Physics 30 Dec 06 Diploma Exam

Neutrinos that are given off in the fission reaction

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Unformatted text preview: ls, several switches, and connecting wires. These components are to be used to construct a hair dryer. Schematics of Hair Dryer Components F 110 V AC Heating coil (two) Electric fan Switch (several) The design requirements for your hair dryer are that the fan is always on when the hair dryer is on and that it has two heat settings: high and low. Written Response — 15% 1. • Draw a schematic diagram of a hair dryer circuit that meets the design requirements. • Based on the circuit diagram you have drawn, analyze the operation of the hair dryer. In your response, explain how the switch settings and their locations in the circuit control the low and high heat settings. Also, explain why the hair dryer should be designed so that the fan is on whenever the hair dryer is on. Note: Marks will be awarded for the physics principles used in your response and for the effective communication of your response. 22 Written-response question 2 begins on the next page. 23 Use the following information to answer the next question. Several Canadian companies are redesigning and testing bulletproof vests. One company does a test that involves firing a target rifle at a crash test dummy wearing a vest. Dummy Regular bullet mBullet = 8.00 g vBullet = 650 m/s Vest The company is testing the vests with both regular bullets and armourpiercing bullets. The armour-piercing bullet travels 1.20 times faster and has 1.20 times the mass of the regular bullet shown above. Written Response — 15% 2. • Quantitatively compare the kinetic energy of the armour-piercing bullet with the kinetic energy of the regular bullet. • How much energy is released by the explosion of the gunpowder if the transfer of energy from the explosion to the regular bullet is 90.0% efficient? • The regular bullet is in the rifle barrel for 1.42 × 10–3 s. What is the average force exerted on the regular bullet by the expanding gases? 24 Use this additional information to answer the next part of the question. A second test performed by the company has the regular bullet strike the vest at a glancing angle. The mass of the vest and the dummy is 56.0 kg. The bullet–vest collision is inelastic. vBullet = 98.0 m/s mBullet = 8.00 g Path of bullet Vest 120° Dummy vBullet = 650 m/s • Determine the resultant speed of the vest and the dummy following the glancing collision shown above. Clearly communicate your understanding of the physics principles that you are using to solve this question. You may communicate this understanding mathematically, graphically, and/or with written statements. 25 You have now completed the examination. If you have time, you may wish to check your answers. 26 Fold and tear along perforation. PHYSICS DATA SHEET CONSTANTS Gravity, Electricity, and Magnetism Trigonometry and Vectors Acceleration Due to Gravity or Gravitational Field Near Earth ............ ag or g = 9.81 m/s or 9.81 N/kg Gravitational Constant ........................ G = 6.67 × 10 Mass of Earth ...................................... Me = 5.98 × 10 kg Radius of Earth ................................... Re = 6.37 × 10 m Coulomb’s Law Constant.................... k = 8.99 × 10 N.m /C Electron Volt....................................... 1 eV = 1.60 × 10 Elementary Charge.............................. e = 1.60 × 10 Index of Refraction of Air................... n = 1.00 Speed of Light in Vacuum .................. c = 3.00 × 10 m/s 2 opposite sin θ = hypotenuse N.m /kg –11 2 2 24 2 –19 –19 R= adjacent cosθ = hypotenuse 6 9 For any Vector 2 tan θ = opposite tan θ = adjacent J 8 2 2 2 Ry Rx Rx = R cos θ a b c = = sin A sin B sin C C 2 Rx + Ry * R Ry = R sin θ 2 c = a + b – 2ab cos C Prefixes Used With SI Units Atomic Physics Energy of an Electron in the 1st Bohr Orbit of Hydrogen...................... E1 = – 2 .18 × 10 –18 Prefix J or –13.6 eV Symbol Exponential Value Prefix Symbol Exponential Value r1 = 5.29 × 10 Rydberg’s Constant for Hydrogen ...... m 7 eV.s –9 giga .............. G .................10 9 micro............ µ ..................10 RH = 1.10 × 10 /m –11 –15 12 –6 mega ............ M .................10 6 kilo .............. k ..................10 3 centi ............. c ..................10 Radius of 1st Bohr Orbit of Hydrogen –34 tera .............. T ..................10 –3 h = 6.63 × 10 pico .............. p...................10 nano ............. n...................10 J.s or 4.14 × 10 Planck’s Constant................................ –12 –2 hecto ............ h ..................10 2 deci .............. d...................10 –1 deka ............. da ................10 1 milli ............. m..................10 Particles Rest Mass Charge Alpha Particle ............... mα = 6.65 × 10 –27 Electron......................... m e = 9.11 × 10 –31 Neutron ......................... m n = 1.67 × 10 –27 Proton........
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