1. E 11. E 21. D 2. C 12. C 22. D 3. B 13. D 23. A 4. D 14. D 24. E 5. A 15. B 25. D 6. B 16. C 26. E 7. B 17. D 27. E 8. D 18. D 28. A 9. E 19. E 29. A 10.A 20. C 30. C 31. u = 0m/s s = 400 t = 17.5s a = ? s = ut + 1/2(at2) 400 = 0 + 0.5xa17.52 a = 400/153.25 a = 2.6m/s2 32. Fslope = mgsinq Fslope = 2.0x9.8sin30o Fslope = 9.8N Fresultant = Fslope - Ffriction Fresultant = 9.8-4.0 Fresultant = 5.8N 33. P1 = 1.6x107Pa V1 = 0.060m3 P2 = 2.5x105Pa V2 = ? P1V1 = P2V2 V2 = P1V1/P2 V2 = 3.84m3 34. E = QV E = 5x12 E = 60J 35. The ratio Vc/Vs will decrease. The current in the circuit increases as the frequency of the supply is increased. This means the potential difference across the resistor(VR) must have increased. Furthermore, as Vc decreases as VR increases, the ratio Vc/Vs must decrease, if Vs is held constant. 36.a. The student requires to know : i. the initial temperature of the lead; ii. the mass of ice melted as a result of placing the lead block in the ice; and iii. the specific latent heat of fusion of water. NB. Assume the lead block cools down to a temperature of 0oC. b. Some of the melted ice is a result of heat from the surroundings and not from the lead. Not allowing for this, by setting up a control experiment, will increase the calculated amount of energy released by the lead. This will in turn lead to a specific heat capacity greater than the accepted value. 37.a. dsinq = nl d = (1/300)mm = 0.00333mm = 3.333x10-6m n = 2 q = 23o l = ? l = dsinq/n l = 3.333x10-6sin23o/2 l = 651nm b. Red 38. Mean t1/2 = Total/N Mean t1/2 = (53.0+54.1+57.5+56.3+55.1)/5 Mean t1/2 = 276/5 Mean t1/2 = 55.2s Random error = (max - min)/N Random error = (57.5 - 53.0)/5 Random error = 4.5/5 Random error = 0.9s t1/2 = (55.2+-0.9)s
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