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(II) Estimate the air pressure inside a category 5 hurricane, where the wind speed is 300 $\mathrm{km} / \mathrm{h}$ (Fig. $53 ) .$

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$$9.7 \times 10^{4} \mathrm{Pa}$$

04:15

Averell Hause

Physics 101 Mechanics

Chapter 13

Fluids

Fluid Mechanics

Cornell University

Simon Fraser University

Hope College

McMaster University

Lectures

03:45

In physics, a fluid is a substance that continually deforms (flows) under an applied shear stress. Fluids are a subset of the phases of matter and include liquids, gases, plasmas and, to some extent, plastic solids.

09:49

A fluid is a substance that continually deforms (flows) under an applied shear stress. Fluids are a subset of the phases of matter and include liquids, gases and plasmas. Fluids display properties such as flow, pressure, and tension, which can be described with a fluid model. For example, liquids form a surface which exerts a force on other objects in contact with it, and is the basis for the forces of capillarity and cohesion. Fluids are a continuum (or "continuous" in some sense) which means that they cannot be strictly separated into separate pieces. However, there are theoretical limits to the divisibility of fluids. Fluids are in contrast to solids, which are able to sustain a shear stress with no tendency to continue deforming.

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in this problem. We're gonna talk about the Bernoulli equation. So what? The burden the equation tells us is that, uh, there is a constant that is conserved through streamline. This constant is the pressure put blows. One half of Rome B squared row is then the fluid V is the speed plus row G age. Tell me why instead of a and this is equal, this is constant. Okay, why is the height off the streamline on this is obtained, considering that role is constant. Okay, eso What we have in our problem is that we must calculate What is the pressure off the air inside a hurricane? Um, Category five hurricane, Actually, that has a speed v of 300 kilometers per hour. Eso I'm gonna consider that Why is constant throughout the streamline, So I'm going to just disregard the term the potential term we have P plus roll. One half of Roe V squared is constant. And and, um, since this is constant, this must be equal to this term outside the hurricane. So this is the term inside and outside the hurricane. Uh, we can assume that we are The air is still at rest so there is only pressure and that is the atmospheric pressure. So I'm gonna call it P 18. This is outside and our goal is to find P inside. So I'm gonna isolated. So be inside is the atmospheric pressure amount is one half of roe v squared a role. The density of the air is 1.2 to 5 kg per cubic meter. So P is the atmospheric pressure of 1.1 times 10 to the fifth. Moscow mine is one half of one point 2 to 5 kg per cubic meter times V squared. Uh, no notice that we have to transform V from kilometers per hour 2 m per second. So v is 300 Columbus per hour divided by 3.53 point six and that is equal to 83. Wait 3 m per second. We have to multiplied by 83.3 m per second, so the pressure is equal to 9.67 times 10 to the fourth was called, which is the answer to our exercise

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