Pressure. Pressure Demos Pencil tip vs. Eraser Dart suction caps Magdeburg Spheres Plunger or Rubber...

Pressure

Pressure Demos

• Pencil tip vs. Eraser• Dart suction caps• Magdeburg Spheres• Plunger or Rubber mat• Lift shot glass with hand• Egg in Bottle/Out• Break Paint Stick• Straw demo• Olive oil can

• Put student in bag, vacuum out air.

• Vacuum pump balloon/peep

• BED OF NAILS• Mini bed and Balloon.• Implode soda can• Balloon into bottle• Upside down cup

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Pressure = Force / Area

• Pressure is the amount of force exerted per unit area.• Pressure = Force / Area

• P = F/A– F is Force and the unit is Newtons (N)– A is Area and the unit is Meters2 (m2)– The unit for Pressure is N/ m2 or a Pascal (Pa)

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Smaller Area

Larger Area

An Equal Force applied over a smaller area increases pressure

Forza Science!!!!!

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Sample Problem 1

• A Force of 15 Newtons is applied over a surface area of 0.6m2. What is the pressure?

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Sample Problem 2

• A Force of 25 Newtons is applied over a surface area of 1.3m2. What is the pressure?

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Sample Problem 3

• A pressure of 50 pascals is pushing down on an area of 5m2. What is the force?

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Manipulating Equations

• P=F/A• A = F/P• F = AP

11P

FA

P]by sidesboth [Divide P

F

P

PA

AP For F AP

A]by sidesboth [Multiply A*A

FP*A

A

FP

Bed of Nails: How does it work

Pressure

• If force is constant but area increases, pressure will _______________

• If force is constant but area decreases, pressure will _______________

• If area is constant but Force increases, pressure will _______________

• If area is constant but Force decreases, pressure will _______________

Pressure

• If force is constant but area increases, pressure will ___DECREASE____

• If force is constant but area decreases, pressure will ___INCREASE___

• If area is constant but Force increases, pressure will ___INCREASE___

• If area is constant but Force decreases, pressure will ___DECREASE____

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Fluids Exert Pressure

• Air Pressure• Water Pressure• Blood Pressure

• PRESSURE IN A FLUID ACTS IN ALL DIRECTIONS!

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Straw Demo

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Olive Oil Demo

• If I fill up an empty container of olive oil with water and put a hole near the bottom of it what happens?

• The weight (Force) of the water pushes down and creates water pressure forcing it out.

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Your head hurts when you dive down in a pool because of the weight of the water above you.

Gas Pressure

• In a container of gas, fluid pressure is the result of the particles bouncing off the sides of the container that is holding them.

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Air Pressure vs Water Pressure

Olive Oil

Water Pressure due to theweight of the water above.

Hole

Air Pressure

Air PressureAir Pressure

Air Pressure and Water Pressure balance when the top is on!

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Air Pressure vs Water Pressure

Olive Oil

Water Pressure due to theweight of the water above and air pressure fromabove.

Hole

Air Pressure

Air PressureAir Pressure

Water comes out when the top is off! Since the container is no longer sealed not only is the weight of the water pushing down but the weight of air in our atmosphere is pushing down on it as well.

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Balloon in a Vacuum Pump

• What happens to the balloon when I remove air?

• What happens to the balloon when you add air?

• Lift the cover!

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Pressure on a Balloon

Air Pressure from the atmosphere is pushing in.

Air Pressure from the compressed air inside the balloon is pushing out.

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Pressure on a Balloon

Removing the inward pressure allows the balloon to expand

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Pressure on a Balloon

Increasing the inward pressure forces the balloon to contract

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Does Air Have Weight

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Force (weight) = mg Pressure = F/AAir has mass so air has weight or air exerts a force on us.Since air exerts a force on us, there is air pressure.

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Air Pressure is 14.7psi

• Air exerts a pressure of 14.7 pounds per square inch!!!!

• CLASS Activity:

• Measure the surface area of an object in in2 and determine the weight of the air pushing own on it.

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Calculate the Pressure on Your Head

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Top Down View of a PersonApproximate the surface area as a rectangle (L/W)

Area Rectangle = LxW

Force = Area x Pressure

Area = ? (measure it) Pressure = 14.7psi

Calculate the weight of air pushing down on your head and shoulders right now.

Why aren’t you getting crushed by the weight of the air?

HeadShoulder Shoulder

Length (inches)

Width (inches)

Mat and Can Crush

Vacuum Pump

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Explain these using Pressure and the Kinetic Theory of Matter

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We aren’t crushed because Pressure acts in all Directions

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Humans have air in our bodies

• That air pushes out with the same pressure that the atmosphere is trying to crush us with.

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Pressure increases with depth in a fluid

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This is why your ears pop in a plane

On the surface of the earth the pressure in your air balances with the air pressure outside of it. As you ascend in a plane the outside air pressure lowers but the pressure remains the same in your ears. It is pushing with a higher force and fortunately, it pops to release the extra pressure…

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Is there Air Pressure in Outer Space?

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Pressure increases with Depth

• P = pgh• p = density (rho)– kg/m3 (water = 1000kg/m3)

• g = gravity (9.81m/s2)• h = height (meters from surface)

• Where is the pressure the greatest? A,B or C.

POOL OF WATER

Pressure Increases

A

B

C

Surface

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Derivation of P=pgh

• P = F/A [Force is weight of the fluid]• P = weight/A [weight = mg]• P = mg/A [D = m/V so m = DV]• P = DVg/A [V = LxWxH and A = LxW • P = Dg (LxWxH) / (LxW) [the LxW cancels]• P = Dgh [ we use p instead of D]• P = pgh

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Problem #1

• P = pgh• p = density (rho)

– kg/m3 – (water = 1000kg/m3)– (salt water = 1040kg/m3)

• g = gravity (9.81m/s2)

• h = height (meters from surface)

• In 1960 two men aboard the Trieste dove down to the deepest part of the ocean in the Marianas Trench in the Pacific Ocean. The reached a depth of over 10,911 meters. Calculate the pressure in pascals pushing down on the top of the submarine at that depth.

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Problem #2

• P = pgh• p = density (rho)– kg/m3 – (water = 1000kg/m3)– (salt water = 1040kg/m3)

• g = gravity (9.81m/s2)

• h = height (meters from surface)

• What is the pressure on a person’s head if she dives down 30 meters in a lake filled with fresh water?

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Problem #3

• P = pgh• p = density (rho)– kg/m3 – (water = 1000kg/m3)– (salt water = 1040kg/m3)

• g = gravity (9.81m/s2)

• h = height (meters from surface)

• What is the pressure on a submarine if it dives down 1000 meters in salt water.

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Problem #4

• P = pgh• p = density (rho)– kg/m3 – (water = 1000kg/m3)– (salt water = 1040kg/m3)

• g = gravity (9.81m/s2)

• h = height (meters from surface)

• At what depth in a pool of fresh water would a person experience a pressure of 29,400 pascals.

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Bernouilli’s Principle

• As the Speed of a moving fluid increases, the pressure of the moving fluid decreases.

• S P Siphoning Demo

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Two Bernoulli’s Principle Demos

Extra Credit!

Bernoulli

Bernoulli

Bernoulli Demos

Invert Cut Bottle!

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Bernouilli’s Principle explains FlightA wing is shaped so that air traveling over the top of it has a longer path and therefore moves faster. This means there is lower Pressure above a wing and higher pressure below it. This is called Lift!

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Bernoulli’s Principle and Newton’s 3rd Law

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Bernoulli's principle Explains why covering part of a hose makes it shoot further

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River Flow Rate: increases as a river narrows and decreases as it widens.

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Mentos Demo

Homework: Explain how a vacuum works and how a curve ball is thrown.

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Pascal’s Principle

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What happens when you squeeze a tube of toothpaste?

Pascal’s Principle: A Change in pressure at any point in an enclosed fluid will be transmitted equally to all parts of the fluid.

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What happens when you squeeze a tube of toothpaste?

Pascal’s Principle: A Change in pressure at any point in an enclosed fluid will be transmitted equally to all parts of the fluid.

P1 = P2

P1 P2

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What happens when you squeeze a tube of toothpaste?

Pascal’s Principle: A Change in pressure at any point in an enclosed fluid will be transmitted equally to all parts of the fluid.

P1 = P2

F1 /A1 F2 /A2

F1 /A1 =F2 /A2

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Hydraulic Device

F1 /A1 = F2 /A2

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Hydraulic Device

A Small amount of Force can lift a very heavy object.

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Pascal’s Principle Problem #1

• A Hydraulic lift uses Pascal’s principle to lift a 19,000N car. If the area of the small piston is (A1) equals 10.5cm2 and the area of the large piston (A2) equals 400cm2, what force needs to be exerted on the small piston to lift the car?

F1 / A1 = F2 /A2

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Pascal’s Principle Problem #2

• If a force of 100lbs is applied to the small piston (A1) which has an area of 12in2 and the area of the large piston (A2) is 300in2, how heavy of an object can be lifted?

F1 / A1 = F2 /A2

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Pascal’s Principle Problem #3

• A Hydraulic lift uses Pascal’s principle to lift a 4000lb car. If the area of the small piston is (A1) equals 4in2 and the area of the large piston (A2) equals 65in2, what force needs to be exerted on the small piston to lift the car?

F1 / A1 = F2 /A2

Hydraulic Brakes / Air Matresses

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Properties of Gas and Gas Laws

• Fills its container• Spreads easily• Mix well (e.g. air)• Low Density• Compressible• Mostly empty space

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Boyle’s Law

• For a fixed amount of gas at constant temperature:

• A Volume increases, pressure decreases.

• As volume decreases, pressure increases.

• P1V1 = P2V2 PHET PHYSICS APP

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P1V1 = P2V2

P is pressure in pascals (pa) or kilopascals (kpa)

V is volume in Liters

• Solve for V1 • Solve for P2

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3 Problems pg 98

1. A flask contains 155ml of hydrogen gas at a pressure of 22,500Pa. Under what pressure would the gas have a volume of 90ml

2. If the pressure exerted on a 300.0ml sample of hydrogen gas at constant temperature is increased from 500Pa 750Pa, what will be the final volume of the sample?

3. A sample of oxygen gas has a volume of 150mL at a pressure of 947Pa. If the temperature remains constant, what will the volume of the gas be at a pressure of 1000Pa?

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3 Problems pg 98

1. A flask contains 155cm3 of hydrogen gas at a pressure of 22.5kPa. Under what pressure would the gas have a volume of 90cm3 (Recall that 1cm3 = 1 ml and 1000ml = 1 L)

2. If the pressure exerted on a 300.0ml sample of hydrogen gas at constant temperature is increased from 0.500Kpa to 0.750 kPa, what will be the final volume of the sample?

3. A sample of oxygen gas has a volume of 150mL at a pressure of 0.947kPa. If the temperature remains constant, what will the colume of the gas be at a pressure of 1.000kPa?

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What Happens to a Bubble blown underwater? Why?

http://www.youtube.com/watch?v=sn67vDBiL04

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Gay-Lussac’s law

• For a fixed volume of gas :

• As temperature increases Pressure increases.

• As temperature decreases Pressure decreases.

• P1/T1 = P2/T2

PHET PHYSICS APP

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DANGER http://www.youtube.com/watch?v=AV8JzWeIuxI

• Exploding Beer Keg Kills Man

• NEW MILFORD, Conn. (AP) - October 24, 2006 -- Investigators were trying to determine who tossed a beer keg into a burning barrel at a party, causing a deadly explosion that sent metal shards slicing through a crowd of people.

• The explosion early Sunday killed Sean M. Caselli, 22, of New Milford. Seven other people were taken to hospitals with burns and shrapnel wounds, police said. Caselli was struck in the neck by a piece of flying metal.

• Sgt. Lee Grabner said investigators interviewed witnesses Sunday to try to identify the person witnesses say threw a quarter-keg of beer into the flames, and to determine whether criminal charges should be filed.

• Fires had been set in several barrels to keep people at the partygoers warm at the outdoors party in western Connecticut, said Police Captain Michael Mrazik.

• "This is a certain tragedy," said Police Chief Colin McCormack. "However, nothing I have been apprised of to this point in this investigation, which I caution is at the very early states, indicates a deliberate act on anyone's part."

• (Copyright ©2013 by The Associated Press. All Rights Reserved.)

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Charle’s Law

• For a fixed amount of gas at constant pressure :

• As Temperature increases, volume also increases.

• As Temperature decreases, the volume also decreases.

• V1/T1 = V2/T2 PHET PHYSICS APP

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Gas Law SummaryBoyle’s Law Gay-Lussac’s Law Charles’ Law

Constant Temperature Constant Volume Constant Pressure

As V↑P↓ As V↓P↑ As T↑P↑ As T↓P↓ As T↑V↑ As T↓V↓

P1V1 = P2V2 P1/T1 = P2/T2 V1/T1 = V2/T2