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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.

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Presentation on theme: "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."— Presentation transcript:

1 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)

2 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).

3 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.

4 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.

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

6 Density – a comparison of an object’s mass to its volume. ρ H20 = 1 g/cm 3 = 1 g/ml ‘roe’ = 1000 kg/m 3 Ex1) Determine if a 500g sealed hollow tube with a radius of 2cm, 30cm length will float in water.

7 Density – a comparison of an object’s mass to its volume. ρ H20 = 1 g/cm 3 = 1 g/ml ‘roe’ = 1000 kg/m 3 Ex2) Determine the density of an unknown solid if its mass is.4kg and it displaces 21 mls of H 2 O.

8 Density – a comparison of an object’s mass to its volume. ρ H20 = 1 g/cm 3 = 1 g/ml = 1000 kg/m 3 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 H 2 O.(Move the decimal 2X for each (21mls = 21cm 3 = m 3 ) dimention.) 21mls

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

10 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 F B

11 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 F B F g - if the object’s weight is greater than it buoyant force, it sinks.

12 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 F B F g - if the object’s weight is equal to its buoyant force, it suspends.

13 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 F B F g - 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

14 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 cm 3. Convert this density to kg/m 3.

15 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 cm 3. Convert this density to kg/m 3.

16 2. A silvered color metal cube measure 1cm on a side and has a mass of 10.5g. Find its density in kg/m 3.

17 2. A silvered color metal cube measure 1cm on a side and has a mass of 10.5g. Find its density in kg/m 3.

18 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/m 3, what is the volume of the ball? F B F g

19 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/m 3, what is the volume of the ball? F B F g

20 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? F B F net = F g – F B F g

21 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? F B F net = F g – F B m. a = 5N – 3N (.5)a = 2 a = 4 m/s 2 F g

22 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: F B F g F g Density than 1000kg/m 3 V > m 3 V < m 3

23 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: F B F g F g Density than 1000kg/m 3

24 Chapter 9.2 – Pressure Ex1) A force of 100N is applied to the head of nail, with an area of 1cm 2. What pressure is exerted to the nail head?

25 Fluid pressure is created by the weight of the fluid Atmospheric Pressure: - 101,300 Pascals (N/m 2 ) - 1 atmosphere mm Hg (10.3m H 2 O) psi (pounds per inch 2 )

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

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

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

29 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

30 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

31 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:

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

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

34 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

35 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

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

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

38 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?

39 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?

40 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?

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

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

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

44 29. Gauge 11km down to bottom of the ocean.

45

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

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

48 Ch9.3 – Fluid Dynamics Pascal’s principle – if a pressure is exerted on a fluid, that pressure is transmitted thru out the entire fluid. Energy is conserved: W in = W out F in. s in = F out. s out

49 Atmospheric pressure -exerts 101,300 N/m 2 at STP Barometer StrawEvaporation and Boiling

50 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. A1A1 A2A2 Volume rate of flow:

51 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. A1A1 A2A2 Volume rate of flow:

52 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. A1A1 A2A2 Volume rate of flow: V Units:

53 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. A1A1 A2A2 Volume rate of flow: V Units: As the cross-sectional area decreases, the velocity increases.

54 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. A1A1 A2A2 Volume rate of flow: V Units: Continuity eqn: Higher velocity Lower velocity As the cross-sectional area decreases, the velocity increases.

55 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 pressure Higher velocity Lower pressure A1A1 A2A2 Volume rate of flow: V Units: Continuity eqn: As the cross-sectional area decreases, the velocity increases. Ch9 HW#3 pg331 30,31,48,51,+ 2 bonus questions

56 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?

57 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?

58 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?

59 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 0.050m 2. 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?

60 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 0.050m 2. 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?

61 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 0.050m 2. 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?

62 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?

63 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?

64 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? A 1 (water) A2A2 25cm 20cm

65 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 100cm 3 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 = v i t + ½at = 4.9t 2 t = 0.23s a) b) A 1 (water) A2A2 25cm 20cm

66 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 100cm 3 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 = v i t + ½at = 4.9t 2 t = 0.23s a) b) A 1 (water) A2A2 25cm 20cm

67 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.

68 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?)

69 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?)

70 Buoyant force – weight of the fluid displaced!

71

72 mass this volume F B = mass of fluid X gravity F B = m. g

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

74 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 Find this volume

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

76 Ex2) A 50g plastic cube measures 4cm per side. When submerged in water, how much F B of it sticks above the surface? F net = F g – F B F g 0 = 0.5N – (1000kg/m 3 )V(9.8m/s 2 ) 0 = 0.5N – (1000kg/m 3 )(l. w. h)(9.8m/s 2 ) 0 = 0.5N – (1000kg/m 3 )[(0.04). (0.04). h](9.8m/s 2 ) 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

77 52. The area of the face of the small piston of a hydraulic press is 10cm 2. 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 cm 2 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 64.0cm 2 and 3200cm 2 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?

78 52. The area of the face of the small piston of a hydraulic press is 10cm 2. 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 cm 2 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 64.0cm 2 and 3200cm 2 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?

79 56. A hydraulic lift consists of two interconnected pistons filled with common working liquid. If the areas of the piston faces are 64.0cm 2 and 3200cm 2 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?

80 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.

81 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. (F B = weight of the fluid displaced.) Volume = 1m x.5m x.6m =.3m 3 F B F g

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

83 Bonus #1) A 100g plastic cube measures 5cm per side. When submerged in water, how much of it sticks above the surface? F B F net = F g – F B F g 0 = 1.0N – (1000kg/m 3 )V(9.8m/s 2 ) 0 = 1.0N – (1000kg/m 3 )(l. w. h)(9.8m/s 2 ) 0 = 1.0N – (1000kg/m 3 )[(0.05). (0.05). h](9.8m/s 2 ) h = 0.041m (4.1cm of the block below the surface) Above the surface: 0.9cm

84 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. r 3

85 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. r 3 F B F net = F g – F B + F T F g F T 0 = 1.80N – (1000kg/m 3 )V(9.8m/s 2 ) + F T 0 = 1.80N – (1000kg/m 3 )(π. r 3 )(9.8m/s 2 ) + F T 0 = 1.80N – (1000kg/m 3 )[4/3π ](9.8m/s 2 ) + F T F T = 2.62N – 1.80N F T = 0.8N

86 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/m 3. a. Calculate the gauge pressure, P G, 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 m 2.

87 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/m 3. a. Calculate the gauge pressure, P G, 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 m 2. = 4x10 7 Pa

88 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/m 3. a. Calculate the gauge pressure, P G, 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 m 2. = 4x10 7 Pa 4x10 7 N/m 2 = (1025 kg/m 3 )(9.8m/s 2 )(h) h = 4155m

89 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/m 3. a. Calculate the gauge pressure, P G, 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 m 2. = 4x10 7 Pa 4x10 7 N/m 2 = (1025 kg/m 3 )(9.8m/s 2 )(h) h = 4155m F = PA = (4x10 7 N/m 2 )(.01m 2 ) =4x10 5 N

90 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.

91 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.

92 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.

93 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

94 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/m 3. a. Calculate the gauge pressure p G 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 1cm 2 section of the gold. P = ρgh

95 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/m 3. a. Calculate the gauge pressure p G 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/m 2 = (13,550kg/m 3 )(9.8m/s 2 )(h) h = 2.29m c. Calculate the magnitude F of the force due to the mercury on a 1cm 2 section of the gold.

96 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/m 3. a. Calculate the gauge pressure p G 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/m 2 = (13,550kg/m 3 )(9.8m/s 2 )(h) h = 2.29m c. Calculate the magnitude F of the force due to the mercury on a 1cm 2 section of the gold.

97 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

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

99 Bernoulli Equation: (In a closed system, pressure stays constant) P total1 = P total2 P 1 + ρgh 1 + ½ρv 1 2 = P 2 + ρgh 2 + ½ρv 2 2

100 Ex1) The large container shown is filled with dont copy red a liquid of density 1.1x10 3 kg/m 3. A small until it changes hole of area 2.5x10 -6 m 2 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 m 3. a. Calculate the volume rate of flow of liquid from the hole in m 3 /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?

101 Ex1) The large container shown is filled with a liquid of density 1.1x10 3 kg/m 3. A small hole of area 2.5x10 -6 m 2 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 m 3. a. Calculate the volume rate of flow of liquid from the hole in m 3 /s. b. Calc the speed of liquid as it exits from the hole.

102 Ex1) The large container shown is filled with a liquid of density 1.1x10 3 kg/m 3. A small hole of area 2.5x10 -6 m 2 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 m 3. a. Calculate the volume rate of flow of liquid from the hole in m 3 /s b. Calc the speed of liquid as it exits from the hole.

103 Ex1) The large container shown is filled with a liquid of density 1.1x10 3 kg/m 3. A small hole of area 2.5x10 -6 m 2 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 m 3. c. Calc the height h of liquid needed above the hole to cause the speed you determined in part (b) P 1 + ρgh 1 + ½ρv 1 2 = P 2 + ρgh 2 + ½ρv 2 2

104 Ex1) The large container shown is filled with a liquid of density 1.1x10 3 kg/m 3. A small hole of area 2.5x10 -6 m 2 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 m 3. c. Calc the height h of liquid needed above the hole to cause the speed you determined in part (b) P 1 + ρgh 1 + ½ρv 1 2 = P 2 + ρgh 2 + ½ρv 2 2 ρgh 1 = ½ρv 2 2 h 1 = 0.29m

105 Ex1) The large container shown is filled with a liquid of density 1.1x10 3 kg/m 3. A small hole of area 2.5x10 -6 m 2 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 m 3. d. Calc the force of the fluid passing thru the hole.

106 Ex1) The large container shown is filled with a liquid of density 1.1x10 3 kg/m 3. A small hole of area 2.5x10 -6 m 2 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 m 3. d. Calc the force of the fluid passing thru the hole.

107 Ex1) The large container shown is filled with a liquid of density 1.1x10 3 kg/m 3. A small hole of area 2.5x10 -6 m 2 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 m 3. 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?

108 Ex1) The large container shown is filled with a liquid of density 1.1x10 3 kg/m 3. A small hole of area 2.5x10 -6 m 2 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 m 3. 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? s y = v iy t + ½at 2 s x =v x t.10 = ½(9.8)t 2 = (0.29m/s)(0.14s) t = 0.14s = 0.34m Ch9 HW#6 Fluid Pressure FRQ

109 Ch9 HW#6 Fluid Pressure Free Response 1. The large container shown is filled with a liquid of density 1000kg/m 3. A small hole of area m 2 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 m 3. a. Calculate the volume rate of flow of liquid from the hole in m 3 /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

110 1. The large container shown is filled with a liquid of density 1000kg/m 3. A small hole of area m 2 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 m 3. a. Calculate the volume rate of flow of liquid from the hole in m 3 /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.

111 1. The large container shown is filled with a liquid of density 1000kg/m 3. A small hole of area m 2 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 m 3. a. Calculate the volume rate of flow of liquid from the hole in m 3 /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)

112 a. Calculate the volume rate of flow of liquid from the hole in m 3 /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) P 1 + ρgh 1 + ½ρv 1 2 = P 2 + ρgh 2 + ½ρv 2 2 ρgh 1 = ½ρv 2 2 h 1 = 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.

113 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 Start at

114 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 Start at

115 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 Start at

116 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 -3 m and is 3.0m below the fountain’s opening. The density of water is 1.0x10 3 kg/m 3. Calculate the gauge pressure in the feeder pipe at this point. v iy v ix 50°

117 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 -3 m and is 3.0m below the fountain’s opening. The density of water is 1.0x10 3 kg/m 3. Calculate the gauge pressure in the feeder pipe at this point. v fy 2 = v iy 2 + 2as y v iy 2 = v i. sinθ 0 = v iy 2 + 2(-9.8)(0.15) v i = 2.24 m/s v iy = 1.71 m/s v iy 50° v ix

118 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 -3 m and is 3.0m below the fountain’s opening. The density of water is 1.0x10 3 kg/m 3. Calculate the gauge pressure in the feeder pipe at this point. v fy 2 = v iy 2 + 2as y v iy 2 = v i. sinθ 0 = v iy 2 + 2(-9.8)(0.15) v i = 2.24 m/s v iy = 1.71 m/s v iy 50° v ix b.

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

120 (The radius of the fountain’s exit hole is 4.0x10 -3 m, v fountain = 2.24 m/s) c. The fountain is fed by a pipe that at one point has a radius of 7.0x10 -3 m and is 3.0m below the fountain’s opening. The density of water is 1.0x10 3 kg/m 3. Calculate the gauge pressure in the feeder pipe at this point. Find height above fountain: P 1 + ρgh 1 + ½ρv 1 2 = P 2 + ρgh 2 + ½ρv 2 2 ρgh f = ½ρv f 2 (9.8). h f = ½(2.4) 2 h f = 0.26m Therefore height to pipe with gauge: h p = = 3.26m Gauge Total pressure to the depth of the pipe: ρgh p = (1000)(9.8)(3.26) = 31,948Pa Gauge would read this if water wasn’t flowing. hfhf hphp

121 (The radius of the fountain’s exit hole is 4.0x10 -3 m, v fountain = 2.24 m/s) c. The fountain is fed by a pipe that at one point has a radius of 7.0x10 -3 m and is 3.0m below the fountain’s opening. The density of water is 1.0x10 3 kg/m 3. Calculate the gauge pressure in the feeder pipe at this point. Find height above fountain: P 1 + ρgh 1 + ½ρv 1 2 = P 2 + ρgh 2 + ½ρv 2 2 ρgh f = ½ρv f 2 (9.8). h f = ½(2.4) 2 h f = 0.26m Therefore height to pipe with gauge: h p = = 3.26m Gauge Total pressure to the depth of the pipe: We need the vel in bottom pipe: ρgh p = (1000)(9.8)(3.26) = 31,948Pa A f. v f = A p. v p Gauge would read this if water wasn’t flowing. π(4x10 -3 ) 2 (2.24) = π(7x10 -3 ) 2 (v p ) Instead it reads ρgh p – ½ρv p 2 v p = 0.73 m/s P g = ρgh p – ½ρv p 2 = 31,948 – ½(1000)(.73) 2 = 31,948 – = Pa Ch9 HW#7 Hydrostatics FRQ hfhf hphp

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

123 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 -3 m and is 2.0m below the fountain’s opening. The density of water is 1.0x10 3 kg/m 3. Calculate the gauge pressure in the feeder pipe at this point. v fy 2 = v iy 2 + 2as y v iy 2 = v i. sinθ 0 = v iy 2 + 2(-9.8)(0.2) v i = v iy = 1.98 m/s v iy 80° v ix b.

124 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 -3 m and is 2.0m below the fountain’s opening. The density of water is 1.0x10 3 kg/m 3. Calculate the gauge pressure in the feeder pipe at this point. v fy 2 = v iy 2 + 2as y v iy 2 = v i. sinθ 0 = v iy 2 + 2(-9.8)(0.2) v i = 2.01 m/s v iy = 1.98 m/s v iy 80° v ix b.

125 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 -3 m and is 2.0m below the fountain’s opening. The density of water is 1.0x10 3 kg/m 3. Calculate the gauge pressure in the feeder pipe at this point. v fy 2 = v iy 2 + 2as y v iy 2 = v i. sinθ 0 = v iy 2 + 2(-9.8)(0.2) v i = 2.01 m/s v iy = 1.98 m/s v iy 80° v ix b.

126 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 -3 m and is 2.0m below the fountain’s opening. The density of water is 1.0x10 3 kg/m 3. Calculate the gauge pressure in the feeder pipe at this point. Height of fountain: ρgh f = ½ρv f 2 h f h f = _______ h p Therefore height to pipe with gauge: h p = _______ Total pressure to the depth of the pipe: We need the vel in bottom pipe: ρgh p = (1000)(9.8)(2.206) A f. v f = A p. v p =______ Pa π(2x10 -3 ) 2 (2.01) = π(5x10 -3 ) 2 (v p ) v p = ____ m/s P g = ρgh p – ½ρv p 2

127 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 -3 m and is 2.0m below the fountain’s opening. The density of water is 1.0x10 3 kg/m 3. Calculate the gauge pressure in the feeder pipe at this point. Height of fountain: ρgh f = ½ρv f 2 (9.8). h f = ½(2.01) 2 h f h f = 0.206m h p Therefore height to pipe with gauge: h p = = 2.206m Total pressure to the depth of the pipe: We need the vel in bottom pipe: ρgh p = (1000)(9.8)(2.206) A f. v f = A p. v p =21,619 Pa π(2x10 -3 ) 2 (2.01) = π(5x10 -3 ) 2 (v p ) v p = 0.32 m/s P g = ρgh p – ½ρv p 2 = 21,619 – ½(1000)(.32) 2 = 21,567.8 Pa

128 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? F net = F g – F B 0 = mg – ρVg 2. A large cruise ship weighs 225,000 tons, or 2x10 8 kg. If density of sea water is 1025kg/m 3, how much sea water does it displace? F net = F gs – F B 0 = mg – ρVg 3. A smaller cruise ship displaces 100,000m 3 of sea water. What is its weight? F net = F gs – F B 0 = F gs – ρVg

129 Ch9 Rev 1. Be prepared to explain Bernoulli’s Effect Lower velocity Higher pressure Higher velocity Lower pressure A1A1 A2A2

130 2. A scuba diver with all his gear has a total mass of 120kg and a total volume of 0.11m 3. 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?

131 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

132 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) A n. v n = A p. v p π(0.0005) 2 (5m/s) = π(0.0013) 2 (v p ) v p = 0.74 m/s b) Height of water tank above nozzle: Height of water above underground pipe: h = 2.28m

133 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) A n. v n = A p. v p π(0.0005) 2 (5m/s) = π(0.0013) 2 (v p ) v p = 0.74 m/s b) Height of water tank above nozzle: ρgh p = ½ρv 2 (1000)(9.8)(h p ) = ½(1000)(5m/s) 2 h p = 1.28m Height of water above underground pipe: h = 2.28m c) Total pressure to underground pipe: Bernoulli: P g =

134 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) A n. v n = A p. v p π(0.0005) 2 (5m/s) = π(0.0013) 2 (v p ) v p = 0.74 m/s b) Height of water tank above nozzle: ρgh p = ½ρv 2 (1000)(9.8)(h p ) = ½(1000)(5m/s) 2 h p = 1.28m Height of water above underground pipe: h = 2.28m Total pressure to underground pipe: ρgh p = (1000)(9.8)(2.28) = 22,300Pa Bernoulli: ρgh p = ½ρv 2 + P gauge 22,300 = ½(1000)(0.74) 2 + P g P g = 22,026 Pa


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