Physics: Post your doubts here!

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_21.pdf

Question 4 b (iii) how???

strange question okay here goes remember that maxima and minima are formed at alternating points for e.g if n=1 is maxima then n=3,5,7 can only be used for the n values the angle and grating are the same so dsinO is a constant and nλ should be equal to it so just put in the n values in that and find the corresponding wavelenght for this question n=3 gives 417nm so 420nm to 2 s.f hope this helped

 

[Thought blocker]

strange question okay here goes remember that maxima and minima are formed at alternating points for e.g if n=1 is maxima then n=3,5,7 can only be used for the n values the angle and grating are the same so dsinO is a constant and nλ should be equal to it so just put in the n values in that and find the corresponding wavelenght for this question n=3 gives 417nm so 420nm to 2 s.f hope this helped

Elaborate a bit more, and have a gaps PLEASE

 

[sameen saeed]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_23.pdf Can anyone help me with question no. 4 b(ii).

 

[DeViL gURl B)]

strange question okay here goes remember that maxima and minima are formed at alternating points for e.g if n=1 is maxima then n=3,5,7 can only be used for the n values the angle and grating are the same so dsinO is a constant and nλ should be equal to it so just put in the n values in that and find the corresponding wavelenght for this question n=3 gives 417nm so 420nm to 2 s.f hope this helped

Thank u .. But a bit of more elaboration would be just perfect! Please
Thank u

 

[MiniSacBall]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_23.pdf Can anyone help me with question no. 4 b(ii).

I was stuck in this for a while because i forgot that i was doing physics, i was using mechanics xD

Well you can use two ways, 2gs=v^2-u^2, or K.E lost = P.E gained (Since air resistance is neglected).
Well in the case of 2gh=v^2 - u^2 , here you can use the values of g = -9.8 [because it is going upwards], v= 0 [because at max height the velocity is 0] , u=16 [given]
And easily get the value of h,

Or
Using K.E = P.E -------- K.E = 8.32, [solved in last question] , and P.E = mgh, with their respective values. g = 9.8 , m = 0.065, Solve and you will get the answer.

 

[MiniSacBall]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s05_qp_2.pdf

q7 b iii
Explain please!

 

[NinjaInPyjamas]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_22.pdf 6 (b)

 

[TheJDOG]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_2.pdf

q7 b iii
Explain please!

Damn thought it was Part ii

iii)
E=IR + Ir
r= internal resistance.

Find Current, V/R= 1.5/2 =0.75 (resistance from previous question.

3=0.75(2) + 0.75r

r= 2 ohms

 

[sameen saeed]

I was stuck in this for a while because i forgot that i was doing physics, i was using mechanics xD

Well you can use two ways, 2gs=v^2-u^2, or K.E lost = P.E gained (Since air resistance is neglected).
Well in the case of 2gh=v^2 - u^2 , here you can use the values of g = -9.8 [because it is going upwards], v= 0 [because at max height the velocity is 0] , u=16 [given]
And easily get the value of h,

Or
Using K.E = P.E -------- K.E = 8.32, [solved in last question] , and P.E = mgh, with their respective values. g = 9.8 , m = 0.065, Solve and you will get the answer.

sorry but i asked for b(ii) part ,the question which asked for ratio of P.E and K.E

 

[_Ahmad]

Thank u .. But a bit of more elaboration would be just perfect! Please
Thank u

since "d" and sin(θ) are constant we can answer this question only by substituting values in "n" an "λ"
In part ii) we have n=2 and λ=625 nm so we can find the value of constant nλ which will help us to answer this question
nλ= 2*625=1250

Now lets put different values of n,
when n=1 we get corresponding wavelength as 1250nm
when n=3, λ=417
when n=4, λ=312.5

Since the wavelength of visible light is from 7*10^-7 to 4*10^-7 m
the only answer in range is 417nm (420nm to 2d.p)

Hope it helps

 

[Haya Ahmed]

Thank u .. But a bit of more elaboration would be just perfect! Please
Thank u

Did you get Q1 part b (iii) in the same paper ? and If yes can you explain ! .. Thanks

 

[Haya Ahmed]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_21.pdf

Q1b (iii)(part1 and 2) and b (iv)
Q2 b (i) (iii)

Please someone help . Thanks !

 

[AhsanAfzal]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_21.pdf
Q no. 5 (d) can anyone explain it to me?

 

[_Ahmad]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_21.pdf
Q no. 5 (d) can anyone explain it to me?

i) Narrower the slits, because wavelength of light is less than microwaves
ii) decrease the distance between the slits, which will increase the fringe width and hence fringes will be more clear.

 

[_Ahmad]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_21.pdf

Q1b (iii)(part1 and 2) and b (iv)
Q2 b (i) (iii)

Please someone help . Thanks !

for q2 bi)

 

[MiniSacBall]

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

 

[Haya Ahmed]

for q2 bi)

Thanks what about the rest of the questions !

 

[princeali97]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_21.pdf

Q1b (iii)(part1 and 2) and b (iv)
Q2 b (i) (iii)

Please someone help . Thanks !

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.

 

[NinjaInPyjamas]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_23.pdf
6 (b) can someone please explain the formulas!

 

[princeali97]

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

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.
Your welcome and Best of luck.