Ch9.1 – States of Matter and Density Kinetic Theory (PME) 1. All matter is made of particles. 2. Particles in constant motion. 3. All collisions are perfectly elastic. (No energy lost)

Kinetic Theory (PME) 1. All matter is made of particles. 2. Particles in constant motion. 3. All collisions are perfectly elastic. (No energy lost) Solids - particles have low KE, so intermolecular forces hold them into a crystalline structure. Definite shape, definite volume (for all practical purposes).

Kinetic Theory (PME) 1. All matter is made of particles. 2. Particles in constant motion. 3. All collisions are perfectly elastic. (No energy lost) Solids - particles have low KE, so intermolecular forces hold them into a crystalline structure. Definite shape, definite volume (for all practical purposes). Liquids - higher KE breaks free of structure, but not enough to completely break free of each other. Definite volume, but takes shape of container.

Kinetic Theory (PME) 1. All matter is made of particles. 2. Particles in constant motion. 3. All collisions are perfectly elastic. (No energy lost) Solids - particles have low KE, so intermolecular forces hold them into a crystalline structure. Definite shape, definite volume (for all practical purposes). Liquids - higher KE breaks free of structure, but not enough to completely break free of each other. Definite volume, but takes shape of container. Gas - high KE, completely break free of each other. Can change volume and shape.

Density – a comparison of an object’s mass to its volume. ‘roe’

Density – a comparison of an object’s mass to its volume. ρH20 = 1 g/cm3 = 1 g/ml ‘roe’ = 1000 kg/m3 Ex1) Determine if a 500g sealed hollow tube with a radius of 2cm, 30cm length will float in water.

Density – a comparison of an object’s mass to its volume. ρH20 = 1 g/cm3 = 1 g/ml ‘roe’ = 1000 kg/m3 Ex2) Determine the density of an unknown solid if its mass is .4kg and it displaces 21 mls of H2O.

Density – a comparison of an object’s mass to its volume. ρH20 = 1 g/cm3 = 1 g/ml = 1000 kg/m3 Ex1) Determine if a 500g sealed hollow tube with a radius of 2cm, 30cm length will float in water. (sinks) Ex2) Determine the density of an unknown solid if its mass is .4kg and it displaces 21 mls of H2O. (Move the decimal 2X for each (21mls = 21cm3 = 0.000021m3) dimention.) 21mls

Buoyant Force – a fluid exerts an overall upward force on an object, caused by the pressure of the fluid.

Buoyant Force – a fluid exerts an overall upward force on an object, caused by the pressure of the fluid. Net effect is an upward force FB

Buoyant Force – a fluid exerts an overall upward force on an object, caused by the pressure of the fluid. Net effect is an upward force FB Fg - if the object’s weight is greater than it buoyant force, it sinks.

Buoyant Force – a fluid exerts an overall upward force on an object, caused by the pressure of the fluid. Net effect is an upward force FB Fg - if the object’s weight is equal to its buoyant force, it suspends.

Buoyant Force – a fluid exerts an overall upward force on an object, caused by the pressure of the fluid. Net effect is an upward force FB Fg - if the object’s weight is less than its buoyant force, it floats. How deep though? That requires us to know more about pressure. Ch9 HW#1 1 – 5

Ch9 HW#1 1 – 5 1. What is the density of an unknown shiny, yellow-ish metal, given that is masses at 38.6g and has a volume of 2 cm3. Convert this density to kg/m3.

Ch9 HW#1 1 – 5 1. What is the density of an unknown shiny, yellow-ish metal, given that is masses at 38.6g and has a volume of 2 cm3. Convert this density to kg/m3.

2. A silvered color metal cube measure 1cm on a side and has a mass of 10.5g. Find its density in kg/m3.

2. A silvered color metal cube measure 1cm on a side and has a mass of 10.5g. Find its density in kg/m3.

3. A solid plastic ball has a mass of 0. 020kg 3. A solid plastic ball has a mass of 0.020kg. It is placed in a bucket of water and stays suspended below the surface. If the density of water is 1000kg/m3, what is the volume of the ball? FB Fg

3. A solid plastic ball has a mass of 0. 020kg 3. A solid plastic ball has a mass of 0.020kg. It is placed in a bucket of water and stays suspended below the surface. If the density of water is 1000kg/m3, what is the volume of the ball? FB Fg

4. The mass of an object is 0.500kg. The buoyant force acting on it is 3.0N. What is the net force acting on it? What is its acceleration? FB Fnet = Fg – FB Fg

4. The mass of an object is 0.500kg. The buoyant force acting on it is 3.0N. What is the net force acting on it? What is its acceleration? FB Fnet = Fg – FB m.a = 5N – 3N (.5)a = 2 a = 4 m/s2 Fg

5. When a person breathes in, they float in fresh water. When they breathe out, they sink. If my mass is 75kg, what volumes do I occupy before and after the breath? Breathe in: Breathe out: FB Fg Fg Density < 1000kg/m3 Density > than 1000kg/m3 V > 0.075 m3 V < 0.075 m3

5. When a person breathes in, they float in fresh water. When they breathe out, they sink. If my mass is 75kg, what volumes do I occupy before and after the breath? Breathe in: Breathe out: FB Fg Fg Density < 1000kg/m3 Density > than 1000kg/m3

Chapter 9.2 – Pressure Ex1) A force of 100N is applied to the head of nail, with an area of 1cm2. What pressure is exerted to the nail head?

Fluid pressure is created by the weight of the fluid Atmospheric Pressure: - 101,300 Pascals (N/m2) - 1 atmosphere - 760 mm Hg (10.3m H2O) - 14.7 psi (pounds per inch2)

Four ways to affect fluid pressure: 1. Increase or decrease volume:

Four ways to affect fluid pressure: 1. Increase or decrease volume: 2. Increase or decrease temperature:

Four ways to affect fluid pressure: 1. Increase or decrease volume: Combined Gas Law: 2. Increase or decrease temperature:

Four ways to affect fluid pressure: 1. Increase or decrease volume: Combined Gas Law: 2. Increase or decrease temperature: 3. Increase or decrease # of particles: R, Ideal Gas Law Constant: moles

Four ways to affect fluid pressure: 1. Increase or decrease volume: Combined Gas Law: 2. Increase or decrease temperature: 3. Increase or decrease # of particles: R, Ideal Gas Law Constant: moles

Four ways to affect fluid pressure: 1. Increase or decrease volume: Combined Gas Law: 2. Increase or decrease temperature: 3. Increase or decrease # of particles: R, Ideal Gas Law Constant: moles 4. The weight of the fluid creates pressure:

Ex2) Prove that P = ρgh is true for any shape container: 1. Cylinder: 2. Rectangular prism:

Prove that P = ρgh is true for any shape container: 1. Cylinder: 2. Rectangular prism: Works for any shape container, and combinations!

Ex3) What total pressure acts on you when you swim to the bottom of a 3m deep pool at sea level? Ch9 HW#2 pg 331, 3,22,23,24,25,29

Ex3) What total pressure acts on you when you swim to the bottom of a 3m deep pool at sea level? Ch9 HW#2 pg 331, 3,22,23,24,25,29

Ch9 HW#2 p331 3,22,23,24,25,29 3. White Dwarf stars small and massive. A 1 in3 chunk on Earth weighs 1 ton! (2000 lbs) Determine density in SI units.

Ch9 HW#2 p331 3,22,23,24,25,29 3. White Dwarf stars small and massive. A 1 in3 chunk on Earth weighs 1 ton! (2000 lbs) Determine density in SI units.

22. How tall should vertical pipe be filled with water if its gauge pressure reads 400 kPa? 23. Swimming pool 5m wide by 10m long filled to 3m depth, what is gauge pressure at bottom?

22. How tall should vertical pipe be filled with water if its gauge pressure reads 400 kPa? 23. Swimming pool 5m wide by 10m long filled to 3m depth, what is gauge pressure at bottom?

22. How tall should vertical pipe be filled with water if its gauge pressure reads 400 kPa? 23. Swimming pool 5m wide by 10m long filled to 3m depth, what is gauge pressure at bottom?

24. Rectangular tank 2m x 2m x 3. 5m tall filled to depth of 2 24. Rectangular tank 2m x 2m x 3.5m tall filled to depth of 2.5m with gasoline ( = 680kg/m3). What is the gauge pressure 2.0m below the surface? 25. O2 tank has an internal pressure of 5X atm. What outward force of each square centimeter of inner wall? 29. Gauge pressure @ 11km down to bottom of the ocean. {

24. Rectangular tank 2m x 2m x 3. 5m tall filled to depth of 2 24. Rectangular tank 2m x 2m x 3.5m tall filled to depth of 2.5m with gasoline ( = 680kg/m3). What is the gauge pressure 2.0m below the surface? 25. O2 tank has an internal pressure of 5X atm. What outward force of each square centimeter of inner wall? {

24. Rectangular tank 2m x 2m x 3. 5m tall filled to depth of 2 24. Rectangular tank 2m x 2m x 3.5m tall filled to depth of 2.5m with gasoline ( = 680kg/m3). What is the gauge pressure 2.0m below the surface? 25. O2 tank has an internal pressure of 5X atm. What outward force of each square centimeter of inner wall? {

29. Gauge pressure @ 11km down to bottom of the ocean.

29. Gauge pressure @ 11km down to bottom of the ocean.

Ch9.3 – Fluid Dynamics Pascal’s principle – if a pressure is exerted on a fluid, that pressure is transmitted thru out the entire fluid.

Ch9.3 – Fluid Dynamics Pascal’s principle - if a pressure is exerted on a fluid, that pressure is transmitted thru out the entire fluid.

Ch9.3 – Fluid Dynamics Win = Wout Fin.sin = Fout.sout Pascal’s principle – if a pressure is exerted on a fluid, that pressure is transmitted thru out the entire fluid. Energy is conserved: Win = Wout Fin.sin = Fout.sout

Barometer Straw Evaporation and Boiling Atmospheric pressure -exerts 101,300 N/m2 at STP Barometer Straw Evaporation and Boiling

Bernoulli Effect – As the cross sectional area decreases, the speed of the fluid increases. As the speed of a fluid increases, the pressure it exerts perpendicular decreases. A2 A1 Volume rate of flow:

Bernoulli Effect – As the cross sectional area decreases, the speed of the fluid increases. As the speed of a fluid increases, the pressure it exerts perpendicular decreases. A2 A1 Volume rate of flow:

Bernoulli Effect – As the cross sectional area decreases, the speed of the fluid increases. As the speed of a fluid increases, the pressure it exerts perpendicular decreases. A2 A1 Volume rate of flow: V Units:

As the cross-sectional area decreases, the velocity increases. Bernoulli Effect – As the cross sectional area decreases, the speed of the fluid increases. As the speed of a fluid increases, the pressure it exerts perpendicular decreases. A2 A1 Volume rate of flow: V Units: As the cross-sectional area decreases, the velocity increases.

As the cross-sectional area decreases, the velocity increases. Bernoulli Effect – As the cross sectional area decreases, the speed of the fluid increases. As the speed of a fluid increases, the pressure it exerts perpendicular decreases. Lower velocity Higher velocity A2 A1 Continuity eqn: Volume rate of flow: V Units: As the cross-sectional area decreases, the velocity increases.

As the cross-sectional area decreases, the velocity increases. Bernoulli Effect – As the cross sectional area decreases, the speed of the fluid increases. As the speed of a fluid increases, the pressure it exerts perpendicular decreases. Lower velocity Higher velocity A1 A2 Higher pressure Lower pressure Continuity eqn: Volume rate of flow: V Units: As the cross-sectional area decreases, the velocity increases. Ch9 HW#3 pg331 30,31,48,51,+ 2 bonus questions

Ch9 HW#3 pg331 30,31,48,51,+ 2 bonus questions 30. In a large apartment house water is stored in a tank on the roof 30.5m above a faucet in the kitchen. What is the gauge pressure at the faucet? 31. How deep must you dive in fresh water before the gauge equals 1.00 atmospheres?

Ch9 HW#3 pg331 30,31,48,51,+ 2 bonus questions 30. In a large apartment house water is stored in a tank on the roof 30.5m above a faucet in the kitchen. What is the gauge pressure at the faucet? 31. How deep must you dive in fresh water before the gauge equals 1.00 atmospheres?

Ch9 HW#3 pg331 30,31,48,51,+ 2 bonus questions 30. In a large apartment house water is stored in a tank on the roof 30.5m above a faucet in the kitchen. What is the gauge pressure at the faucet? 31. How deep must you dive in fresh water before the gauge equals 1.00 atmospheres?

48. Consider a narrow horizontal cylindrical chamber permanently sealed at one end and closed off at the other end with a tightly fitted moveable piston having an area of 0.050m2. The chamber is filled with oil, and a compressive force of 1000N is applied to the liquid via the piston. Determine the pressure read by a sensor (a) at the flat far end and (b) halfway in along the wall. piston 51. A hydraulic press consists of two connected cylinders: one 8.00cm in diameter, the other 20.00cm. The cylinders are sealed with moveable pistons and the whole system is filled with oil. If a force of 600N is then applied to the smaller piston, what force will be exerted on the larger piston?

48. Consider a narrow horizontal cylindrical chamber permanently sealed at one end and closed off at the other end with a tightly fitted moveable piston having an area of 0.050m2. The chamber is filled with oil, and a compressive force of 1000N is applied to the liquid via the piston. Determine the pressure read by a sensor (a) at the flat far end and (b) halfway in along the wall. piston 51. A hydraulic press consists of two connected cylinders: one 8.00cm in diameter, the other 20.00cm. The cylinders are sealed with moveable pistons and the whole system is filled with oil. If a force of 600N is then applied to the smaller piston, what force will be exerted on the larger piston?

48. Consider a narrow horizontal cylindrical chamber permanently sealed at one end and closed off at the other end with a tightly fitted moveable piston having an area of 0.050m2. The chamber is filled with oil, and a compressive force of 1000N is applied to the liquid via the piston. Determine the pressure read by a sensor (a) at the flat far end and (b) halfway in along the wall. piston 51. A hydraulic press consists of two connected cylinders: one 8.00cm in diameter, the other 20.00cm. The cylinders are sealed with moveable pistons and the whole system is filled with oil. If a force of 600N is then applied to the smaller piston, what force will be exerted on the larger piston?

Bonus ? #1) Water is flowing at 5 m/s through a garden hose with a radius of 1cm. A nozzle is attached to the hose, reducing the radius of the hose down to 0.25cm. What speed does the water flow out the nozzle?

Bonus ? #1) Water is flowing at 5 m/s through a garden hose with a radius of 1cm. A nozzle is attached to the hose, reducing the radius of the hose down to 0.5cm. What speed does the water flow out the nozzle?

Bonus ? #2) Water is flowing through a garden hose with a radius of 1cm. A nozzle is attached to the hose, reducing the radius of the hose down to 0.25cm. The water emerges from the nozzle and lands in a bucket 20cm away, as shown. The bucket collects 100ml of water in 10s. a) What is the volume rate of flow from the nozzle? b) What is the velocity of the water in the garden hose? A2 A1 (water) 25cm 20cm

A nozzle is attached to the hose, reducing the radius of the hose down Bonus ? #2) Water is flowing through a garden hose with a radius of 1cm. A nozzle is attached to the hose, reducing the radius of the hose down to 0.25cm. The water emerges from the nozzle and lands in a bucket 20cm away, as shown. The bucket collects 100cm3 of water in 10s. a) What is the volume rate of flow from the nozzle? b) What is the velocity of the water in the garden hose? s = vit + ½at2 0.25 = 4.9t2 t = 0.23s a) b) A2 A1 (water) 25cm 20cm

A nozzle is attached to the hose, reducing the radius of the hose down Bonus ? #2) Water is flowing through a garden hose with a radius of 1cm. A nozzle is attached to the hose, reducing the radius of the hose down to 0.25cm. The water emerges from the nozzle and lands in a bucket 20cm away, as shown. The bucket collects 100cm3 of water in 10s. a) What is the volume rate of flow from the nozzle? b) What is the velocity of the water in the garden hose? s = vit + ½at2 0.25 = 4.9t2 t = 0.23s a) b) A2 A1 (water) 25cm 20cm

Ch9.4 - Buoyancy Buoyant force – force of a fluid acting on an object. The fluid will make the object lighter by an amount equal to the weight of the fluid it displaces. The net force of the fluid on the object pushes the object upward.

Ex1) A rock is lowered into a beaker of water, displacing 120ml of water. If the rock weighs 4N dry, what is its weight while immersed in water? (Basically, what is the tension in the rope?)

Ex1) A rock is lowered into a beaker of water, displacing 120ml of water. If the rock weighs 4N dry, what is its weight while immersed in water? (Basically, what is the tension in the rope?)

Buoyant force – weight of the fluid displaced!

Buoyant force – weight of the fluid displaced!

FB = m.g Buoyant force – weight of the fluid displaced! mass this volume FB = mass of fluid X gravity FB = m.g

FB = m.g FB = ρV.g Buoyant force – weight of the fluid displaced! mass this volume FB = mass of fluid X gravity FB = m.g FB = ρV.g

If an object weighs more than the weight of Find this volume If an object weighs more than the weight of the volume of water it displaces, it sinks. If an object weighs less that the weight of the volume of water it displaces, it floats. If an object weighs the same as the volume of water displaced it will remain suspended. Find this volume

Ex2) A 50g plastic cube measures 4cm per side. When submerged in water, how much of it sticks above the surface?

0 = 0.5N – (1000kg/m3)[(0.04).(0.04).h](9.8m/s2) Ex2) A 50g plastic cube measures 4cm per side. When submerged in water, how much FB of it sticks above the surface? Fnet = Fg – FB Fg 0 = 0.5N – (1000kg/m3)V(9.8m/s2) 0 = 0.5N – (1000kg/m3)(l.w.h)(9.8m/s2) 0 = 0.5N – (1000kg/m3)[(0.04).(0.04).h](9.8m/s2) h = 0.032m (3.2cm of the block below the surface) Above the surface: 0.8cm Ch9 HW#4 pg331 52,56,59, + 2 Bonus ?’s

Ch9 HW#4 pg331 52,56,59, + 2 Bonus ?’s 52. The area of the face of the small piston of a hydraulic press is 10cm2. An input force of 100N is applied to that piston and we wish to have the large piston exert a corresponding output force of 9600N. What must be the area in cm2 of the face of the larger piston? 56. A hydraulic lift consists of two interconnected pistons filled with common working liquid. If the areas of the piston faces are 64.0cm2 and 3200cm2 and if a 900kg car rests on the latter, how much force must be exerted to raise the vehicle very slowly? If the car must be raised 2.00m, how far must the input piston be depressed?

52. The area of the face of the small piston of a hydraulic press is 10cm2. An input force of 100N is applied to that piston and we wish to have the large piston exert a corresponding output force of 9600N. What must be the area in cm2 of the face of the larger piston? 56. A hydraulic lift consists of two interconnected pistons filled with common working liquid. If the areas of the piston faces are 64.0cm2 and 3200cm2 and if a 900kg car rests on the latter, how much force must be exerted to raise the vehicle very slowly? If the car must be raised 2.00m, how far must the input piston be depressed?

56. A hydraulic lift consists of two interconnected pistons filled with common working liquid. If the areas of the piston faces are 64.0cm2 and 3200cm2 and if a 900kg car rests on the latter, how much force must be exerted to raise the vehicle very slowly? If the car must be raised 2.00m, how far must the input piston be depressed?

59. What is the buoyant force exerted on a sunken treasure chest that has come to rest on a few small rocks at the bottom of a fresh-water lake? The chest is 1.00-m long, 0.50-m wide, and 0.60-m high and contains 10kg of pure gold.

59. What is the buoyant force exerted on a sunken treasure chest that has come to rest on a few small rocks at the bottom of a fresh-water lake? The chest is 1.00-m long, 0.50-m wide, and 0.60-m high and contains 10kg of pure gold. (FB = weight of the fluid displaced.) Volume = 1m x .5m x .6m = .3m3 FB Fg

Bonus #1) A 100g plastic cube measures 5cm per side. When submerged in water, how much of it sticks above the surface? FB Fnet = Fg – FB Fg

0 = 1.0N – (1000kg/m3)[(0.05).(0.05).h](9.8m/s2) Bonus #1) A 100g plastic cube measures 5cm per side. When submerged in water, how much of it sticks above the surface? FB Fnet = Fg – FB Fg 0 = 1.0N – (1000kg/m3)V(9.8m/s2) 0 = 1.0N – (1000kg/m3)(l.w.h)(9.8m/s2) 0 = 1.0N – (1000kg/m3)[(0.05).(0.05).h](9.8m/s2) h = 0.041m (4.1cm of the block below the surface) Above the surface: 0.9cm

Bonus #2) A 180g plastic ball has a diameter of 4cm Bonus #2) A 180g plastic ball has a diameter of 4cm. It is attached by a string to a gold plate, causing it to sink to the bottom of a fresh water lake, as shown. What is the tension in the string? Vol = 4/3π.r3

0 = 1.80N – (1000kg/m3)(π.r3)(9.8m/s2) + FT Bonus #2) A 180g plastic ball has a diameter of 4cm. It is attached by a string to a gold plate, causing it to sink to the bottom of a fresh water lake, as shown. What is the tension in the string? Vol = 4/3π.r3 FB Fnet = Fg – FB + FT Fg FT 0 = 1.80N – (1000kg/m3)V(9.8m/s2) + FT 0 = 1.80N – (1000kg/m3)(π.r3)(9.8m/s2) + FT 0 = 1.80N – (1000kg/m3)[4/3π.0.043](9.8m/s2) + FT FT = 2.62N – 1.80N FT = 0.8N

Ch9.5 – Buoyant Force FRQ Ex1) While exploring a sunken ocean liner, a researcher found the absolute pressure on the robot observation submarine at the level of the ship to be about 413 atmospheres. The density of seawater is 1025 kg/m3. a. Calculate the gauge pressure, PG, on the sunken ocean liner. b. Calculate the depth D of the sunken ocean liner. c. Calculate the magnitude F of the force due to the water on a viewing port of the submarine at this depth if the viewing port has a surface area of 0.0100m2.

Ch9 – Buoyant Force FRQ Ex1) While exploring a sunken ocean liner, a researcher found the absolute pressure on the robot observation submarine at the level of the ship to be about 413 atmospheres. The density of seawater is 1025 kg/m3. a. Calculate the gauge pressure, PG, on the sunken ocean liner. b. Calculate the depth D of the sunken ocean liner. c. Calculate the magnitude F of the force due to the water on a viewing port of the submarine at this depth if the viewing port has a surface area of 0.0100m2. = 4x107Pa

Ch9 – Buoyant Force FRQ Ex1) While exploring a sunken ocean liner, a researcher found the absolute pressure on the robot observation submarine at the level of the ship to be about 413 atmospheres. The density of seawater is 1025 kg/m3. a. Calculate the gauge pressure, PG, on the sunken ocean liner. b. Calculate the depth D of the sunken ocean liner. c. Calculate the magnitude F of the force due to the water on a viewing port of the submarine at this depth if the viewing port has a surface area of 0.0100m2. = 4x107Pa 4x107N/m2 = (1025 kg/m3)(9.8m/s2)(h) h = 4155m

Ch9 – Buoyant Force FRQ Ex1) While exploring a sunken ocean liner, a researcher found the absolute pressure on the robot observation submarine at the level of the ship to be about 413 atmospheres. The density of seawater is 1025 kg/m3. a. Calculate the gauge pressure, PG, on the sunken ocean liner. b. Calculate the depth D of the sunken ocean liner. c. Calculate the magnitude F of the force due to the water on a viewing port of the submarine at this depth if the viewing port has a surface area of 0.0100m2. = 4x107Pa 4x107N/m2 = (1025 kg/m3)(9.8m/s2)(h) h = 4155m F = PA = (4x107N/m2)(.01m2) =4x105N

Suppose the ocean liner came to rest at the surface of the ocean before it started to sink. Due to the resistance of the seawater, the sinking ocean liner then reached a terminal velocity of 10.0 m/s after falling for 30 s. d. Determine the magnitude a of the average acceleration of the ocean liner during this period of time. e. Assuming the acceleration was constant, calculate the distance d below the surface at which the ocean liner reached this terminal velocity. f. Calculate the time t it took the ocean liner to sink from the surface to the bottom of the ocean.

Suppose the ocean liner came to rest at the surface of the ocean before it started to sink. Due to the resistance of the seawater, the sinking ocean liner then reached a terminal velocity of 10.0 m/s after falling for 30 s. d. Determine the magnitude a of the average acceleration of the ocean liner during this period of time. e. Assuming the acceleration was constant, calculate the distance d below the surface at which the ocean liner reached this terminal velocity. f. Calculate the time t it took the ocean liner to sink from the surface to the bottom of the ocean.

Suppose the ocean liner came to rest at the surface of the ocean before it started to sink. Due to the resistance of the seawater, the sinking ocean liner then reached a terminal velocity of 10.0 m/s after falling for 30 s. d. Determine the magnitude a of the average acceleration of the ocean liner during this period of time. e. Assuming the acceleration was constant, calculate the distance d below the surface at which the ocean liner reached this terminal velocity. f. Calculate the time t it took the ocean liner to sink from the surface to the bottom of the ocean.

Suppose the ocean liner came to rest at the surface of the ocean before it started to sink. Due to the resistance of the seawater, the sinking ocean liner then reached a terminal velocity of 10.0 m/s after falling for 30 s. d. Determine the magnitude a of the average acceleration of the ocean liner during this period of time. e. Assuming the acceleration was constant, calculate the distance d below the surface at which the ocean liner reached this terminal velocity. f. Calculate the time t it took the ocean liner to sink from the surface to the bottom of the ocean. Ch9 HW#5 Buoyant Force FRQ

Ch9 HW#5 – Buoyant Force Free Response 1. While conducting a strange pressure experiment a gold bar is placed in a large container of Mercury, and is allowed to sink to the bottom. A researcher found the absolute pressure at the bottom of the container to be about 4 atm. The density of mercury is 13,550 kg/m3. a. Calculate the gauge pressure pG of a pressure gauge lowered to the bottom of the container. b. Calculate the depth D of the sunken gold bar. c. Calculate the magnitude F of the force due to the mercury on a 1cm2 section of the gold. P = ρgh

Ch9 HW#5 – Buoyant Force Free Response 1. While conducting a strange pressure experiment a gold bar is placed in a large container of Mercury, and is allowed to sink to the bottom. A researcher found the absolute pressure at the bottom of the container to be about 4 atm. The density of mercury is 13,550 kg/m3. a. Calculate the gauge pressure pG of a pressure gauge lowered to the bottom of the container. b. Calculate the depth D of the sunken gold bar. P = ρgh 303,900N/m2 = (13,550kg/m3)(9.8m/s2)(h) h = 2.29m c. Calculate the magnitude F of the force due to the mercury on a 1cm2 section of the gold.

Ch9 HW#5 – Buoyant Force Free Response 1. While conducting a strange pressure experiment a gold bar is placed in a large container of Mercury, and is allowed to sink to the bottom. A researcher found the absolute pressure at the bottom of the container to be about 4 atm. The density of mercury is 13,550 kg/m3. a. Calculate the gauge pressure pG of a pressure gauge lowered to the bottom of the container. b. Calculate the depth D of the sunken gold bar. P = ρgh 303,900N/m2 = (13,550kg/m3)(9.8m/s2)(h) h = 2.29m c. Calculate the magnitude F of the force due to the mercury on a 1cm2 section of the gold.

Ch9 HW#3 – Buoyant Force Free Response Suppose that the gold bar was held at the surface of the mercury then released. Due to the resistance of the fluid, the sinking gold bar reached a terminal velocity of 2.0 m/s after falling for 1.5s. d. Determine the magnitude a of the average acceleration of the bar during this period of time. e. Assuming the acceleration was constant; calculate the distance d below the surface at which the bar reached this terminal velocity. f. Calculate the time t it tool the bar to sink from the surface to the bottom of the container. d = 2.3m – 1.5m = .79m

Ch9.6 Fluid Dynamics and Hydrostatics Equations You Should Know! (You are authorized to highlight your cheat sheet.)

Bernoulli Equation: (In a closed system, pressure stays constant) Ptotal1 = Ptotal2 P1 + ρgh1 + ½ρv12 = P2 + ρgh2 + ½ρv22

Ex1) The large container shown is filled with dont copy red a liquid of density 1.1x103 kg/m3. A small until it changes hole of area 2.5x10-6 m2 is opened in the h side of the container a distance h below the liquid surface, which allows a stream of liquid to flow through the hole and into d a beaker placed to the right of the container. At the same time, liquid is added to the x container at an appropriate rate so that h remains constant. The amount of liquid collected in the beaker in 2.0min is 7.2x10-4 m3. a. Calculate the volume rate of flow of liquid from the hole in m3/s. b. Calc the speed of liquid as it exits from the hole. c. Calc the height h of liquid needed above the hole to cause the speed you determined in part (b) d. Calc the force of the fluid passing thru the hole. e. If the height, d, of the hole to the base of the container, is 10cm, how far away should a beaker be placed to catch the fluid?

Ex1) The large container shown is filled with a liquid of density 1.1x103 kg/m3. A small hole of area 2.5x10-6 m2 is opened in the h side of the container a distance h below the liquid surface, which allows a stream of liquid to flow through the hole and into d a beaker placed to the right of the container. At the same time, liquid is added to the x container at an appropriate rate so that h remains constant. The amount of liquid collected in the beaker in 2.0min is 7.2x10-4 m3. a. Calculate the volume rate of flow of liquid from the hole in m3/s. b. Calc the speed of liquid as it exits from the hole.

Ex1) The large container shown is filled with a liquid of density 1.1x103 kg/m3. A small hole of area 2.5x10-6 m2 is opened in the h side of the container a distance h below the liquid surface, which allows a stream of liquid to flow through the hole and into d a beaker placed to the right of the container. At the same time, liquid is added to the x container at an appropriate rate so that h remains constant. The amount of liquid collected in the beaker in 2.0min is 7.2x10-4 m3. a. Calculate the volume rate of flow of liquid from the hole in m3/s b. Calc the speed of liquid as it exits from the hole.

Ex1) The large container shown is filled with a liquid of density 1.1x103 kg/m3. A small hole of area 2.5x10-6 m2 is opened in the h side of the container a distance h below the liquid surface, which allows a stream of liquid to flow through the hole and into d a beaker placed to the right of the container. At the same time, liquid is added to the x container at an appropriate rate so that h remains constant. The amount of liquid collected in the beaker in 2.0min is 7.2x10-4 m3. c. Calc the height h of liquid needed above the hole to cause the speed you determined in part (b) P1 + ρgh1 + ½ρv12 = P2 + ρgh2 + ½ρv22

ρgh1 = ½ρv22 h1 = 0.29m Ex1) The large container shown is filled with a liquid of density 1.1x103 kg/m3. A small hole of area 2.5x10-6 m2 is opened in the h side of the container a distance h below the liquid surface, which allows a stream of liquid to flow through the hole and into d a beaker placed to the right of the container. At the same time, liquid is added to the x container at an appropriate rate so that h remains constant. The amount of liquid collected in the beaker in 2.0min is 7.2x10-4 m3. c. Calc the height h of liquid needed above the hole to cause the speed you determined in part (b) P1 + ρgh1 + ½ρv12 = P2 + ρgh2 + ½ρv22 ρgh1 = ½ρv22 h1 = 0.29m

Ex1) The large container shown is filled with a liquid of density 1.1x103 kg/m3. A small hole of area 2.5x10-6 m2 is opened in the h side of the container a distance h below the liquid surface, which allows a stream of liquid to flow through the hole and into d a beaker placed to the right of the container. At the same time, liquid is added to the x container at an appropriate rate so that h remains constant. The amount of liquid collected in the beaker in 2.0min is 7.2x10-4 m3. d. Calc the force of the fluid passing thru the hole.

Ex1) The large container shown is filled with a liquid of density 1.1x103 kg/m3. A small hole of area 2.5x10-6 m2 is opened in the h side of the container a distance h below the liquid surface, which allows a stream of liquid to flow through the hole and into d a beaker placed to the right of the container. At the same time, liquid is added to the x container at an appropriate rate so that h remains constant. The amount of liquid collected in the beaker in 2.0min is 7.2x10-4 m3. d. Calc the force of the fluid passing thru the hole.

Ex1) The large container shown is filled with a liquid of density 1.1x103 kg/m3. A small hole of area 2.5x10-6 m2 is opened in the h side of the container a distance h below the liquid surface, which allows a stream of liquid to flow through the hole and into d a beaker placed to the right of the container. At the same time, liquid is added to the x container at an appropriate rate so that h remains constant. The amount of liquid collected in the beaker in 2.0min is 7.2x10-4 m3. e. If the height, d, of the hole to the base of the container, is 10cm, how far away should a beaker be placed to catch the fluid?

Ex1) The large container shown is filled with a liquid of density 1.1x103 kg/m3. A small hole of area 2.5x10-6 m2 is opened in the h side of the container a distance h below the liquid surface, which allows a stream of liquid to flow through the hole and into d a beaker placed to the right of the container. At the same time, liquid is added to the x container at an appropriate rate so that h remains constant. The amount of liquid collected in the beaker in 2.0min is 7.2x10-4 m3. e. If the height, d, of the hole to the base of the container, is 10cm, how far away should a beaker be placed to catch the fluid? sy = viyt + ½at2 sx=vxt .10 = ½(9.8)t2 = (0.29m/s)(0.14s) t = 0.14s = 0.34m Ch9 HW#6 Fluid Pressure FRQ

Ch9 HW#6 Fluid Pressure Free Response 1. The large container shown is filled with a liquid of density 1000kg/m3. A small hole of area 0.0000010m2 is opened in the side of the container a distance h below the liquid surface, which allows a stream of liquid to flow through the hole and into a beaker placed to the right of the container. At the same time, liquid is added to the container at an appropriate rate so that h remains constant. The amount of liquid collected in the beaker in 1.0 minute is 0.00040m3. a. Calculate the volume rate of flow of liquid from the hole in m3/s. b. Calculate the speed of liquid as it exits from the hole. c. Calculate the height h of liquid needed above the hole to cause the speed you determined in part (b) d. Suppose that there is now less liquid in the beaker so that the height h is reduced to h/2. In relation to the beaker, where will the liquid hit the tabletop? ___left of the beaker ___In the beaker ___right of the beaker Justify your answer. 2. Prove, that the pressure at the bottom of a rectangle tank is Start at 3. Prove, that the pressure at the bottom of a triangular tank is Start at

1. The large container shown is filled with a liquid of density 1000kg/m3. A small hole of area 0.0000010m2 is opened in the side of the container a distance h below the liquid surface, which allows a stream of liquid to flow through the hole and into a beaker placed to the right of the container. At the same time, liquid is added to the container at an appropriate rate so that h remains constant. The amount of liquid collected in the beaker in 1.0 minute is 0.00040m3. a. Calculate the volume rate of flow of liquid from the hole in m3/s. b. Calculate the speed of liquid as it exits from the hole. c. Calculate the height h of liquid needed above the hole to cause the speed you determined in part (b) d. Suppose that there is now less liquid in the beaker so that the height h is reduced to h/2. In relation to the beaker, where will the liquid hit the tabletop? ___left of the beaker ___In the beaker ___right of the beaker Justify your answer.

1. The large container shown is filled with a liquid of density 1000kg/m3. A small hole of area 0.0000010m2 is opened in the side of the container a distance h below the liquid surface, which allows a stream of liquid to flow through the hole and into a beaker placed to the right of the container. At the same time, liquid is added to the container at an appropriate rate so that h remains constant. The amount of liquid collected in the beaker in 1.0 minute is 0.00040m3. a. Calculate the volume rate of flow of liquid from the hole in m3/s. b. Calculate the speed of liquid as it exits from the hole. c. Calculate the height h of liquid needed above the hole to cause the speed you determined in part (b)

ρgh1 = ½ρv22 h1 = 2.23m a. Calculate the volume rate of flow of liquid from the hole in m3/s. b. Calculate the speed of liquid as it exits from the hole. c. Calculate the height h of liquid needed above the hole to cause the speed you determined in part (b) P1 + ρgh1 + ½ρv12 = P2 + ρgh2 + ½ρv22 ρgh1 = ½ρv22 h1 = 2.23m d. Suppose that there is now less liquid in the beaker so that the height h is reduced to h/2. In relation to the beaker, where will the liquid hit the tabletop? ___left of the beaker ___In the beaker ___right of the beaker Justify your answer.

2. Prove, that the pressure at the bottom of a rectangle tank is P = ρgh Start at 3. Prove, that the pressure at the bottom of a triangular tank is P = ρgh

2. Prove, that the pressure at the bottom of a rectangle tank is P = ρgh Start at 3. Prove, that the pressure at the bottom of a triangular tank is P = ρgh

2. Prove, that the pressure at the bottom of a rectangle tank is P = ρgh Start at 3. Prove, that the pressure at the bottom of a triangular tank is P = ρgh

Ch9.7 More Hydrostatics Ex1) A drinking fountain projects water at an initial angle of 50° above the horizontal, and the water reaches a maximum height of 0.150m above the point of exit. Air resistance is negligible. a. Calc the speed at which the water leaves the fountain. b. The radius of the fountain’s exit hole is 4.0x10-3 m. Calc the volume rate of flow. c. The fountain is fed by a pipe that at one point has a radius of 7.0x10-3m and is 3.0m below the fountain’s opening. The density of water is 1.0x103kg/m3. Calculate the gauge pressure in the feeder pipe at this point. viy vix 50°

0 = viy2 + 2(-9.8)(0.15) vi = 2.24 m/s viy = 1.71 m/s Ex1) A drinking fountain projects water at an initial angle of 50° above the horizontal, and the water reaches a maximum height of 0.150m above the point of exit. a. Calc the speed at which the water leaves the fountain. b. The radius of the fountain’s exit hole is 4.0x10-3 m. Calc the volume rate of flow. c. The fountain is fed by a pipe that at one point has a radius of 7.0x10-3m and is 3.0m below the fountain’s opening. The density of water is 1.0x103kg/m3. Calculate the gauge pressure in the feeder pipe at this point. vfy2 = viy2 + 2asy viy2 = vi.sinθ 0 = viy2 + 2(-9.8)(0.15) vi = 2.24 m/s viy = 1.71 m/s viy 50° vix

0 = viy2 + 2(-9.8)(0.15) vi = 2.24 m/s viy = 1.71 m/s Ex1) A drinking fountain projects water at an initial angle of 50° above the horizontal, and the water reaches a maximum height of 0.150m above the point of exit. a. Calc the speed at which the water leaves the fountain. b. The radius of the fountain’s exit hole is 4.0x10-3 m. Calc the volume rate of flow. c. The fountain is fed by a pipe that at one point has a radius of 7.0x10-3m and is 3.0m below the fountain’s opening. The density of water is 1.0x103kg/m3. Calculate the gauge pressure in the feeder pipe at this point. vfy2 = viy2 + 2asy viy2 = vi.sinθ 0 = viy2 + 2(-9.8)(0.15) vi = 2.24 m/s viy = 1.71 m/s viy 50° vix b.

(The radius of the fountain’s exit hole is 4. 0x10-3 m, vfountain = 2 (The radius of the fountain’s exit hole is 4.0x10-3 m, vfountain = 2.24 m/s) c. The fountain is fed by a pipe that at one point has a radius of 7.0x10-3m and is 3.0m below the fountain’s opening. The density of water is 1.0x103kg/m3. Calculate the gauge pressure in the feeder pipe at this point. Gauge

(The radius of the fountain’s exit hole is 4. 0x10-3 m, vfountain = 2 (The radius of the fountain’s exit hole is 4.0x10-3 m, vfountain = 2.24 m/s) c. The fountain is fed by a pipe that at one point has a radius of 7.0x10-3m and is 3.0m below the fountain’s opening. The density of water is 1.0x103kg/m3. Calculate the gauge pressure in the feeder pipe at this point. Find height above fountain: P1 + ρgh1 + ½ρv12 = P2 + ρgh2 + ½ρv22 ρghf = ½ρvf2 (9.8).hf = ½(2.4)2 hf = 0.26m Therefore height to pipe with gauge: hp = 0.26 + 3.0 = 3.26m Gauge Total pressure to the depth of the pipe: ρghp = (1000)(9.8)(3.26) = 31,948Pa Gauge would read this if water wasn’t flowing. hf hp

ρghp = (1000)(9.8)(3.26) = 31,948Pa Af.vf = Ap.vp hf hp (The radius of the fountain’s exit hole is 4.0x10-3 m, vfountain = 2.24 m/s) c. The fountain is fed by a pipe that at one point has a radius of 7.0x10-3m and is 3.0m below the fountain’s opening. The density of water is 1.0x103kg/m3. Calculate the gauge pressure in the feeder pipe at this point. Find height above fountain: P1 + ρgh1 + ½ρv12 = P2 + ρgh2 + ½ρv22 ρghf = ½ρvf2 (9.8).hf = ½(2.4)2 hf = 0.26m Therefore height to pipe with gauge: hp = 0.26 + 3.0 = 3.26m Gauge Total pressure to the depth of the pipe: We need the vel in bottom pipe: ρghp = (1000)(9.8)(3.26) = 31,948Pa Af.vf = Ap.vp Gauge would read this if water wasn’t flowing. π(4x10-3)2(2.24) = π(7x10-3)2(vp) Instead it reads ρghp – ½ρvp2 vp = 0.73 m/s Pg = ρghp – ½ρvp2 = 31,948 – ½(1000)(.73)2 = 31,948 – 266.5 = 31681.5 Pa Ch9 HW#7 Hydrostatics FRQ hf hp

Lab9.1 – Hydrostatics - due tomorrow - Ch9 HW#4 due @ beginning of period - Ch9 Lab HW 1 – 3 due tomorrow

0 = viy2 + 2(-9.8)(0.2) vi = viy = 1.98 m/s Ch9 HW#7 1. A drinking fountain projects water at an initial angle of 80° above the horizontal, and the water reaches a maximum height of 0.20m above the point of exit. Air resistance is negligible. a. Calc the speed at which the water leaves the fountain. b. The radius of the fountain’s exit hole is 2.0x10-3 m. Calc the volume rate of flow. c. The fountain is fed by a pipe that at one point has a radius of 5.0x10-3m and is 2.0m below the fountain’s opening. The density of water is 1.0x103kg/m3. Calculate the gauge pressure in the feeder pipe at this point. vfy2 = viy2 + 2asy viy2 = vi.sinθ 0 = viy2 + 2(-9.8)(0.2) vi = viy = 1.98 m/s viy 80° vix b.

0 = viy2 + 2(-9.8)(0.2) vi = 2.01 m/s viy = 1.98 m/s viy 80° vix b. Ch9 HW#7 1. A drinking fountain projects water at an initial angle of 80° above the horizontal, and the water reaches a maximum height of 0.20m above the point of exit. Air resistance is negligible. a. Calc the speed at which the water leaves the fountain. b. The radius of the fountain’s exit hole is 2.0x10-3 m. Calc the volume rate of flow. c. The fountain is fed by a pipe that at one point has a radius of 5.0x10-3m and is 2.0m below the fountain’s opening. The density of water is 1.0x103kg/m3. Calculate the gauge pressure in the feeder pipe at this point. vfy2 = viy2 + 2asy viy2 = vi.sinθ 0 = viy2 + 2(-9.8)(0.2) vi = 2.01 m/s viy = 1.98 m/s viy 80° vix b.

0 = viy2 + 2(-9.8)(0.2) vi = 2.01 m/s viy = 1.98 m/s Ch9 HW#7 1. A drinking fountain projects water at an initial angle of 80° above the horizontal, and the water reaches a maximum height of 0.20m above the point of exit. Air resistance is negligible. a. Calc the speed at which the water leaves the fountain. b. The radius of the fountain’s exit hole is 2.0x10-3 m. Calc the volume rate of flow. c. The fountain is fed by a pipe that at one point has a radius of 5.0x10-3m and is 2.0m below the fountain’s opening. The density of water is 1.0x103kg/m3. Calculate the gauge pressure in the feeder pipe at this point. vfy2 = viy2 + 2asy viy2 = vi.sinθ 0 = viy2 + 2(-9.8)(0.2) vi = 2.01 m/s viy = 1.98 m/s viy 80° vix b.

ρghp = (1000)(9.8)(2.206) Af.vf = Ap.vp 1b. The radius of the fountain’s exit hole is 2.0x10-3 m. Calc the volume rate of flow. c. The fountain is fed by a pipe that at one point has a radius of 5.0x10-3m and is 2.0m below the fountain’s opening. The density of water is 1.0x103kg/m3. Calculate the gauge pressure in the feeder pipe at this point. Height of fountain: ρghf = ½ρvf2 hf hf = _______ hp Therefore height to pipe with gauge: hp = _______ Total pressure to the depth of the pipe: We need the vel in bottom pipe: ρghp = (1000)(9.8)(2.206) Af.vf = Ap.vp =______ Pa π(2x10-3)2(2.01) = π(5x10-3)2(vp) vp = ____ m/s Pg = ρghp – ½ρvp2

ρghp = (1000)(9.8)(2.206) Af.vf = Ap.vp 1b. The radius of the fountain’s exit hole is 2.0x10-3 m. Calc the volume rate of flow. c. The fountain is fed by a pipe that at one point has a radius of 5.0x10-3m and is 2.0m below the fountain’s opening. The density of water is 1.0x103kg/m3. Calculate the gauge pressure in the feeder pipe at this point. Height of fountain: ρghf = ½ρvf2 (9.8).hf = ½(2.01)2 hf hf = 0.206m hp Therefore height to pipe with gauge: hp = 0.206 + 2.0 = 2.206m Total pressure to the depth of the pipe: We need the vel in bottom pipe: ρghp = (1000)(9.8)(2.206) Af.vf = Ap.vp =21,619 Pa π(2x10-3)2(2.01) = π(5x10-3)2(vp) vp = 0.32 m/s Pg = ρghp – ½ρvp2 = 21,619 – ½(1000)(.32)2 = 21,567.8 Pa

Ch9.2 Lab HW 1. A lab group makes a boat out of clay. The clay has a mass of 44 g. How much water must be displaced to allow the boat to float? Fnet = Fg – FB 0 = mg – ρVg 2. A large cruise ship weighs 225,000 tons, or 2x108kg. If density of sea water is 1025kg/m3, how much sea water does it displace? Fnet = Fgs – FB 3. A smaller cruise ship displaces 100,000m3 of sea water. What is its weight? 0 = Fgs– ρVg

1. Be prepared to explain Bernoulli’s Effect Ch9 Rev 1. Be prepared to explain Bernoulli’s Effect Lower velocity Higher velocity A1 A2 Higher pressure Lower pressure

2. A scuba diver with all his gear has a total mass of 120kg and a total volume of 0.11m3. The diver is walking along the bottom of the ocean, tethered to a boat above, as shown. The tension in the tether is 50N. The diver accidentally steps on a sea urchin. With what force does the urchin push back?

3. Fresh water is in a holding tank on a hill, above the nozzle of a hose. The water is squirting out the nozzle at a speed of 5m/s. The radius of the nozzle is 0.5cm. The water is piped to the hose through an underground pipe 1m below the nozzle. The pipe has a radius of 1.3cm. a) What speed does water flow through the underground pipe? b) What is the height of the water level above the underground pipe? c) If there is a pressure gauge on the pipe, what does it read? ?m 1m

a) An.vn = Ap.vp π(0.0005)2(5m/s) = π(0.0013)2(vp) 3. Fresh water is in a holding tank on a hill, above the nozzle of a hose. The water is squirting out the nozzle at a speed of 5m/s. The radius of the nozzle is 0.5cm. The water is piped to the hose through an underground pipe 1m below the nozzle. The pipe has a radius of 1.3cm. a) What speed does water flow through the underground pipe? b) What is the height of the water level above the underground pipe? c) If there is a pressure gauge on the pipe, what does it read? 1m a) An.vn = Ap.vp π(0.0005)2(5m/s) = π(0.0013)2(vp) vp = 0.74 m/s b) Height of water tank above nozzle: Height of water above underground pipe: h = 2.28m

a) An.vn = Ap.vp π(0.0005)2(5m/s) = π(0.0013)2(vp) 3. Fresh water is in a holding tank on a hill, above the nozzle of a hose. The water is squirting out the nozzle at a speed of 5m/s. The radius of the nozzle is 0.5cm. The water is piped to the hose through an underground pipe 1m below the nozzle. The pipe has a radius of 1.3cm. a) What speed does water flow through the underground pipe? b) What is the height of the water level above the underground pipe? c) If there is a pressure gauge on the pipe, what does it read? ?m 1m a) An.vn = Ap.vp π(0.0005)2(5m/s) = π(0.0013)2(vp) vp = 0.74 m/s b) Height of water tank above nozzle: ρghp = ½ρv2 (1000)(9.8)(hp) = ½(1000)(5m/s)2 hp = 1.28m Height of water above underground pipe: h = 2.28m c) Total pressure to underground pipe: Bernoulli: Pg =

Bernoulli: ρghp = ½ρv2 + Pgauge 3. Fresh water is in a holding tank on a hill, above the nozzle of a hose. The water is squirting out the nozzle at a speed of 5m/s. The radius of the nozzle is 0.5cm. The water is piped to the hose through an underground pipe 1m below the nozzle. The pipe has a radius of 1.3cm. a) What speed does water flow through the underground pipe? b) What is the height of the water level above the underground pipe? c) If there is a pressure gauge on the pipe, what does it read? a) An.vn = Ap.vp π(0.0005)2(5m/s) = π(0.0013)2(vp) vp = 0.74 m/s b) Height of water tank above nozzle: ρghp = ½ρv2 (1000)(9.8)(hp) = ½(1000)(5m/s)2 hp = 1.28m Height of water above underground pipe: h = 2.28m Total pressure to underground pipe: ρghp = (1000)(9.8)(2.28) = 22,300Pa Bernoulli: ρghp = ½ρv2 + Pgauge 22,300 = ½(1000)(0.74)2 + Pg Pg = 22,026 Pa