Truly, I am slow when it comes to physics. Any help is appreciated! Three cups i

Question

Truly, I am slow when it comes to physics. Any help is appreciated!

Three cups inverted into water arc shown below. Cup 1 has a water level inside that is below the water level exposed to the atmosphere. Cup 2 has a water level inside that is above the water level exposed to the atmosphere. Cup 3 has a layer of oil (which is less dense than water) atop the water level inside of it. Cup 2 and cup 3 have the same volume of air trapped in them. All three cups have air trapped inside of them. Apply the energy density model to each cup so you can rank the pressures (P1, P2, P3) of the air pockets trapped inside them, from least to greatest pressure. There are many interesting applications of our energy density model to the flow of blood in the human circulatory system. One interesting phenomenon is an aneurysm, a sudden abnormal enlargement of a section of an artery due to a weakening of the arterial wall. If the blood flow rate remains constant through the artery, how does the pressure in the enlarged section (the aneurysm) compare to the pressure in the rest of the artery? Neglect any affects of resistance to flow in your explanation.

Explanation / Answer

5.1-5

Energy density is just a fancy name for pressure

Pressure is same at the bottom of the cups (same level-Pascal's law)

thus, Air pressure 1 + h1d1g = Air pressure 2 + h2d1g

= Air pressure 3 + (h2-h1)d2g +h1d1g

from the first 2, we get that since h2>h1, AP2<AP1

from the next 2, we get that since d2<d1, AP3>AP2

from first and third, we get that AP1>AP3

thus, finally AP1>AP3>AP2

5.1-6

for fluids flowing in tubes (blood vessel in this case)

P+0.5dv^2 + gh is constant (also called the bernoulli equation

for the same blood vessel, the heights remain same i.e h1=h2

for same flow rate, inc in area decreases the speed at which the blood flows as vA must remain same

hence, P increases due to enlargement

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