Physics: Post your doubts here!

[Ikram Khaliq]

which physical state has the greatest potential energy?

 

[sma786]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_11.pdf
Question 10 and 13 please?

 

[Yousif Mukkhtar]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf
Question 10 and 13 please?

Q10: {INITIAL=mu) [FINAL= -mu ( because it moving in opposite direction)]
Change in momentum = Final momentum-Initial Moment
=-mu-mu
=-2mu

Q13
Anticlockwise moments= Clockwise moments
20 x 9.81 x60= 100x 10x 9.81 + 50x9.81 x
11772-9810=50x9.81x
x=4 cm clockwise
40+4=44 cm

 

[Yousif Mukkhtar]

Can some one explain these please q12
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_11.pdf

Q20
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w12_qp_11.pdf

 

[sma786]

Q10: {INITIAL=mu) [FINAL= -mu ( because it moving in opposite direction)]
Change in momentum = Final momentum-Initial Moment
=-mu-mu
=-2mu

Q13
Anticlockwise moments= Clockwise moments
20 x 9.81 x60= 100x 10x 9.81 + 50x9.81 x
11772-9810=50x9.81x
x=4 cm clockwise
40+4=44 cm

Thanks !

 

[sma786]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_11.pdf
question 30. how do we find the mass?

 

[sma786]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_11.pdf
question 31. why C?

 

[Yousif Mukkhtar]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf
question 31. why C?

q30

F=qV/d
F=(1.6x10^-19 x 12 x10^3)/25 x10^-3
F=7.7 x 10^-14 N

q31
10 A pass each through each cross section area as given
So 10 A means 10 C per second. So to find electrons.
no of electrons=10 C/1.6 x10^-19
n=6.3x10^19 electrons

 

[Monojit Saha]

Can some one explain these please q12
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_11.pdf

Q20
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_11.pdf

12: Its more of a mechanics question.. First find the acceleration from the two equations. . i- T-800= 8a and
ii) 1200-0T=1200a .. u get a= 2ms-2.. we kno that for a pulley,, the two components travels the same distance and with teh same speed.. so the two bodies will meet half way between 18 m, thus 9 m. So, use v2=u2+2as for anonw of the bodies. taking the s as 9 .. u get the v as 6 ms-1.

20: its simple.. just find the K E developed i.e 1/2 mv2

 

[Usman04]

hi guys i have many problems to discuss i am writing all the problems below

firsly i have seen many mcqs about how many maximas minimas or bright fringes or observed in differaction grating and number of line per mm is D or N. i remeber one mcq in past paper plzz solve it oct/nov 2012 paper 12 que 30

second problem mcqs 13 from oct/2012
oct/nov 2011 paper variant 12 q26,31 why c is not write in q 31
oct/nov 2011 variant 11 q20 may june 2011 varient 11 q 27

 

[syed1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_11.pdf
question 30. how do we find the mass?

Hey how are ya?

Mass? I think you have linked the wrong paper :\

E=V/d
E=4.8x10^5 NC^-1

The Value of Charge on an electron is .. 1.6 * 1o^-19 C .. from the data booklet..

E=F/Q
F=EQ
F=4.8x10^5 * 1.6*10^-19
F=7.68*10^-14 N

So answer is D.

 

[Nab900]

Assalam O Alaikum,
Can someone please explain q 36 of s12_qp_13
Why is the answer C
Why isnt the answer D
and q 6 of w12_qp_11
please explain these two questions as deeply as you can





Thanks in advance

 

[white rose]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_12.pdf
can anyone explain me q. 15, 16 & 26...ur help is highly appreciated..

 

[Manobilly]

May june2003 q40

 

[gary221]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_12.pdf
can anyone explain me q. 15, 16 & 26...ur help is highly appreciated..

15) When the force = F, and distance = s, work done = Fs.
Since work done = K.E. gain → Fs = 8-4 = 4J
Fs = 4J

So, when force = 2F, and distance = 2s, work done = 2F * 2s →4Fs.
Since Fs = 4J, 4Fs = 4 * 4 = 16 J.
So gain in K.E. = 16 J.
And since the kinetic energy already equals 4 J....the new K.E. = 4 + 16 → 20 J
Ans = B

Hope u get it!
All credit to Nibz.

 

[gary221]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_12.pdf
can anyone explain me q. 15, 16 & 26...ur help is highly appreciated..

16) As i already said, work done = amount of energy gained.
So, since the K.E. of the vehicle = 4.5 * 10^5 J
and we know that work done = Fs.
So, Fs = 4.5 * 10^5
s = (4.5 * 10^5)/6000
s = 75 m.

Ignore the mass, thts just to confuse u.
Hope u gt it!
All credit to Nibz.

 

[Manobilly]

M/j 2007 q13.

 

[gary221]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_12.pdf
can anyone explain me q. 15, 16 & 26...ur help is highly appreciated..

26) As shown in the fig, at time = 18 s, phase diff btw the 2 waves = 180° or π.
1 wavelength = 360°
So, 1/8 of a wavelength = 360°/8 → 45°.

Phase diff = 180° at 18 s.
So, phase diff = 45° at x.
x = (18 * 45)/180
x = 4.5 s

Ans = B

Hope u get it!
All credit to Nibz.

 

[gary221]

M/j 2007 q13.

ppr link pls?

 

[Manobilly]

ppr link pls?

Can't upload the paper ,here is th question though.