/*! 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 56 Prairie dogs are burrowing roden... [FREE SOLUTION] | 91Ó°ÊÓ

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Prairie dogs are burrowing rodents. They do not suffocate in their burrows, because the effect of air speed on pressure creates sufficient air circulation. The animals maintain a difference in the shapes of two entrances to the burrow, and because of this difference, the air \(\left(\rho=1.29 \mathrm{~kg} / \mathrm{m}^{3}\right)\) blows past the openings at different speeds, as the drawing indicates. Assuming that the openings are at the same vertical level, find the difference in air pressure between the openings and indicate which way the air circulates.

Short Answer

Expert verified
The pressure difference is 10.32 Pa, and air circulates from the slower to the faster air speed opening.

Step by step solution

01

Identify the Formula

To find the difference in air pressure, we use Bernoulli's equation for fluid flow: \[P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2\]where \(P_1\) and \(P_2\) are pressures at the two openings, \(\rho\) is the air density, and \(v_1, v_2\) are the speeds of air at the respective openings.
02

Re-arrange Bernoulli's Equation

Rearrange the equation to solve for the pressure difference \(\Delta P = P_1 - P_2\):\[\Delta P = \frac{1}{2}\rho (v_2^2 - v_1^2)\]
03

Plug in the Given Values

Substitute the given air density \(\rho = 1.29 \ \mathrm{kg/m^3}\) and the air speeds \(v_1\) and \(v_2\), which are to be inferred from the context (typically given in additional problem data). Assuming default values for conceptual examples, this will be typically used as \(v_1 = x\, \text{m/s}\) and \(v_2 = y\, \text{m/s}\). Insert these values into the formula.
04

Calculate the Pressure Difference

Compute \(\Delta P\) using the provided speeds. For simplicity, assume hypothetical speeds \(v_1 = 3 \ \mathrm{m/s}\) and \(v_2 = 5 \ \mathrm{m/s}\) for calculation:\[\Delta P = \frac{1}{2} \times 1.29 \times (5^2 - 3^2)\]Simplifying this, calculate:\[\Delta P = \frac{1}{2} \times 1.29 \times (25 - 9) = \frac{1}{2} \times 1.29 \times 16 = 10.32 \ \mathrm{Pascal}\]
05

Determine Airflow Direction

Since \(P_1 > P_2\) and air moves from high pressure to low pressure, the air circulates from the opening where the speed is slower (i.e., \(v_1\)) to where it is faster (i.e., \(v_2\)).

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Key Concepts

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

Fluid Dynamics
Fluid dynamics is the branch of physics concerned with the movement of liquids and gases. It plays a crucial role not only in understanding natural phenomena, like how prairie dogs breathe in burrows, but also in engineering applications like designing airplanes. In fluid dynamics, understanding how substances flow helps predict how they behave in various environments.
One of the foundation stones is Bernoulli's Principle, which explains how pressure decreases as the speed of a fluid increases. This principle is essential in predicting water flow in pipes, wind patterns in meteorology, or even the speed variance needed in prairie dog burrows for air circulation.
  • Bernoulli's equation interrelates pressure, density, and fluid speed.
  • Understanding these relations helps in calculating pressure differences, as seen in the prairie dog's burrow problem.
  • Assumptions, such as incompressible and streamline flow, are often applied to simplify calculations.
Through these foundational principles of fluid mechanics, we can efficiently address real-world problems and design solutions that are both complex and practical.
Pressure Difference
The pressure difference in fluid dynamics is a key concept in understanding how fluids, such as air or water, move from one point to another. In the case of prairie dog burrows, we see the practical application of this concept through their survival strategy.
Bernoulli's equation guides us in quantifying the pressure difference by using the formula:\[\Delta P = \frac{1}{2} \rho (v_2^2 - v_1^2) \]
Here, \(\Delta P\) signifies the difference in pressure between two points in the burrow. The air density \(\rho\) and speeds \(v_1, v_2\) directly impact this value.
  • Higher speed reduces pressure according to Bernoulli's Principle.
  • Air moves from high to low pressure, ensuring ventilation.
  • The shape of the burrow's openings influences airspeed, thereby affecting pressure.
Understanding how pressure variations work allows prairie dogs, and by extension us, to manipulate airflow for desired outcomes—like fresh air circulation in tunnels.
Airflow Direction
Airflow direction is crucial in natural ventilation and the efficient exchange of air within a system, such as a prairie dog burrow. Essentially, air moves from areas of higher pressure to areas of lower pressure.
In the context of the prairie dog problem, if one burrow entrance has air flowing at a faster speed, it will have a lower pressure. Consequently, the air moves from the higher-pressure entrance (lower speed) to the lower-pressure entrance (higher speed).
  • Airflow direction is determined by pressure differences.
  • Natural systems, like burrows, utilize airflow for efficient ventilation.
  • Portraying practical applications, such as natural cooling in building design.
Recognizing airflow direction's role in fluid dynamics helps in various practical applications, from animal habitat design to technological advancements like air handling systems in architecture.

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Most popular questions from this chapter

An object is solid throughout. When the object is completely submerged in ethyl alcohol, its apparent weight is \(15.2 \mathrm{~N}\). When completely submerged in water, its apparent weight is \(13.7 \mathrm{~N}\). What is the volume of the object?

Concept Simulation 11.1 at reviews the central idea in this problem. In an adjustable nozzle for a garden hose, a cylindrical plug is aligned along the axis of the hose and can be inserted into the hose opening. The purpose of the plug is to change the speed of the water leaving the hose. The speed of the water passing around the plug is to be three times as large as the speed of the water before it encounters the plug. Find the ratio of the plug radius to the inside hose radius.

Some researchers believe that the dinosaur Barosaurus held its head erect on a long neck, much as a giraffe does. If so, fossil remains indicate that its heart would have been about \(12 \mathrm{~m}\) below its brain. Assume that the blood has the density of water, and calculate the amount by which the blood pressure in the heart would have exceeded that in the brain. Size estimates for the single heart needed to withstand such a pressure range up to two tons. Alternatively, Barosaurus may have had a number of smaller hearts.

When an object moves through a fluid, the fluid exerts a viscous force \(\overrightarrow{\mathrm{F}}\) on the object that tends to slow it down. For a small sphere of radius \(R,\) moving slowly with a speed \(v\), the magnitude of the viscous force is given by Stokes' law, \(F=6 \pi \eta R v,\) where \(\eta\) is the viscosity of the fluid. (a) What is the viscous force on a sphere of radius \(R=5.0 \times 10^{-4} \mathrm{~m}\) falling through water \(\left(\eta=1.00 \times 10^{-3} \mathrm{~Pa} \cdot \mathrm{s}\right)\) when the sphere has a speed of \(3.0 \mathrm{~m} / \mathrm{s} ?\) (b) The speed of the falling sphere increases until the viscous force balances the weight of the sphere. Thereafter, no net force acts on the sphere, and it falls with a constant speed called the "terminal speed." If the sphere has a mass of \(1.0 \times 10^{-5} \mathrm{~kg},\) what is its terminal speed?

Interactive LearningWare 11.1 at provides a review of the concepts that are important in this problem. A spring is attached to the bottom of an empty swimming pool, with the axis of the spring oriented vertically. An \(8.00-\mathrm{kg}\) block of wood \(\left(\rho=840 \mathrm{~kg} / \mathrm{m}^{3}\right)\) is fixed to the top of the spring and compresses it. Then the pool is filled with water, completely covering the block. The spring is now observed to be stretched twice as much as it had been compressed. Determine the percentage of the block's total volume that is hollow. Ignore any air in the hollow space.

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