If velocity of A relative to ground = velocity of B relative to ground = the velocity of A relative to B =

Transcription

1 L Physis MC nswers Year:1989 Question Number: 3,0,,4,6,9,30,31,36,40,4 1989MC (3) If eloity of relatie to ground = and eloity of relatie to ground =, then the eloity of relatie to = X X Y Y Suppose that X is stationary and Y moes to the right. So relatie to Y, X appears to moe to the left. In addition, X moes upwards, so X's eloity is to the "left and up", as seen by Y. 1989MC (0) rotating dis is seen to be stationary when Flashing rate = freq of the rotating dis (one reolution between two flashes) Flashing rate = (freq of rotating dis)/ (two reolutions between two flashes) Flashing rate = (freq of rotating dis)/3 (three reolutions between two flashes) so on. rotating dis appears to moe bakward slowly when the rotating freq of the dis is slightly lower than one of the freezing frequenies, e.g. First flash 0 0 Seond flash (not omplete a full reolution) Third flash First sight is at 0 0, seond sight is at 0, third sight is at 4 0. so it appears to moe bakwards. Page 1

2 Now, the flashing rate is 50 Hz. Dis rotating at 50 Hz, 100 Hz, 150 Hz,..an be frozen by the flash. The answer in 1989() is 98 Hz, beause it is slightly lower than 100 Hz MC () dsin = m 10 3 p lines per mm, is the angle when m =, so sin (1) p 3p lines per mm, when m=1 and the aelength is 5/4, so sin = 1( ) () 3p 4 Diide (1) by () 3sinsin = 8/5 or sin = (15sin/8 1989MC (4) u The moing objet reeie wae of freq f ' f The wae is refleted, the wall now ats a moing soure. u stationary obserer detet f '' f ' f u u The aboe formulae are not stritly orret for EM waes (radar - mirowaes) eause u << The apparent waelength /f'' = u u f u 1 u MC (6) 1 u u 1 1 u u( ) f u u u spherial ondutor (essentially any type of ondutor) an be regarded as a apaitor. apaitor onsists of two ondutors, but a sphere doesn t. How ome? Page

3 No wonder. The other ondutor is infinity. To any system, Q is proportional to V and hene the proportionality onstant is defined as the apaitane (Q = CV). Q If the potential at infinity is zero, then the potential at the spherial surfae is V =, where a is 4 a its radius. V is in fat the p.d. between the onduting sphere and infinity. Therefore the apaitane of a onduting sphere is C = The energy stored in a apaitor is Q V Q QV CV. C 1 So the energy stored in a harged sphere is (4 0 a) V ( 0a) V This is the total work in bringing all the harges from infinity to the surfae of the sphere. a 1989MC (9) D R C Q S P Current enters pass through P ( to )------exit at, or urrent enters pass through Q ( to D)------path D -----exit at, or urrent enters -----path C -----pass through S (C to )------exit at, or Current enters -----path C ----pass through R (C to D)------exits at. In other words, a urrent entering at will be diided into four parts and eah will pass through P, Q, R or S, they join and leae at. So, the four resistors are effetiely joined in PRLLEL. 1989MC (30) F= qe. q is negatie, so F is opposite to E. F is upward, so aeleration is also upwards.. Page 3

4 Path III and IV are impossible. t their "highest points", the ertial eloity is zero. n upward aeleration will not produe a downward eloity at a later time! 1989MC(31) X is situated at the mid-point of, R X = 5, Total resistane between X is 10/3 and the resistane between X is 5 X and X are is series, the "10V" is diided into two parts in the ratio 10/3 : 5 Final p.d. aross X = 10 ( / ) = 4V 1989MC (36) Without going into details, we will say immediately that the fore ating on at the moment when the swith is losed must be repulsie. Why? efore the swith is losed; no flux passing through. The flux now floods into, what is the response of? It "runs away" in order to restore the "zero field" state Lenz's law. The fore is repulsie when the urrents in two neighboring straight wires are opposite in diretion. The urrent in the red wire is upward, so the urrent in the blue wire must be downwards. So the urrent in is antilokwise. a rat [The field at produed by the rising urrent in is downward So the field produed by the indued urrent is upwards. Suh a urrent must be antilokwise] Page 4

5 1989MC (40) It is a ery diffiult question. I just explain why "D" is the orret answer. Y-input earth R R 1 I I E X-input We are going to plot I C against I. CRO measures oltage, not urrent, so we need to find oltages whih are proportional to I C and I.. The major diffiulty is to find a suitable ommon ground The transistor onduts only when the urrents flows as shown in the aboe figure. X-input measure the p.d. aross R 1, whih is proportional to I. "X-input" is positie relatie to "earth", so the trae appears on the right of the enter of the sreen (positie x). Y-input measures the p.d. aross R 1 and R. Sine I E >> I, so the p.d. aross R 1 an be negleted, as ompared with that aross R. Y-input essentially measures the p.d. aross R, whih is proportional to I E. I E is approximately equal to I C. So the signal fed into Y-input is basially the same waeform as I C. Y -input is positie with respet to earth, so the signal is displayed on the side of positie y. Why are there so many approximations? The answer is that we annot find a ommon ground better than that. Page 5

6 1989MC(4) bsorption spetrum n atom an absorb a photon of energy whose alue is just the right amount to raise the atom to a higher energy leel. When white light, whih ontains all waelengths, is passed through a ooler gas or apour, photons of those energy that orrespond to transitions between energy leels are absorbed. The resulting exited atoms re-radiate their exitation energy almost at one, but these photons ome off in random diretions with only a few in the same diretion as the original beam of white light. This auses dark lines in the spetrum of the original beam. The dark lines are not ompletely dark. They appear dark by ontrast with the bright bakground. The absorption spetrum of any element is idential with its emission spetrum (1) (3) Iodine apour absorbs the light whose frequenies will be emitted by it. () ll the absorbed energy will be emitted (in all diretions) 1989MC (44) 6V = state 1 0V = state 0 From the oltage graphs, the truth table of the logi gate is obtained: IN1 IN IN 1 output IN output Page 6

7 s ompared with the standard logi gates: OR gate ND NOR gate NND Obiously, the logi gate is a NND gate. (nswer E) Page 7