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Physics: Post your doubts here!
- Thread starter XPFMember
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for q2 bi)
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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_qp_2.pdf
Q7 a)
My only doubt is when S1 was closed the resistance was 15 ohm, so it means that A had a resistance of 15 ohm, isn't it. Then when S2 only was closed then resistance was 30 ohm. So it means that B has resistance of 15 ohm, because total resistance is 30 ohm, so 15 +15. And then when S2 and S3 were closed the resistance is 15 ohm, how come, Isn't it supposed to be 30 ohm. i mean like if C was not faulty then the resistance should have been 30+ , but if it is faulty then should be 30 ohm. Help me please.
And b)
Is lamp a also shorted, it says in mark scheme
I have never understood electricity part, of physics. I am always confused until today, hope fully, some day i will know everything, i just don't understand voltage and current itself. I just know they are some random I and V. And some formulas linking them. I just don't have any concept in it.
Help me please,
Thanks & God Bless You
Q7 a)
My only doubt is when S1 was closed the resistance was 15 ohm, so it means that A had a resistance of 15 ohm, isn't it. Then when S2 only was closed then resistance was 30 ohm. So it means that B has resistance of 15 ohm, because total resistance is 30 ohm, so 15 +15. And then when S2 and S3 were closed the resistance is 15 ohm, how come, Isn't it supposed to be 30 ohm. i mean like if C was not faulty then the resistance should have been 30+ , but if it is faulty then should be 30 ohm. Help me please.
And b)
Is lamp a also shorted, it says in mark scheme
I have never understood electricity part, of physics. I am always confused until today, hope fully, some day i will know everything, i just don't understand voltage and current itself. I just know they are some random I and V. And some formulas linking them. I just don't have any concept in it.
Help me please
,Thanks & God Bless You
Last edited:
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Q1 b(iii) 1. From the graph u can see that the speed starts decrease from C.This means that frictional force has increased.Previously(before C) the speed was constant and the parachutist was at terminal velocity which means that frictional force was equal to weight.
2.From C to E u can see that speed first decreases at a constant rate i.e constant deceleration and then it decreases non-uniformly until its constant.So Since F=ma.Acceleration was constant for the part CD so frictional force will be constant.Then for part DE u can see that the rate of decrease i.e deceleration decreases.So frictional force decreases.
iv) 1.Average acceleration=total change in velocity/Total time. So from graph u can see that at t=17 , v=20 and t=50,v=50.So 20-50/2 will give u -15ms-2.
2.So u use newtons 2nd law here. W-F=ma (u can see that weight is greater than frictional force which opposes the motion)
95g - F=ma
F=2357N
2bi (iii) R=1/Gradient of graph.this is the formula.Taking the final and initial points we have : gradient=(0.275-0.05)/5=0.045.
R=1/0.045=22.2ohms.
If u have any problems then please ask me.
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http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_23.pdf
6 (b) can someone please explain the formulas!
6 (b) can someone please explain the formulas!
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Hey calm down brother.Its an easy concept and yes i sometimes too find the electricity part a bit irritating.But heres the logic in this one.You just didnt read the question properly.It says "three similar lamps A, B and C".So all of them should have 15ohm.But u can see that when S1 and S2 are closed the resistance is 30 ohm which is wrong it should have been 22.5 ohm as B and C are in parallel.But its coming greater.So fault is in lamp C becuz other lamps are showing 15ohms resistance as u already figured out.http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_qp_2.pdf
Q7 a)
My only doubt is when S1 was closed the resistance was 15 ohm, so it means that A had a resistance of 15 ohm, isn't it. Then when S2 only was closed then resistance was 30 ohm. So it means that B has resistance of 15 ohm, because total resistance is 30 ohm, so 15 +15. And then when S2 and S3 were closed the resistance is 15 ohm, how come, Isn't it supposed to be 30 ohm. i mean like if C was not faulty then the resistance should have been 30+ , but if it is faulty then should be 30 ohm. Help me please.
And b)
Is lamp a also shorted, it says in mark scheme
I have never understood electricity part, of physics. I am always confused until today, hope fully, some day i will know everything, i just don't understand voltage and current itself. I just know they are some random I and V. And some formulas linking them. I just don't have any concept in it.
Help me please
Thanks & God Bless You
Your welcome and Best of luck.
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Formula to be used, is : d sin theta = n times lambda...
Where d , n and lambda stands for Distance b/w adjacent lines of grating, order of maximum and wavelength respectively.
Hope you now understood the mark scheme
(Y).
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_22.pdf
can anyone please explain Q3 b ii)
can anyone please explain Q3 b ii)
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That helps a lot, thanks
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Tension = Force
Rule
for calculating the force down the slope is Wcos(teta)
for calculating the force above the slope is WSin(teta)
so the tension in the wire is up the slope so you'll use the formula
Force(tension) = WSin(teta) = W=weight = (450 x 9.81 x sin(12) )
= 917.8+640(as frictional force)
=1570
Hope I helped
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Constant velocity means that the upward and downward forces on the slope are equal. The two downward forces are weight, mg sin theeta, which always acts down the slope, and friction, which would now act in the opposite direction to the motion of the log. Thus the two forces down the slope would equal the tension which are mg sin theeta + 650
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Well S1 and S2 are not closed together in the question. I think you meant S2 closed only. When S2 was closed, it gave us the correct reading of 30 ohm, as both the lamps are in series. However, when S2 and S3 are closed, it should have given us 22.5 as you stated. However, it gave us 15 ohm only. That is why the problem is in lamp 3. I think you just confused the switches with one another while explaining it to himHey calm down brother.Its an easy concept and yes i sometimes too find the electricity part a bit irritating.But heres the logic in this one.You just didnt read the question properly.It says "three similar lamps A, B and C".So all of them should have 15ohm.But u can see that when S1 and S2 are closed the resistance is 30 ohm which is wrong it should have been 22.5 ohm as B and C are in parallel.But its coming greater.So fault is in lamp C becuz other lamps are showing 15ohms resistance as u already figured out.
Your welcome and Best of luck.
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http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_21.pdf
can someone please draw!
can someone please draw!
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I don't get how suddenly sin theta/n lambda! :S
Can you please explain the difference between this question and Q2 b i) from http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_23.pdf
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For calculating the maximum number of orders, use d sin 90 = n lamda.... We use 90 (from what i can understand) because that is the maximum possible angle at which the light can diffract. However, it would give you a value in decimals, and you will round it off to the previous value.
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But that is the question i explained to you
s
so can you please tell the difference between Q3 b ii) from http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_22.pdf
and Q2 b i) from http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_23.pdf
Sorry, but the question you explain was from variant 22 of the same yearBut that is the question i explained to you
so can you please tell the difference between Q3 b ii) from http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_22.pdf
and Q2 b i) from http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_23.pdf
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Oops sorry. I just saw the diagram and thought it was the same question.
The difference here is that the block is accelerating. Thus the force up the slope would be greater than the two forces down the slope, unlike the previous question where they were equal. Use Newton's second law
F - 525 - mgsin theeta = ma
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Yes oh im sorry.I wrote the wrong lamp number.My apologies!