1. C 11. E 21. C 2. C 12. D 22. E 3. B 13. A 23. A 4. D 14. D 24. B 5. B 15. D 25. E 6. B 16. C 26. A 7. E 17. A 27. A 8. C 18. E 28. B 9. C 19. E 29. B 10.C 20. C 30. E 31. Step one : calculate the time the arrow takes to reach the target using the fact that the horizontal velocity is constant. t = shor/vhor t = 30/100 t = 0.3s Step two : calculate the vertical displacement of the ball in this time. uver= 0m/s t = 0.3s a = -9.8m/s/s sver = ? sver = ut + 1/2(at2) sver = 0 + 1/2(-9.8x0.32) sver = -0.441m Step three : calculate the radius of the target. r = 1.5-0.9 r = 0.6m As the arrow falls less than the radius of the target it hits the target. 32.a.i. a.ii. Frope = Tension(T) As the rope is stationary the forces acting on the buoy are balanced. Fup = w + T Fup = 50 + 1200 Fup = 1250N b. The buoyancy force is not dependent on depth. Therefore, the buoyancy force is as calculated in part a.ii.(1250N). 33. In circuit 1 the total resistance in the circuit is equal to the sum of the internal resistor(r) and the resistance of the bulb (Rbulb). Rtotal = r + Rbulb In circuit 2 the total resistance in the circuit is equal to the sum of the internal resistor(r) and the resistance of the two bulbs in parallel. Rtotal = r + Rbulb/2 The total resistance in circuit 2 is less than that in circuit 1. Thus the current(I) flowing through the internal resistor in circuit 2 is greater. The "lost" voltage (Vlost=Ir) across the internal resistor is therefore greater in circuit 2. This means the voltage across the lamps (Vlamp = E -Ir) is less in circuit 2 reducing the brightness of the lamps. 34.a. Vrms = Vpk/21/2 Vrms = 12/1.414 Vrms = 8.5V b. P = Irms2 Irms = Vrms/R Irms = 8.5/4 Irms = 2.12A P = 2.122x4 P = 18W 35. dsinq = nl d = 1/2.5x105 = 4x10-6m l = 600x10-9m n = 1 q = ? sinq = nl/d sinq = 1x600x10-9/4x10-6 sinq = 0.15 q = 8.63o 36.a. Possible transitions : E3 -> E2 E3 -> E1 E3 -> E0 E2 -> E1 E2 -> E0 E1 -> E0 b. A total of 6 lines are in the spectrum. Low frequency radiation will be produced when the transition is between energy levels with a small energy gap. The smallest energy gap is between levels E3 and E2. c. Transitions between certain states will be more frequent than others. The more frequent a transition, the more intense the spectral line will be. 37. Ek = Ephoton - Ework function Ephoton = hf = hc/l Ephoton = (6.63x10-34x3x108)/5.4x10-7 Ephoton = 3.68x10-19J Ek = 3.7x10-19 - 2.9x10-19 Ek = 0.8x10-19
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