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Ch. 19 Liquids. Molecules flow, moving/flowing over one another. Takes the shape of its container.

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Presentation on theme: "Ch. 19 Liquids. Molecules flow, moving/flowing over one another. Takes the shape of its container."— Presentation transcript:

1 Ch. 19 Liquids

2 Molecules flow, moving/flowing over one another. Takes the shape of its container.

3 Pressure is defined as force per unit of area.

4 Each book exerts the same force, but which one exerts a greater pressure?

5 Equation Force Pressure = Area

6

7 Units of Pressure English: lb / ft 2, lb / in 2, or p.s.i. Metric: N/m 2 or Pascal (Pa)

8 Ex 1: How much pressure is applied by a car jack that applies a force of 500 N, if the area in contact is 0.35 m 2 ?

9 G: A = 0.35 m 2, F = 500 N U: P = ? E: P = F/A S: P = 500 N/ 0.35 m 2 S: P = N/ m 2 or Pa

10 Ex 2: If an airplane window that is 40 cm by 60 cm feels a pressure of 40,000 Pa. How much force is acting on the window?

11 G: L = 40 cm = 0.4 m w = 60 cm = 0.6 m P = 40,000 Pa U: F = ? E: F = PA

12 We don’t have area (A), A = Lw A = (.4)(.6) A = 0.24 m 2

13 F = (40,000)(.24) F = 9600 N

14 Pressure increases with depth. The deeper the body of water or you go, the more water there is on top of you per square inch.

15

16 Pressure depth relationship Pressure=weight density x height P =  x h

17 Weight density of water is always 62.4 lb / ft 3.

18 Ex 3: A woman broke the world’s free dive record by going just a little over 400 ft under water with no breathing apparatus, what pressure did her body feel?

19 G: h = 400 ft,  62.4 lb/ft 3. U: P = ? E: P =  x h S: P = (62.4 lb/ft 3 )(400 ft) S: P = 24,960 lb/ft 2

20 The pressure of a liquid is the same at any given depth regardless the shape of the container.

21 The pressure P is the same on the bottom of each vessel.

22 The forces that produce a pressure are exerted equally in all directions.

23 Buoyancy The apparent loss of weight of an object that is submerged in a fluid.

24 Buoyant Force Is the upward force equal to the weight of the fluid displaced.

25 Pressure is greater at the bottom of the object because the depth is greater.

26 Buoyant force is due to pressure difference between top and bottom of object. Pressure difference does not depend on depth

27 Density of object determines the depth of submersions:

28 1.If denser than the fluid, it sinks. 2.If less dense than the fluid, it will float. 3.If the same density, it will neither float nor sink.

29 Archimedes Principles The buoyant force is equal to the weight of the liquid displaced. Immersion: all or part under water. Submersion: completely under water

30 An object completely submerged always displaces a volume equal to its own.

31 Buoyant force = weight of liquid displaced

32

33 Specific Gravity Is the ratio of any substance’s density to the density of water.

34 Density of Specific a substance Gravity = Density of water

35 Specific Gravity is equal to its density numerically, but has no units.

36 Principle of Floatation A floating object will displace a weight of fluid equals to its own weight.

37

38 Pascal’s Principle Pressure applied to a fluid in a closed container is transmitted equally

39 to every point of the fluid and to the walls of the container.

40 When the area of the 2 “pistons” are the same the input force is the same as the output force.

41 What if you increased the area of the 2 nd “piston” by 25? By Pascal’s Principle: P 1 = P 2

42 By the definition of Pressure: F 1 /A 1 = F 2 /A 2 So if the 2 nd area is increased by a factor of 25.

43 In order to maintain equality, the force has to be increased by the same factor.

44 Pascal Principle

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46 Distance each piston moved is different. F 1 d 1 = F 2 d 2

47 Ex 4: The small piston on a hydraulic lift has an area of 0.20 m 2. A car weighing 1.2 x 10 4 N sits on the rack with a large piston. If the area of the large piston is 0.9 m 2, how much force must be applied to the small piston to support the car?

48 G: A 1 = 0.2 m 2, F 2 = 1.2 x 10 4 N, A 2 =0.9 m 2 U: F 1 = ? E: F 1 /A 1 = F 2 /A 2 F 1 = F 2 A 1 /A 2

49 S: F 1 =(1.2 x 10 4 )(0.2)/(0.9) S: F 1 = 2666.7 N

50 Ex 5: How far will the 2 nd piston rise if the 1 st piston moves 1.2 m?

51 The distance the 2 nd piston moves is equal to the reciprocal of the factor of the ↑ or ↓ in area times the distance the 1 st piston moves.

52 Pascal’s Principle applies to all fluids, gases as well liquids.


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