11/13/2015 9:49 AM 1 Chapter 9 Fluids and Buoyant Force In Physics, liquids and gases are...

04/20/23 16:42 1

Chapter 9 Fluids and Buoyant Force

In Physics, liquids and gases are collectively called fluids.

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Fluids and Buoyant Force

A fluid is a nonsolid state of matter in which the atoms or molecules are free to move past each other, as in a gas or a liquid.

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Liquids differ from gases in that liquids have a definite volume, gases do not.

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Recall, volume is the amount of space that an object occupies.

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Of course, whether an object floats or sinks has little to do with size. It is the object’s mass density that matters.

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Mass density (or just density) is the mass per unit volume of a substance.

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The Mass density of fresh water at 4oC and 1 atm of pressure is 1.00 g/cm3 or 1.00X103kg/m3

Recall, 1 ml = 1 cm3

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That means that one liter (1000 ml) of fresh water at 4oC and 1 atm of pressure has a mass of one kilogram (1000 g).

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Scuba divers, submarines and many types of fish control their density to sink or rise in the water.

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Formula for Mass Density

volume

mass density

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Example 1. If gold has a mass density of 19.3 g/cm3 at STP (standard temperature and pressure), how large would a 3.75 g sample of pure gold be?

Given:

Solution:

= 19.3 g/cm3 m = 3.75 g

3cm 0.194

/cmg 19.3

g 3.75

3cm 0.194

/cmg 19.3

g 3.75

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You may have noticed that objects feel lighter in the water than they do in the air.

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This is because the water exerts a buoyant force in the opposite direction as gravity.

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Buoyant Force is a force that acts upward on an object submerged in a liquid or floating on a liquid’s surface.

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Do you know the story of Archimedes, and how he discovered what has come to be known as “Archimedes Principle”?

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Archimedes (287-212 BC) is considered to be one of the greatest mathematicians of all times, and he is sometimes called the “father of mathematical physics.”

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Hiero, the king of Syracuse, commissioned a craftsman to fashion a crown, and provided he with an exact amount of gold to make it.

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The crown that Heiro received weighed the same amount as the gold he had provided, but he suspected the craftsman had used some silver instead of pure gold, and he asked Archimedes to prove it.

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Archimedes was contemplating the problem while in the bath. He realized that by getting in the bath, he displaced an amount of water that was proportional to the part of his body that he submerged.

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He is said to have become so excited that he ran naked through the streets, shouting “Eureka! Eureka!” (this translates to, “I have found it!”)

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He formulated what is now known as Archimedes principle, and a way to test the composition of the crown.

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Archimedes Principle – any object that is completely or partially submerged in a fluid experiences an upward buoyant force equal in magnitude to the weight of the fluid displaced by the object.

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Formula for Buoyant Force

gm fluid) (displacedF F fwB

displaced fluid of weight forcebuoyant of magnitude

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Fluids and Buoyant ForceExample 2. A piece of an unknown mineral that is dropped into mercury displaces 0.057 m3 of Hg. Find the buoyant force on the mineral. (note, of Hg = 13.6 x 103 kg/m3 at STP.)Given:

= 13.6 x 103 kg/m3 V = 0.057 m3

m so V m so gV gm F fffB ρ

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Fluids and Buoyant ForceExample 2. A piece of an unknown mineral that is dropped into mercury displaces 0.057 m3 of Hg. Find the buoyant force on the mineral. (note, of Hg = 13.6 x 103 kg/m3 at STP.)Given:

Find: Solution:

= 13.6 x 103 kg/m3 V = 0.057 m3

N 7600 )m/s )(9.81m )(0.057mkg/ 10 x (13.6

gV gm F2333

N 7600 )m/s )(9.81m )(0.057mkg/ 10 x (13.6

gV gm F2333

N 7600 )m/s )(9.81m )(0.057mkg/ 10 x (13.6

gV gm F2333

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If the buoyant force on an object is greater than the weight of the object, the object will bob to the surface.

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For an object that is floating, the buoyant force is equal and opposite to the weight of the object.

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In other words, if an object is less dense than the fluid that it rests in, it will float.

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The density of the object determines how much of the object’s volume will be submerged below the surface.

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Formula for Determining what Portion of a Floating Object will

Submerge

objectsub V

Sub=submerged V=Volume of entire object

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Fluids and Buoyant ForceExample 3. A 2.00 m3 block of wood has a density of 0.500 x 103 kg/m3. If this object is placed in fresh water at 4oC and 1 atm of pressure, what volume will be below the surface of the water? Given:

Solution:

object = 0.500 x 103 kg/m3 V = 2.00 m3

fluid = 1. 00 x 103 kg/m3 Vsub

objectsub m 1.00

kg/m 10 x 1.00

)kg/m 10 x )(0.500m (2.00

objectsub m 1.00

kg/m 10 x 1.00

)kg/m 10 x )(0.500m (2.00

objectsub m 1.00

kg/m 10 x 1.00

)kg/m 10 x )(0.500m (2.00

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Notice, if an object is ½ as dense as water, then ½ of it will be under the water.

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If an object is 90% as dense as water, than 90% of the object will be below the surface (AS IN AN ICEBERG).

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If the density of the object is greater than the density of the fluid, it will continue to sink.

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The apparent weight of a submerged object depends upon its density.

This, so called “apparent weight” is really just the net force (Fnet) in the y-axis.

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Formulas for Determining the FNet of a Submerged Object

WBNet F F F

gm gm F ofNet

gV gV F ooffNet

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Fluids and Buoyant ForceExample 4. Gold has a density of 19.3 x 103 kg/m3. If a piece of gold with a volume of 0.450 m3 is completely submerged in water, how much will it seem to weigh?

Given:

object = 19.3 x 103 kg/m3 Vo = 0.450 m3

fluid = 1.00 x 103 kg/m3 Vf = 0.450 m3

Hint: The Volume of the water is the same as the volume of the object.

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Fluids and Buoyant ForceGiven:

object = 19.3 x 103 kg/m3 Vo = 0.450 m3

fluid = 1.00 x 103 kg/m3 Vf = 0.450 m3

Solution:)Vg - ( gV gV F ofooffNet

Because Vo = Vf, we can rewrite our equation.

N 10x 8.08 4- N 80785.35-

)m/s )(9.81m )(0.450kg/m x1019.3 - kg/m 10 x (1.00

)Vg - ( F233333

DOWN (it will sink, not float!)

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Another useful formula is Apparent weight = Fg – FB

Remember…FW and Fg are the same…

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Chapter 9 Section 9-1Fluids and Buoyant Force

Summary of Formulas

substance gr. sp.

gm fluid) (displacedF F fwB

objectsub V

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Chapter 9 Section 9-1Fluids and Buoyant Force

Summary of Formulas.

Apparent weight = Fg – FB

WBNet F F F gV gV F ooffNet

gm gm F ofNet f

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Fluid Pressure and Temperature

An object that is submerged in water, or any other fluid, experiences pressure on all sides. You have likely noticed the effects of this pressure while swimming.

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Gases also exert pressure. Your body feels the weight of the atmosphere on it. Your ears also “pop” when you change altitude quickly.

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Pressure is the magnitude of the force on a surface per unit area.

Movie Clip

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Formula for Pressure

force pressure

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Can you explain why a finger does not pop a balloon even when you exert more force on the balloon than you do with a needle?

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The SI unit of pressure is the pascal (Pa), which is equal to 1 N/m2.

F p Conversion Factor

1 Pa = 1 N/m2

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Fluid Pressure and TemperatureExample 1. A cylinder-shaped column with a height of 2.0 m and a base radius of 0.50 m has a mass of 1250 kg. Find the pressure it exerts on the ground.Given:

Solution:

m = 1250 kg h = 2.0 m r = 0.50 m

Pa 10 x 1.6 4 m) (0.50

)m/s kg)(9.81 (1250

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There are many other common units of pressure, including the atmosphere (atm) and millimeters of mercury (mm of Hg).

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Important Conversion Factors

1 atm = 760 mm of Hg = 760 torr = 1.013 x 105 Pa

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Fluid Pressure and TemperatureExample 2. Convert 2.30 atm to Pa.

2.30 atm Pa=atm 1.00

Pa 10 x 1.013x

2.33 x 105

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A barometer is an instrument that is used to measure atmospheric pressure.

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You have probably noticed that when you squeeze a balloon, it will bulge outwards in a different area. This can be explained by Pascal’s Principle.

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Pascal’s Principle – Pressure applied to a fluid in a closed container is transmitted equally to every point of the fluid and to the walls of the container.

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Formula for Pascal’s Principle

1increase A

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Example 3. In a hydraulic car lift, the compressed air exerts a force on a piston with a base radius of 5.00 cm. The pressure is transmitted to a second piston with a base radius of 35.0 cm. How much force must be exerted on the first piston to lift a car with a mass of 1.40 x 103 kg on the second piston?

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Fluid Pressure and TemperatureExample 4. In a hydraulic car lift, the compressed air exerts a force on a piston with a base radius of 5.00 cm. The pressure is transmitted to a second piston with a base radius of 35.0 cm. How much force must be exerted on the first piston to lift a car with a mass of 1.40 x 103 kg on the second piston?

Given:

Equation:

r1 = 5.00 cm r2 = 35.0 cm m2 = 1.40 x 103 kg

gm x r

A F 22

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Given:

Equation:

r1 = 5.00 cm r2 = 35.0 cm m2 = 1.40 x 103 kg

gm x r

A F 12

Solution:

N 028 )m/s kg)(9.81 10 x (1.40 x cm) (35.0

cm) (5.00

gm x r

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Lesson 9-2Fluid Pressure and Temperature

Pressure varies with depth in a fluid. That is why your ears “pop” when you go up in an airplane or down in the water.

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Lesson 9-2Fluid Pressure and TemperatureAtmospheric pressure (patm) is the pressure exerted by the atmosphere. It varies from day to day and from location to location.

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Lesson 9-2Fluid Pressure and TemperatureStandard atmospheric pressure is equal to 1.0 atm or approximately 1.013 x 105 Pa.

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Lesson 9-2Fluid Pressure and TemperatureGauge pressure (pgauge = gh) is the pressure exerted by just the water, or other fluid. When calculating gauge pressure, leave the atmospheric pressure out of the calculation.

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Gauge Pressure as a Function of Depth

gh pgauge

Where = density of fluid, g = acceleration due to gravity and h = depth

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Lesson 9-2Fluid Pressure and TemperatureExample 5. Calculate the gauge pressure at a point 10.0 m below the surface of the ocean. (note, sea water = 1.025 x 103 kg/m3)

Given:

Solution:

= 1.025 x 103 kg/m3 h = 10.0 m g = 9.81 m/s2

pgauge

Pa 10 x 1.0 m) )(10.0m/s )(9.81kg/m 10 x (1.025 gh p 5233gauge

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Lesson 9-2Fluid Pressure and TemperatureAbsolute pressure (p) is the total pressure exerted by the water (or other fluid) and the atmosphere above the water.

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Absolute Pressure as a Function of Depth

gh p p 0 Where p = absolute pressure, p0 = pressure at point of comparison (usually atmospheric pressure), = density of fluid, g = acceleration due to gravity and h = depth

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Lesson 9-2Fluid Pressure and TemperatureExample 6. Calculate the absolute pressure experienced by a scuba diver 20.0 m below the sea ( = 1.025 x 103 kg/m3). Assume patm = 1.0 x 105 Pa

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Lesson 9-2Fluid Pressure and TemperatureExample 6. Calculate the absolute pressure experienced by a scuba diver 20.0 m below the sea ( = 1.025 x 103 kg/m3). Assume patm = 1.0 x 105 PaGiven:

Find: Solution:

p0 = 1.0 x 105 Pa h = 20.0 m = 1.025 x 103 kg/m3

Pa 10 x 3.0

m) )(20.0m/s )(9.81kg/m 10 x (1.025 Pa) 10 x (1.0

gh p p

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Summary of Formulas

1increase A

gh p p 0

gh pgauge

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Summary of Formulas

1 atm = 760 mm of Hg = 760 torr = 1.013 x 105 Pa

K 732 K C0

K 732 C K 0

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Fluids in Motion

The motion of a fluid can be described as either laminar (smooth) or turbulent (irregular).

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Fluids in Motion

When studying fluids, is helps to consider the behaviors of an ideal fluid.

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Fluids in MotionAn ideal fluid is an imaginary fluid that has no internal friction or viscosity, and is incompressible.

The viscosity of a fluid is a measure of its internal resistance, or its resistance to flow.

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Fluids in Motion

Fluids with high viscosity, like molasses, flow slowly, due to internal friction.

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Fluids in Motion

The law of conservation of mass explains that the mass entering a section of pipe or tube must be the same as the mass that leaves.

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Fluids in Motion

The Continuity Equation

2211 νA νA area x speed in region 1 = area x speed in region 2

The Continuity Equation is an expression of the conservation of mass.

sec)/3(νA mflowrate

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Fluids in Motion

2211 νA νA

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Fluids in Motion

Example 1. An ideal fluid is moving at 3.0 m/s through a section of pipe with a radius of 0.35 m. How fast will it flow through another section of the pipe with a radius of 0.20 m?

HINT: A = r2

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Fluids in Motion

Example 1. An ideal fluid is moving at 3.0 m/s through a section of pipe with a radius of 0.35 m. How fast will it flow through another section of the pipe with a radius of 0.20 m?

Given: 1 = 3.0 m/s r1 = 0.35 m r2 = 0.20 m

Formula:

Solution:

1 νr νr

m/s 9.2 m) (0.20

m/s) (3.0m) (0.35

νr ν

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Fluids in Motion

From example 1, we see that as the area of the pipe decreases, the velocity of the fluid becomes much greater.

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Fluids in Motion

Another property of fluids explains how the wings of an airplane produce lift.

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Fluids in Motion

To demonstrate Bernoulli’s principle for yourself, try blowing air over the top of a piece of paper.

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Fluids in Motion

Bernoulli’s Principle – The pressure in a fluid decreases as the fluid’s velocity increases.

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Fluids in Motion

The wings of an airplane are designed to allow air to flow quicker over the top than the bottom, resulting in lower pressure above the wings.

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Fluids in Motion

Bernoulli’s Equation

constant gh P 2

pressure + kinetic energy per unit volume + gravitational potential energy per unit volume = constant along a given streamline.

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Fluids in Motion

Bernoulli’s Equation for Comparing a Fluid at Two Different Points

pressure + kinetic energy per unit volume + gravitational potential energy per unit volume = pressure + kinetic energy per unit volume + gravitational potential energy per unit volume .

gh P gh P 22

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Important to know!!!!!

IF there is no height difference, you can eliminate the

If there is no area change, velocity stays the same

IF there is no pressure change, P1 = P2

1gh 2ghand

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Fluids in Motion

Example 2. Water is circulated through a system. If the water is pumped with a speed of 0.45 m/s under a pressure of 2.2 x 105 Pa from the first floor through a 6.0-cm diameter pipe, what will the pressure be on the next floor 4.0 m above?

HINT: The key to this question is that the diameter of the pipe doesn’t change and, therefore, the velocity of the water remains constant.

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Fluids in MotionIf 1 = 2 , then we can simplify our equation.

gh P gh P 22

gh P gh P 2211

)h-g(h P P 2112

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Fluids in MotionExample 2. Water is circulated through a system. If the water is pumped with a speed of 0.45 m/s under a pressure of 2.2 x 105 Pa from the first floor through a 6.0-cm diameter pipe, what will the pressure be on the next floor 4.0 m above?Given: P1 = 2.2 x 105 Pa h1- h2 = -4.0 m

water = 1000 kg/m3 g = 9.81 m/s2

Find: Formula:P2 )h-g(h P P 2112

Pa 10x 1.9 5 m) )(-4.0m/s )(9.81kg/m (1000 Pa 10 x 2.2

)h-g(h P P235

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Fluids in Motion

Example 3. A water tank has a water level of 1.2 m. The spigot is located 0.40 m above the ground. If the spigot is opened fully, how fast will water come out of the spigot?

HINT: The container is open to the atmosphere, so the pressure will be the same in both spots. We can assume that the velocity of the water at the top (v2) is essentially zero.

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Fluids in MotionThe pressure of the fluid in both areas is the same, P1 = P2 so they cancel out. Also, the velocity at the top (2) is zero, so we cross out that expression.

gh P gh P 22

gh gh 212

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Fluids in MotionNow, we can divide all expressions by (the fluid doesn’t change, so density is the same throughout) and continue to isolate 1, the unknown.

gh gh 212

1 gh gh 21

1 gh - gh

)h - g(h 122

1 )h - 2g(h 121

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Fluids in Motion

Example 3. A water tank has a water level of 1.2 m. The spigot is located 0.40 m above the ground. If the spigot is opened fully, how fast will water come out of the spigot?

Given: h1 = 0.40 m h2 = 1.2 m g = 9.81 m/s2

Find: Formula:2 )h - 2g(h 121 Solution:

m/s 4.0 )m 0.40 - m )(1.2m/s 2(9.81 21

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Summary of Formulas

Fluids in Motion

2211 νA νA

gh P gh P 22

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The Kinetic Molecular Theory (KMT) of Gases also relates the temperature of a gas to the motion of its particles.

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Temperature is a measure of the average kinetic energy of the particles of a substance.

The SI unit of temperature is the Kelvin (K).

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Converting between Celsius and Kelvin

K 732 C K 0

K 732 K C0

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Lesson 9-2Fluid Pressure and TemperatureExample 7. Convert 22.0 oC to Kelvin.

K 295 K 273 22.0 K 732 C K 0

11/13/2015 9:49 AM 1 Chapter 9 Fluids and Buoyant Force In Physics, liquids and gases are...

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