g} \]Substitute \(\Delta P = 1733.186 \text{ Pa}\), \(\rho = 1.29 \text{ kg/m}^3\), and \(g = 9.81 \text{ m/s}^2\):\[ h = \frac{1733.186}{1.29 \times 9.81} \]
06

Compute the Height of the Building

Calculate the height:\[ h \approx \frac{1733.186}{12.6549} \approx 137.0 \text{ meters} \]

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Key Concepts

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

Barometric Pressure
Barometric pressure, also known as atmospheric pressure, is the force exerted by the atmosphere at a given point. It is typically measured using a barometer, which can be filled with mercury or other fluids. These measurements are often presented in millimeters of mercury (mmHg).

Understanding barometric pressure is essential for many scientific applications, such as meteorology, where it helps predict weather changes. In the context of our exercise, variations in barometric pressure allowed us to determine the height of a building. The difference between the barometric readings at the roof and the ground forms the basis of our calculation.
  • Barometric pressure can vary with altitude, weather, and local conditions.
  • These variations are vital for calculations involving heights and altitudes.
  • Barometers provide a straightforward method to measure atmospheric pressure.
Density of Air
The density of air is a measure of how much mass of air is contained in a given volume. This property is crucial in many physical and engineering calculations, such as aerodynamics and building design. For our exercise, the density of air is assumed to be constant at 1.29 kg/m鲁, simplifying our calculations.

Air density can vary with temperature, pressure, and humidity. These factors all change the mass of air within a specific area, affecting calculations that rely on this property. In physics problems like ours, assuming a constant air density helps isolate other variables, making complex problems more manageable.
  • Density of air affects buoyancy and the behavior of objects in the atmosphere.
  • It is a key factor in calculating pressure differences and understanding atmospheric phenomena.
  • In real-world applications, variations in air density must be accounted for.
Pressure Conversion
Pressure conversion is essential when dealing with different measurement units. In our problem, we converted the pressure difference from mmHg (millimeters of mercury) to Pascals (Pa) to apply the relevant formulas correctly.

Each measurement unit can suit different scientific and practical needs. Knowing how to convert between these units is important for solving physics problems across various settings. The conversion factor for mercury to Pascals, where 1 mmHg equals 133.322 Pa, was vital for our solution.
  • Pressure is frequently measured in units not directly compatible, necessitating conversions.
  • Understanding these conversion factors ensures accurate and sensible results.
  • This knowledge is crucial in both academic exercises and real-world applications, such as meteorology and engineering.

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Most popular questions from this chapter

(a) The volume flow rate in an artery supplying the brain is \(3.6 \times 10^{-6} \mathrm{m}^{3} / \mathrm{s}\). If the radius of the artery is \(5.2 \mathrm{mm},\) determine the average blood speed. (b) Find the average blood speed at a constriction in the artery if the constriction reduces the radius by a factor of \(3 .\) Assume that the volume flow rate is the same as that in part (a).

A water bed for sale has dimensions of \(1.83 \mathrm{m} \times 2.13 \mathrm{m} \times 0.229 \mathrm{m}\) The floor of the bedroom will tolerate an additional weight of no more than \(6660 \mathrm{N}\). Find the weight of the water in the bed and determine whether the bed should be purchased.

Water is circulating through a closed system of pipes in a two-floor apartment. On the first floor, the water has a gauge pressure of \(3.4 \times 10^{5} \mathrm{Pa}\) and a speed of \(2.1 \mathrm{m} / \mathrm{s}\). However, on the second floor, which is \(4.0 \mathrm{m}\) higher, the speed of the water is \(3.7 \mathrm{m} / \mathrm{s}\). The speeds are different because the pipe diameters are different. What is the gauge pressure of the water on the second floor?

One way to administer an inoculation is with a "gun" that shoots the vaccine through a narrow opening. No needle is necessary, for the vaccine emerges with sufficient speed to pass directly into the tissue beneath the skin. The speed is high, because the vaccine \(\left(\rho=1100 \mathrm{kg} / \mathrm{m}^{3}\right)\) is held in a reservoir where a high pressure pushes it out. The pressure on the surface of the vaccine in one gun is \(4.1 \times 10^{6}\) Pa above the atmospheric pressure outside the narrow opening. The dosage is small enough that the vaccine's surface in the reservoir is nearly stationary during an inoculation. The vertical height between the vaccine's surface in the reservoir and the opening can be ignored. Find the speed at which the vaccine emerges.

A suitcase (mass \(m=16 \mathrm{kg}\) ) is resting on the floor of an elevator. The part of the suitcase in contact with the floor measures \(0.50 \mathrm{m} \times 0.15 \mathrm{m}\) The elevator is moving upward with an acceleration of magnitude \(1.5 \mathrm{m} / \mathrm{s}^{2}\) What pressure (in excess of atmospheric pressure) is applied to the floor beneath the suitcase?
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