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- 1. SOLID & FLUIDS BB101 – ENGINEERING SCIENCE
- 2. Introduction
- 3. MATTER Definition :Matter is anything that has mass and occupies space (volume). Example of matter : Chair, books, water, wood & others
- 4. Objective: State the characteristics of solid, liquid and gas.
- 5. Characteristic of Matter Objective: State the characteristics of solid, liquid and gas. CHARACTERIS TIC Arrangement of particles SOLID LIQUID Very closely packed Closely Packed Shapes & Volume Fix Shape & Volume Force between Particles Very strong forces GAS Widely Spaced Fixed volume, but not fixed shape Takes the shapes & volume of its container Not have fixed shape and volume Takes the shape & volume of its container Weak forces Very weak forces or negligible Move freely and Movement Vibrate & spin around their position Vibrate & move randomly but not freely randomly in all directions. High speed & colliding
- 6. Objective: State the characteristics of solid, liquid and gas.
- 7. Objective: State the characteristics of solid, liquid and gas.
- 8. Objective: State the characteristics of solid, liquid and gas.
- 9. Objective: State the characteristics of solid, liquid and gas.
- 10. Objective: State the characteristics of solid, liquid and gas.
- 11. Objective: State the characteristics of solid, liquid and gas.
- 12. Objective: State the characteristics of solid, liquid and gas.
- 13. DENSITY OF MATTER Definition of density : is defined as mass per unit volume Formulae The SI unit is kg/m3 and g/cm3 (special case)
- 14. Example 1 An Object has a mass of 750g and a volume of 5.0 x 10-4m3 . Solution m : 750g mass 0.75kg ρ= volume 0.75 = 5 × −4 10 =1500kg / m 3 @ m : 750g mass ρ= volume 750 = 5 × −4 10 =0.015 g / m 3
- 15. Watch it Sinking Objects.flv
- 16. Relative Density Relative density also known as a specific gravity of matter . To compare the densities of two materials, we compare each with the densities of water. Formulae Relative Density ρ material σ = ρ water Relative Density didn’t have SI unit.
- 17. Substances Density(kg/m3) Subtances Solids Densities (kg /m3) Gases Copper 8890 Air 1.29 Iron 7800 Carbon Dioxide 1.96 Lead 11300 Carbon Monoxide 1.25 Aluminium 2700 Helium 0.178 Ice 917 Hydrogen Wood,white pain 420 Oxygen 1.43 2300 Nitrogen 1.25 Concrete Cork 240 Liquids Propane Water 1000 Seawater 1025 Oil Ammonium 870 Mercury 13600 Alcohol 790 0.0899 0.760 2.02 Objective: Define density and its unit
- 18. Example 2 Find the relative density if Copper is 8890 kgm-3 and water is 1000 kgm-3 . Relative Density ρ material σ= ρ water 8890kg / m −3 σ= 1000kg / m −3 = 8.89 ( NoUnit )
- 19. Pressure
- 20. Situation of Pressure Using hand Using nail Situation 1 (Increasing the pressure by reducing the area) WHICH BALLOON POP EASIER?
- 21. Example 3 Objective: Define pressure and its unit
- 22. Situation 2 Formula The gauge pressure at any depth from the surface of a fluid; Pressure in Liquids. Pressure, P = ρ g h whereas : ρ = density of liquid h = depth P =ρ g h also known as the hydrostatic pressure.
- 23. Pressure depends on depth & density Have you ever noticed? That’s why; the dams are built much thicker at the base than at the top, because, the pressure exerted by the water increases with depth.
- 24. Example 4 Figure below show a Barometer mercury. Find the pressure? (p mercury = 13600 kgm-3) vacuum P atm 76cm P atm mercury
- 25. Solution
- 26. Pascal‘s Principle Piston is pushed in The transmission of pressure in liquid The figure show that when the plunger is pushed in, the pressure of water at the end of the plunger will cause water to spurt out in directions. Pascal’s principle states that the pressure exerted on a confined liquid is transmitted equal in all direction
- 27. Watch it Pascal's Principle.mp4
- 28. In a hydraulic system, pressure on both piston is equal
- 29. Applications of Pascal’s Principle Hydraulic Brakes of a car Hydraulic Brake Hydraulic Jack
- 30. Hydraulic system Shows a simple hydraulics system built according to Pascal Principle Input Force Area large piston Area small piston Fluid Output Force
- 31. Example 5 The cylindrical piston of a hydraulic jack has a cross sectional area of 0.06m2 and the plunger has a cross-sectional area of 0.002m2. a) The upward force for lifting a load placed on top the large piston 9000N. Calculate the downward force on the plunger required to lift this load assuming a 100% work efficiency. b) If the distance moved by the plunger is 75cm,what is the distance moved by the large piston?
- 32. solution Objective: Application of Pascal Principle
- 33. Archimedes Principle Displaced Volume
- 34. Watch it ....Archimedes' principle.mp4
- 35. Example 6 Figure below shows the weight of a mass in the air is 15N. The mass is immersed in water which has density of 1000 kg/m³. Calculate: a) The buoyant force b) The weight of water displaced c) The volume of the immersed body
- 36. Solution
- 37. Activity in group 1.Find the volume of copper of mass 200g if density the cooper is 8890kg/m3?Answer 2. The mass of a proton is 1.67x 10-27kg and it can be considered to be a sphere of roughly 1.35 x10-15m radius. What its density? Answer 3.Find the density of alcohol if 307g occupies 855cm3?Answer
- 38. Activity in group 4. A fruit seller uses a knife with a sharp edge and a cross-sectional area of 0.5cm2 to cut open a watermelon. Answer a) If the force applied on the knife is 18N,what is the pressure exerted by the knife on the watermelon b) After that, he cuts open a papaya using the same knife by exerting a pressure 2.7 x 105 Pa. Calculate the magnitude of force applied to cut the papaya
- 39. Activity in group 5.From the figure below, force input is given by 4000 N and diameter at small piston is 100 cm. If the diameter at large piston given by 250 cm, find the maximum mass of the car than can lift by the force input 4000N.Answer
- 40. Activity in group 6.A concrete slab weighs is 150N. When it fully submerged under the sea, its apparent weight is 102N. answer a) Calculate the buoyant force by concrete slab when immersed in sea water. b)Calculate the density of the sea water in kg/m3 if the volume of the sea water displaced by the concrete slab is 4800cm3
- 41. Answer
- 42. Answer
- 43. Answer
- 44. Answer 5. Force input = 4000 N Area input = π (100/100)2 = 3.1416 m2. Area input = π (250/100)2 = 19.635 m2. (F1/A1) = (F2/A2) (4000 / 3.1416) = ( F2 / 19.635 ) F2 = (4000 / 3.1416) x 19.635 = 25000 N Mass of the car = 25000 / 9.81 = 2548.42 kg.
- 45. Answer 6. a)Buoyant force= Actual weight- Apparent weight = 150N-102N =48N b) Bouyant force= Weight of sea water displaced 48N = p x (4800 x 10-6) x 9.81 N kg-1 p= 1020kgm-3 F=p x V x g
- 46. Conclusion of this solid & fluid
- 47. Reference Longman Essential Physics SPM Yap Eng Keat& Khoo Goh Kow 2012 Pelangi STPM Physics volume2 Poh Liong Yong 2014 JMSK Engineering science BB101 Fourth Ed. 2012 http://hyperphysics.phy-astr.gsu.edu/hbase/pasc.html http://physics.tutorvista.com/fluiddynamics/archimedes-principle.html
- 48. Thank you

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