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Transcript
F.5 Physics Mock Examination (09-10) Paper 2 Answers 1-5 B A A C A 6-10 11-15 16-20 21-25 26-30 31-35 36-40 41-45 C A B C C B A C B A D D A D B A A A C A B B C D C A B A B C C C D B B A A B C A Explanations to selected MC 1. Inertia depends on mass only, but not on velocity. 2. In horizontal direction, F = ma 80 cos 60o – f = 15 x 2 where f = friction f = 10 N 3. R – mg = ma (since R > mg) 550 – 45x10 = 45a => a = 2.22 m s-2 Acceleration is directed upwards because the net force is upward (since R > mg), so the lift is either moving upwards and accelerating at 2.22 m s-2 or moving downwards and decelerating at 2.22 m s-2. 5. Potential energy can be negative because of the choice of reference level. However, it is not a vector. That is, negative value does not mean opposite direction. 6. 0.5 x 70 v2 = 70 x 10 x 6 => v = 10.95 m s-1 7. F - mg sin - f = ma 10 – 0.8 sin 30o – f = 0.8 x 5 f=2N 8. Pt – energy loss = mcT 100 x 10 x 60 – 1000 = 2 c (30-20) c = 2950 J kg-1 oC-1 9. Statement 2, specific heat capacity of liquid X is larger than that of liquid Y because the slope of line corresponding to liquid X is less steep, indicating that X requires longer time to heat up to the same temperature. 10. Note that image height is double that of object height. So image distance should also be double of object distance. Hence lens position should be at g. 11. Consider reversibility of light. 13. f = 5 Hz, T = 1/f = 0.2 s => 0.55 s = 2.75 T 16. Closing switch K does not affect the current of another branch provided that the cell has negligible internal resistance. 17. The voltmeter, being very high resistance, will take up most of the battery voltage (voltmeter and resistor are connected in series.). 18. After S is closed, L1 goes out because there is short circuit across L1, so brightness decreases. P.d. across R = V/3 where V is the battery voltage So old power = (V/3)2 / R = V2 / 9R New power = V2 / R = 9 old power Original current = V (R/2 + R), new current = V / (R/2), new current > old current Hence power delivered by battery increases. 19. RA = V2 / P = 1102 / 20 = 605 RB = V2 / P = 2202 / 40 = 1210 = 2 RA (2) is incorrect because the power of A should be 80 W. (Since P is proportional to V 2) (3) is incorrect. Reason is trivial. 20. Power P (in kW) is given by P x 1800 = 85 Average cost = Pt x 0.9 = (85/1800) x 180 x 0.9 = $7.65 21. Recall opposite direction => repel 26. When a ball is thrown upwards, the net force remains unchanged --- mg. 28. The acceleration decreases but remains positive, indicating that the velocity increases but the rate of increase is decreasing. Finally the acceleration becomes zero and the object travels with constant velocity by Newton’s first law. So the answer is B. 30. Since the block is at rest, the net force acting on the block is zero. 31. mX = 2 mY By conservation of momentum, mX (6) = (mX + mY) v => v = 4 m/s 1 1 m X 6 2 m X mY 4 2 2 Percentage of energy loss = 2 33.3% 1 2 mX 6 2 32. For all the steam to condense completely, energy released = 2.26 x 106 m To melt all the ice completely, energy required = ml = 3.34 x 105 m To raise all the ice completely to 100oC, energy required = m (4200)(100) = 4.2 x 105 m To raise all the water at 50oC to 100oC, energy required = m (4200) (100-50) = 210000 m Since 334000 + 420000 + 210000 = 964000 < 2260000, so only part of the steam condenses and part of the steam remains at 100oC. 33. For A to melt completely, heat required = m x 900 x (50-20) + m x 220000 = 247000 m For B to melt completely, heat required = m x 500 x (60-20) + m x 160000 = 180000 m For C to melt completely, heat required = m x 100 x (70-20) + m x 200000 = 205000 m For D to melt completely, heat required = m x 300 x (80-20) + m x 180000 = 198000 m So the answer is B. 34. n = sin 60o / sin r = 1.44 (since sin r 7.5 10 7.52 2 0.6 ) 35. At z, angle of incidence = 90o – sin-1 0.6 = 53.1o Critical angle c = sin-1 (1/1.44) = 44.0o Since i > c and the light ray travels from optically denser medium to a less dense one, total internal reflection takes place. 36. Statement (1), normally speaking, bringing an alpha source will ionize the surrounding air molecules, and the opposite ions (opposite charges with the charges on the spheres) will neutralize the charged spheres. However, if the activity of the source is very low, only a few particles are emitted, the ions produced may not fully neutralize the charges on the sphere. For statement (3), the force acting on A by B must be equal to the force acting on B by A in magnitude because these two forces are an action and reaction pair. 37. The resultant magnetic field due to the upper two currents is directed downward by symmetry. Check this by using right hand grip rule and find the vector sum. The resultant magnetic field due to the lower two currents is directed upward by symmetry but the magnitude is smaller because the distance is larger. As a result, the total force is directed downwards. 38. Power output = 4 x 40 = 160 W Power input = IV = 0.8 x (10 x 22) = 176 W Hence efficiency = 160 / 176 = 90.9 % 39. P.d. across cable resistance = IR = 40 Power loss in cables = I2 R = 80 Combining, we have IR x I = 80 => 40 I = 80 Thus R = 40/I = 40/2 = 20 => I = 2A Voltage for transmitting power V is given by V I = P V (2) = 200 kW => V = 100 kV 40. Remember: Never SMOKE. 41. (1) Path difference = 10 cm = 2.5 . So destructive interference takes place at X. (2), the amplitude of wave X is zero only when there is complete cancellation. (3), if frequency is doubled, the wavelength is halved, i.e. = 2 cm, then path difference = 10 cm = 5 , so constructive interference will take place at X.