Physics: Post your doubts here!

[Thought blocker]

Why not .. Of course YES !

Will upload it by 9PM

 

[Thought blocker]

HEre you go, its a shorter, I have long one, will uploadit later

 

[DeViL gURl B)]

Okii

busy, will upload 'em later

okkki thank you

 

[DeViL gURl B)]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_21.pdf

Question 4 b (iii) how???

 

[Haya Ahmed]

HEre you go, its a shorter, I have long one, will uploadit later

Thanks A LOT !! .. God Blessings ! ^^ .. Inbox me once you uploaded them aryt ?

 

[Thought blocker]

Thanks A LOT !! .. God Blessings ! ^^ .. Inbox me once you uploaded them aryt ?

Sure

 

[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.