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Chapter 13 States of Matter

Chapter 13 States of Matter. Fluid - A material flows and have no definite shape of their own. Pascal’s Principle: The change in pressure applied at any point in a confined fluid is transmitted undiminished throughout the fluid. Examples Toothpaste Hydraulics . How much can be lifted?.

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Chapter 13 States of Matter

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  1. Chapter 13States of Matter Fluid - A material flows and have no definite shape of their own. Pascal’s Principle: The change in pressure applied at any point in a confined fluid is transmitted undiminished throughout the fluid. • Examples • Toothpaste • Hydraulics

  2. How much can be lifted? F1/A1 =F2/A2 F2 = F1A2/A1 F2 = (20 N)(.1 m2)/(.05 m2) = 40 N

  3. If the little piston moves 1 meter, how far does the big one move? V1=V2 A1H1=A2H2 H2= (.05 m2)(1m)/(.1 m2) H2 = A1H1/A2 H2= .5 m

  4. Chapter 13States of Matter Swimming under pressure But d = m/v or m = dv P = hdg

  5. Chapter 13States of Matter Taking P = hdg and multiplying both sides by A gives PA = Ahdg or F = vdg Where F = vdg is the buoyant force

  6. Chapter 13States of Matter • Archimedes’ Principle - An object immersed in a fluid has an upward force on it equal to the weight of the fluid displaced by the object. • A body sinks if the weight of the fluid it displaces is less than the weight of the body. • A submerged body remains in equilibrium if the weight of the fluid it displaces exactly equals its own weight. • A body floats if it displaces a weight greater than that of its own weight

  7. A B C

  8. Chapter 13 States of Matter A block of wood has a volume of 100 cm3 and a mass of 85 grams. Will it float in water water = 1000 kg/m3 ? Will it float in gas gas = 700 kg/m3 YES NO d= m/v = 85g/100 cm3 = .85 g/cm3= 850 kg/m3

  9. vdg mg Chapter 13States of Matter What is the weight of a rock submerged in water if the rock weighs 30 newtons and has a volume of .002 m3? V = .002 m3 W = 30 N water = 1000 kg/m3 Fnet = Weight - buoyant force Fnet = mg - vdg Fnet = 30 N - (.002 m3)(1000 kg/m3)(9.8 m/s2) Fnet = 30 N - 19.6 N = 10.4 N The acceleration of the rock will be a = F/m A = 10.4 N/3.06 kg = 3.2 m/s2

  10. vdg mg Chapter 13States of Matter What the maximum weight a helium balloon of volume 2 m3 can support in air? V = 2 m3 air = 1.2 kg/m3 helium = .177 kg/m3 Fnet = Weight - buoyant force Fnet = mg - vdg Fnet = (2m3)(.177kg/m3)(9.8m/s2)-(2m3)(1.2 kg/m3)(9.8 m/s2) It can support 20 N Fnet = 3.462 N - 23.52 N

  11. Chapter 13States of Matter Bernouilli’s Principle For the horizontal flow of a fluid through a tube, the sum of the kinetic energy per unit volume and the pressure is a constant.

  12. Chapter 13States of Matter • Cohesion: The force of attraction between like particles. • Adhesion: The force of attraction between unlike particles. • Capillary action: The rise of a liquid in a narrow tube because the adhesive force is stronger than the cohesive force. • Volatile Liquid: A liquid that evaporates quickly.

  13. Chapter 13States of Matter • Adhesion • Cohesion

  14. Chapter 13 States of Matter Surface Tension: The tendency of the surface of a liquid to contract to the smallest possible area

  15. Chapter 13States of Matter Sublimation Melting Vaporization Solid Liquid Gas Freezing Condensation Supercooled

  16. Chapter 13States of Matter Thermal Expansion: The increase in length or volume of a substance when heated. Chart Pg 317 Linear expansion L2 = L1+αL1(T2-T1) Volume expansion V2 = V1+βV1(T2-T1)

  17. Chapter 13States of Matter A iron bar is 3 m long at 21ºC. What is the length of the bar at 100º C? Linear expansion L2 = L1+αL1(T2-T1) L2 = 3 m+(12 x 10 –6 (ºC-1)(3 m)(100ºC- 21ºC) L2 = 3 m +.002844 m = 3.002844 m

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