# UB, Phy101: Chapter 11, Pg 1 Physics 101: Chapter 11 Fluids l Textbook Sections 11.1-11.6 è Density è Pressure è Pascal’s Principle l Textbook Sections.

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UB, Phy101: Chapter 11, Pg 1 Physics 101: Chapter 11 Fluids l Textbook Sections 11.1-11.6 è Density è Pressure è Pascal’s Principle l Textbook Sections 11.6-11.10 è Archimedes Principle & Buoyancy è Fluids in motion: Continuity & Bernoulli’s equation è Some problems, homework hints

UB, Phy101: Chapter 11, Pg 2

UB, Phy101: Chapter 11, Pg 3 l Density = Mass/Volume è  = M/V è units = kg/m 3 l Densities of some common things (kg/m 3 ) è Water 1000 è ice 917(floats on water) è blood 1060(sinks in water) è lead 11,300 è Copper 8890 è Mercury 13,600 è Aluminum 2700 è Wood 550 è air 1.29 è Helium 0.18 Physics 101: Density

UB, Phy101: Chapter 11, Pg 4 Qualitative Demonstration of Pressure y

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UB, Phy101: Chapter 11, Pg 8 Atmospheric Pressure normal atmospheric pressure = 1.01 x 10 5 Pa (14.7 lb/in 2 )

UB, Phy101: Chapter 11, Pg 9 Chapter 11, Preflight You buy a bag of potato chips in Urbana Illinois and forget them (un-opened) under the seat of your car. You drive out to Denver Colorado to visit a friend for Thanksgiving, and when you get there you discover the lost bag of chips. The odd thing you notice right away is that the bag seems to have inflated like a balloon (i.e. it seems much more round and bouncy than when you bought it). How can you explain this ?? Due to the change in height, the air is much thinner in Denver. Thus, there is less pressure on the outside of the bag of chips in Denver, so the bag seems to inflate because the air pressure on the inside is greater than on the outside. PUPU PUPU In Urbana PUPU PDPD In Denver

UB, Phy101: Chapter 11, Pg 10

UB, Phy101: Chapter 11, Pg 11 Pressure and Depth Barometer: a way to measure atomospheric pressure p 2 = p 1 +  gh p atm =  gh Measure h, determine p atm example--Mercury  = 13,600 kg/m 3 p atm = 1.05 x 10 5 Pa  h = 0.757 m = 757 mm = 29.80” (for 1 atm) h p 2 =p atm p 1 =0

UB, Phy101: Chapter 11, Pg 12 Chapter 11, Preflights Suppose you have a barometer with mercury and a barometer with water. How does the height h water compare with the height h mercury ? 1. h water is much larger than h mercury 2. h water is a little larger than h mercury 3. h water is a little smaller than h mercury 4. h water is much smaller than h mercury CORRECT water is much less dense than mercury, so the same amount of pressure will move the water farther up the column. h p 2 =p atm p 1 =0

UB, Phy101: Chapter 11, Pg 13 Chapter 11, Preflights Is it possible to stand on the roof of a five story (50 foot) tall house and drink, using a straw, from a glass on the ground? 1. No 2. Yes Even if a person could completely remove all of the air from the straw, the height to which the outside air pressure moves the water up the straw would not be high enough for the person to drink the water. CORRECT

UB, Phy101: Chapter 11, Pg 14 Chapter 11, Preflight Evacuate the straw by sucking How high will water rise? no more than h = P a /  g = 10.3m = 33.8 feet no matter how hard you suck! h p a p=0

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UB, Phy101: Chapter 11, Pg 22 Summary Density Pressure P 2 = P 1 +  gh Pascal’s Principle

UB, Phy101: Chapter 11, Pg 23 l Textbook Sections 11.6-11.10 è Archimedes Principle & Buoyancy è Fluids in motion: Continuity & Bernoulli’s equation è Some problems, homework hints Note: Everything we do assumes fluid is non-viscous and incompressible.

UB, Phy101: Chapter 11, Pg 24

UB, Phy101: Chapter 11, Pg 25

UB, Phy101: Chapter 11, Pg 26 Archimedes Principle (summary) l Buoyant Force (B) è weight of fluid displaced è B =  fluid x V displ g è W = Mg =  object V object g è object sinks if  object >  fluid è object floats if  object <  fluid è Eureka! l If object floats…. è B=W è Therefore  fluid g V displ. =  object g V object è Therefore V displ. /V object =  object /  fluid

UB, Phy101: Chapter 11, Pg 27 Suppose you float a large ice-cube in a glass of water, and that after you place the ice in the glass the level of the water is at the very brim. When the ice melts, the level of the water in the glass will: 1. Go up, causing the water to spill out of the glass. 2. Go down. 3. Stay the same. CORRECT Chapter 11, Preflight B =  W g V displaced W =  ice g V ice   W g V Must be same!

UB, Phy101: Chapter 11, Pg 28 Chapter 11, Preflight Which weighs more: 1. A large bathtub filled to the brim with water. 2. A large bathtub filled to the brim with water with a battle-ship floating in it. 3. They will weigh the same. Tub of water Tub of water + ship Overflowed water CORRECT Weight of ship = Buoyant force = Weight of displaced water

UB, Phy101: Chapter 11, Pg 29 Chapter 11: Fluid Flow (summary) Mass flow rate:  Av (kg/s) Volume flow rate: Av (m 3 /s) Continuity:  A 1 v 1 =  A 2 v 2 i.e., mass flow rate the same everywhere e.g., flow of river For fluid flow without friction: Bernoulli: P 1 + 1 / 2  v 1 2 +  gh 1 = P 2 + 1 / 2  v 2 2 +  gh 2 A 1 P 1 A 2 P 2 v1v1 v2v2 

UB, Phy101: Chapter 11, Pg 30 Chapter 11, Preflight A stream of water gets narrower as it falls from a faucet (try it & see). Explain this phenomenon using the equation of continuity hmm, gravity stretches it out A1A1 A2A2 V1V1 V2V2 The water's velocity is increasing as it flows down, so in order to compensate for the increase in velocity, the area must be decreased because the density*area*speed must be conserved First off the mass of the water is conserved, and water is a incompressible fluid. The equation of continuity states that A1V1= A2V2. We know that the velocity or V2 will be increasing due to gravity so in order for the equation to work the area must decrease or get narrower.

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UB, Phy101: Chapter 11, Pg 32 Chapter 11, Preflight A large bucket full of water has two drains. One is a hole in the side of the bucket at the bottom, and the other is a pipe coming out of the bucket near the top, which bent is downward such that the bottom of this pipe even with the other hole, like in the picture below: Though which drain is the water spraying out with the highest speed? 1. The hole 2. The pipe 3. Same The speed of the water depends on the pressure at the opening through which the water flows, which depends on the depth of the opening from the surface of the water. Since both the hole and the opening of the downspout are at [the same] depth below the surface, water from the two will be discharged with the same speeds. CORRECT

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UB, Phy101: Chapter 11, Pg 34

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