/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 A mercury barometer reads 747.0 ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A mercury barometer reads 747.0 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 building height is approximately 136.51 meters.

Step by step solution

01

Understanding the Problem

The problem involves two mercury barometer readings: 747.0 mm at the roof and 760.0 mm at the ground. We need to find the height of the building given the air density is 1.29 kg/m³.
02

Using Barometric Formula

To find the height, we use the barometric formula which connects pressure difference, air density, and height. The pressure difference between two heights is due to the weight of a column of air and is given by \( \Delta P = \rho g h \), where \( \rho \) is the air density, \( g \) is the acceleration due to gravity, and \( h \) is the height.
03

Calculate Pressure Difference

Convert the mercury readings from mm to pressure in pascals (Pa). The change in barometer reading is \( 760.0 \, \mathrm{mmHg} - 747.0 \, \mathrm{mmHg} = 13.0 \, \mathrm{mmHg} \). Using \( 1 \, \mathrm{mmHg} = 133.322 \, \mathrm{Pa} \), the pressure difference \( \Delta P = 13.0 \, \mathrm{mmHg} \times 133.322 = 1733.186 \, \mathrm{Pa} \).
04

Solve for Building Height

Using the barometric formula \( \Delta P = \rho g h \), solve for \( h \): \( h = \frac{\Delta P}{\rho g} \). Substituting the values: \( \rho = 1.29 \, \mathrm{kg/m^3} \), \( g = 9.81 \, \mathrm{m/s^2} \), \( \Delta P = 1733.186 \, \mathrm{Pa} \), we find \( h = \frac{1733.186}{1.29 \times 9.81} \approx 136.51 \, \mathrm{m} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Air Pressure
Air pressure is the force exerted by the weight of air molecules on a surface. It is one of the most important factors affecting weather patterns on Earth. As altitude increases, air pressure decreases since there are fewer air molecules pushing down from above.

In this context, when moving from the ground to the roof of a building, the air pressure is lower because of the decreased mass of air above. This concept is key when using instruments like barometers to measure pressure changes accurately in various elevations.

Air pressure is measured in units such as pascals (Pa) or millimeters of mercury (mmHg). The standard atmospheric pressure at sea level is 101,325 Pa or 760 mmHg, and any deviation from this is significant for determining altitude or depth.
Barometric Formula
The barometric formula is a crucial tool for calculating changes in air pressure with altitude. It describes how atmospheric pressure decreases exponentially with an increase in altitude. The formula is useful for determining how far above ground level a particular point is, by considering pressure differences.

In simple terms, the formula can be written as: \[\Delta P = \rho g h\]where
  • \(\Delta P\) is the change in atmospheric pressure,
  • \(\rho\) is the density of air at a specific temperature and humidity level,
  • \(g\) is the acceleration due to gravity (approximately 9.81 m/s²),
  • \(h\) is the height or altitude difference being calculated.
This formula allows for the calculation of the height of a building by knowing the pressure difference from two different elevations, such as the ground and the roof.
Mercury Barometer
A mercury barometer is a device used to measure atmospheric pressure. It consists of a glass tube filled with mercury, inverted in a dish of mercury. Changes in atmospheric pressure cause the mercury to rise or fall in the tube, providing a measurement.

The height of the mercury column in mm (millimeters) changes with air pressure. This is why when moving to higher altitudes or in different weather conditions, the mercury barometer reading will vary. A rise in mercury level indicates an increase in pressure, while a drop signifies a decrease. Due to precision and reliability, mercury barometers are commonly used for scientific experiments and weather stations.

When solving problems like determining the height of a building using barometer readings, understanding how these measurements correlate to pressure differences is vital.
Density of Air
The density of air is a measure of how much air mass is contained in a given volume. It's often expressed in units such as kilograms per cubic meter (kg/m³). Air density greatly influences air pressure and is affected by temperature, altitude, and humidity.

For the exercise involving barometer reading changes, knowing the approximate constant air density (in this case, 1.29 kg/m³) allows for calculations using the barometric formula. This value assumes standard conditions, which is crucial for achieving accurate results.

Fluctuations in air density can alter the precision of barometric calculations if not accounted for. Nevertheless, using an average or assumed constant value simplifies calculations while still providing reasonable estimates in everyday or educational problem-solving scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A siphon tube is useful for removing liquid from a tank. The siphon tube is first filled with liquid, and then one end is inserted into the tank. Liquid then drains out the other end, as the drawing illustrates. (a) Using reasoning similar to that employed in obtaining Torricelli’s theorem (see Example 16), derive an expression for the speed \(v\) of the fluid emerging from the tube. This expression should give \(v\) in terms of the vertical height \(y\) and the acceleration due to gravity \(g\) . (Note that this speed does not depend on the depth \(d\) of the tube below the surface of the liquid.) (b) At what value of the vertical distance y will the siphon stop working? (c) Derive an expression for the absolute pressure at the highest point in the siphon (point \(A )\) in terms of the atmospheric pressure \(P_{0},\) the fluid density \(\rho, g,\) and the heights \(h\) and \(y\) (Note that the fluid speed at point \(A\) is the same as the speed of the fluid emerging from the tube, because the cross-sectional area of the tube is the same everywhere.)

A patient recovering from surgery is being given fluid intravenously. The fluid has a density of \(1030 \mathrm{kg} / \mathrm{m}^{3},\) and \(9.5 \times 10^{-4} \mathrm{m}^{3}\) of it flows into the patient every six hours. Find the mass flow rate in \(\mathrm{kg} / \mathrm{s}\) .

A \(0.10-\mathrm{m} \times 0.20-\mathrm{m} \times 0.30-\mathrm{m}\) block is suspended from a wire and is completely under water. What buoyant force acts on the block?

A pressure difference of \(1.8 \times 10^{3}\) Pa is needed to drive water \(\left(\eta=1.0 \times 10^{-3} \mathrm{Pa} \cdot \mathrm{s}\right)\) through a pipe whose radius is \(5.1 \times 10^{-3} \mathrm{m} .\) The volume flow rate of the water is \(2.8 \times 10^{-4} \mathrm{m}^{3} / \mathrm{s} .\) What is the length of the pipe?

A 1.00-m-tall container is filled to the brim, partway with mercury and the rest of the way with water. The container is open to the atmosphere. What must be the depth of the mercury so that the absolute pressure on the bottom of the container is twice the atmospheric pressure?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.