Chp. 12 Homework pg. 292-292. Total 26 points. Vocab. #1-15,17-18 Concepts #1-5,7,10,12 Problems #2-3,5-8,10,12. Vocab. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 17. 18. physical properties chemical properties amorphous brittleness crystalline ductility elasticity hardness malleability strength tensile strength thermal expansion viscosity fluid pressure buoyancy Archimede’s Principle Concepts 1. A material is most dense as a solid and least dense as a gas. The exception is water Water’s solid state is less dense than it’s liquid state. 2. Physical properties can be observed directly, without changing the substance into another substance. Example: melting and boiling point. Chemical properties can only be observed when a substance is changed into another substance. Example: iron reacting with oxygen to form iron oxide (rust). 3. Density = mass / volume. Density depends on the mass of each atom that make up the material and how tightly packed those atoms or molecules are. 4. Amorphous – atoms arranged randomly. Crystalline solid – arranged in orderly pattern. 5. a. b. c. d. e. f. P C P P P C 7. a. b. c. 2. Ductility 3. Elasticity 1. Brittleness 10. Air flow over the top of the curved wing moves faster than the air below the wing, causing lower pressure on the top than the bottom. This pushes the wing up into the air. 12. Pascal’s principle states that the pressure applied to an incompressible fluid in a closed container is transmitted equally to all parts of the fluid. A force applied to a small area (small hydraulic cylinder) will produce a larger force in a cylinder with a larger area. 13. The marble is more dense than water. 16. A steel ship has a lower average density than water. 17. The buoyant force is the same because they both have the same volume (both displace the same volume of water). Problems 2. The platinum atoms are packed more tightly together than the copper atoms. 3. P = F/A = 810 N / (9 m2) = 90 N/m2 = 90 Pa 5. P = F/A A = F / P = 4000/35 = 114.3 in2 total, so 28.6 in2 per tire 6. Pressure stays the same in the smaller cylinder as the larger cylinder. So, first, calculate the pressure in the input cylinder: P = F/A = 15 / 0.1 = 150 Pa Use that pressure to find the output force: F = P x A = (150 Pa) x (5 m2) = 750 N 7. You need to find the density of air (pg. 36) to determine that 6000 mL of air has a mass of 6 grams (or 0.006 kg). Buoyant force is equal to the weight of a fluid displaced. So if 0.006 kg of air is displaced, you can find the weight of the air displaced: Weight due to gravity = mass x 9.8 m/s2. W = (0.006 kg)(9.8) = 1.96 N. 8. 45 N – 22 N = 23 N 10. 6.5 N – 5 N = 1.5 N 12. less so the weight pushing down of the object would not “overcome” the buoyancy force pushing up on the object.