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nov 02 question 35 plzzzz Rahma Abdelrahman h4rriet plzzz
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nov 02 question 35 plzzzz Rahma Abdelrahman h4rriet plzzz
why does resistance have to increase? if voltage across x y decreases shouldnt it move near x as voltsge is propotional to length???You need to know that the potentiometer circuit shown in the question is the same as a circuit of 2 resistors in series. One resistor is the variable resistor & the other resistor is the wire of length XN.
If the resistance of the variable resistor increases, it will take more voltage and the XN resistor will take less voltage. So the voltage across XY decreases and the position of the moveable contact will have to be nearer to Y so that the resistance increases (because R is proportional to the length of the resistor).
why does resistance have to increase? if voltage across x y decreases shouldnt it move near x as voltsge is propotional to length???
so u mean both the resistors should have same resistance for zero deflection... is that like a rule cuz i dont remember studying it?The V across XY decreases when the R of the variable resistor increases. So, for XY to take more V, it should increases it's R, and to increase its R, it should increase its length.
If the contact moves closer to X, the voltage will decrease even more. You need to get back the old V to get zero deflection once again.
so u mean both the resistors should have same resistance for zero deflection... is that like a rule cuz i dont remember studying it?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_13.pdfhttp://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf Q13 Q22 Q29 Q33 Q37CAN ANYONE SOLVETHANKS
VERY HELPFUL THANKShttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_13.pdf
Q.2 first of all, you have to calculate the time taken to produce one complete wave which can be done by using the rate given(3000 revolution per minute)
my method:
3000 revo takes 60s
1 revo takes = 60/3000 = 0.02s
The question is asking which time base setting will separate the pulses on the screen. Notice that the screen of the c.r.o is 10cm long so we have to make full use of the screen. Which also means that your time base setting cannot be too far away from 0.02s(Calculated value) otherwise it will be very difficult to see the pulses.
(a) 1cm ---> 1s
10cm ---> 1 x 10=10s(Difference is too big! cannot show clearly on the screen)
Difference= 1 - 0.02 = 0.98s
(b) 1cm ---> 10ms
10cm ---> 10 x 10 = 100ms(0.1s) (Not too bad~ can be considered)
Difference= 0.1 - 0.02 = 0.08s
(c) 1cm ---> 100 x 10^-6s
10cm ---> 1 x 10^-3 s (THIS IS SMALLER THAN 0.02s! Cannot even show one complete pulse!)
(d) 1cm ---> 1 x 10^-6
10cm ---> 10 x 1 x 10^-6 = 0.00001s (Cannot even show one complete wave)
I believed the best answer will be B.
Q.17. The question is asking what is the rate does the energy provided to the system by the motor. So this shows that the motor is actually supplying some of its power. You must also know that the mass of the elevator is greater that the mass of the heavy weight because the question already stated "the elevator is partly counterbalanced by a heavy weight." So (m1 > m2) but if that is the case then how can the elevator move up? Yes. The motor and the heavy weight(m2) are providing the energy required to move the elevator up. So I came up with a formula: Power by m2 + power by motor = Power required to move m1. And we will have to use a formula P=FV where F can be the weight and V can be the velocity/speed. So substitute this formula into the earlier formula and we will get: m2gv + Power by motor = m1gv. Rearrange the formula: Power by motor = m1gv - m2gv. Simplify further and we will get: Power by motor = (m1 - m2)gv
Q.18. All calculations. First thing you have to do is to find the resistance of the relay(voltage=16v and current=0.6A) so resistance=16/0.6 = 26.66666(80/3)Ohm. Next we will look at the second scenario where we have two wires being connected to the power supply and also the relay now. So we are given resistance of wire in 1m (0.0050Ohm) and we also know that the distance between the relay and the power supply is 800m so we have to calculate the new resistance of the wire: New resistance = 0.0050 x 800 = 4Ohm. We are not finished yet~ Since we have two of these wires so the resistance has to be multiplied by 2 (4 x 2 = 8Ohm) so overall resistance of 2 wires will be 8Ohm. In order to solve this question you have to imagine it's a series circuit so the current will still be 0.60A. Total resistance in the circuit = Resistance in the 2 wires + Resistance of the relay( (80/3) + 8 = 104/3Ohm) and finally we can use the formula R=V/I to determine the output e.m.f where your R=104/3Ohm and I=0.6A. So final calculation: V= (104/3) x 0.6 = 20.8V) and the answer is C.
Q.35. This is another tricky question. The voltmeter reading will be constant. Why? Because we are not measuring the voltage of the potentiometer P only but we are measuring the voltage for the whole circuit(see the connection of the voltmeter) so even though you slide the potentiometer along x to y it will not have any effect. However, if we are measuring the voltage of the potentiometer only, then the voltage will change when you slide the slider along x to y. So the answer will be A.
Phew~ I hope it's not too complicatedand reply if you have any doubts. Thanks~
VERY HELPFUL THANKS
CAN U SOLVE PLEASE
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf Q13 Q27
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf Q13 Q22 Q29 Q33 Q37
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf Q18 Q29
CAN ANYONE SOLVE
THANKS
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_11.pdfVERY HELPFUL THANKS
CAN U SOLVE PLEASE
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_11.pdf Q13 Q27
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_12.pdf Q13 Q22 Q29 Q33 Q37
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_11.pdf Q18 Q29
CAN ANYONE SOLVE
THANKS
can anyone please help me solve and understand Q 10,12,21 and 40 of s_07 qp_01?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdfcan anyone please help me solve and understand Q 10,12,21 and 40 of s_07 qp_01?
ummmm... regarding the question 33, I think we shared a different opinion.13. You can make a closed triangle of sides 3 cm, 4 cm and 5 cm. Forces in equilibrium can be represented by a closed triangle.
27. WD=qV=the same for both.
13. The acceleration of the box = the acceleration of the 2 kg mass. You must consider them to be one thing. Act like the whole system is one 10 kg mass. 2x9.81 acts downwards, 6 N acts upwards. a=F/m.
22. Volume of one molecule = mass given / density given. Cube root of answer = molecular spacing.
29. v=f x lambda; v=1/t x lambda, so t = lambda/v. Number of wavefronts from XY to P = 3.
33. A is wrong; charged particles are not provided by the supply. B is wrong, current is not a measure of the speed of electrons; it is the number of coulombs passing per second; C is correct, a larger diameter means a smaller resistance; D is wrong, a smaller diameter means a larger resistance.
37. You can redraw the circuits so that it becomes clear, because the way they deliberately drew it so that it is confusing:
18.
29. A stationary wave in a closed end tube always has an antinode at the open end, so the fundamental mode will be 1/4th of a lambda. A stationary wave in an open end tube too always has an antinode, this time at both ends, so the fundamental mode will be 1/2 a lambda. For tube P: If 1 lambda = 20cm, 1/4 lambda = x. x = 5. That's a multiple of 35, so stationary waves can be formed in P. For tube Q: If 1 lambda = 20cm, 1/2 lambda = x. x = 10. That's a multiple of 50, so stationary waves can be formed in Q.
ummmm... regarding the question 33, I think we shared a different opinion.
(a)Will be wrong because the speed of the charged particle not only depends on the e.m.f of the power supply but also the potential difference across the wire as well.
(b) Wrong because the speed of the charged particle does not depends on the current flow.
(c) Larger diameter means lower resistance so the potential difference across the wire will also be lower and since the speed of the charged particle depends on the potential difference, the charged particle will move slower because of lower P.D.
(d) Smaller diameter will result in greater resistance which will later gives bigger P.D across the wire. The bigger the P.D the faster the charged particle will move.
My answer is D.
I know it's rude but I just want to know more. So kindly explain to me your thoughts please???~~~
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s07_qp_1.pdf
Q.10. The formula that links momentum with force will be F x (Change in t) = (Change in momentum). So first we will determine the change in momentum. Change in momentum= P1 - P2 (Not P2 - P1 because if you look at the answers, all are P1-P2 and P1+P2 so we have no choice but to use P1 - P2). Change in time is simple which can be done by simply minus t2 by t1(t2 - t1) so the average force will be: Ave. force = (p1 - p2)/(t2 - t1). So answer will be B.
Q.12. During this collision, the momentum of the system must be conserved(As stated by the principle of conservation of linear momentum) so we will use the formula m1u1 + m2u2 = m1v1 + m2v2. By performing some calculations:
(2)(4) + (4)(1) = (4+2)(Velocity after the collision) *Note that I add the mass of both trolley together because they move off together after collision.
So velocity after the collision will be 2ms^-1.
To find K.E, use the formula: 1/2mv^2 and do calculation: 1/2(4+2)(2)^2 = 12J so final answer is B.
Q.21. This is simple. The only thing that you have to know is only transverse wave can be polarised.(That's what required in our syllabus) (a) wrong since it's a longitudinal wave. (b) Still longitudinal so wrong. (c) Correct because it's a transverse wave. (d) Sound wave cannot be a transverse wave so answer will be C.
Q.40. Question is asking which of the particular nuclei will travel at the lowest speed so it must have something to do with K.E. What I did was determine the number of protons and also the mass number in the nuclei. Mass number is labelled on the top part of the nuclei where as the proton number is labelled at the bottom part. My results (u = atomic mass constant(1.66x10^-27) given in exam)
(a) mass number = 1u
proton number = 1
(b) mass number = 4u
proton number = 2
(c) mass number = 7u
proton number = 3
(d) mass number = 9u
proton number = 4
Now I will introduce one infamous formula which is: 1/2mv^2=eV (eV=electronvolt). Why do we have to use this weird looking formula? Because this question involved the potential difference. The calculation part was quite lengthy so..
(* I will not show the V here because potential difference is the same anyway and e here is the elementary charge(1.6 x 10^-19))
(a) 1/2 (1u)v^2 = e
1/2(1.66x10^-27)v^2 = 1.6 x 10^-19
v^2 = 192771084..
v= 13884.20ms^-1
(b)1/2(4u)v^2 = 2e (*need to multiply by 2 because of 2 protons)
1/2(4(1.66x10^-27)v^2 = 2(1.6x10^-19)
v^2 = 96385542.1
v= 9817.61ms^-1
(c)1/2(7u)v^2 = 3e
1/2(7(1.66x10^-27)v^2 = 3(1.6x10^-19)
v^2 = 82616179
v=9089.344..ms^-1
(d)1/2(9u)v^2 = 4e
1/2(9(1.66 x 10^-27))v^2 = 4(1.6 x 10^-19)
v^2 = 85676037..
v=9256.135ms^-1
Phew~ Overall the answer should be C.
I know it's hard~ But take a deep breath and chill yourself. You can do it~
10. Force = rate of change of momentum, which is p/t. The gradient of a p/t graph is the force. To find the gradient, we use y2-y1/x2-x1.
12. The collision is inelastic. The initial momentum = 12. The final too must = 12. The final mass = 6. The final v=p/m=12/6=2. The final k.e.=1/2x6x(2^2).
21. A and B is ruled out, because only transverse waves can be polarised. D is wrong because sound waves are not transverse. That leaves C.
40. a=F/m. F=Eq. E is the same for all; q & m are different. For A q is 1, B 2, C 3 and D 4. So acceleration for A=1E/1, for B=2E/4=1/2E, for C=3E/7 and D=4E/9.
nov 02 question 35 plzzzz Rahma Abdelrahman h4rriet plzzz
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