Physics: Post your doubts here!

[usama321]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_21.pdf

How to solve question Q4 c ii

look u calculated the speed above (i got 15.49)
and the electrical energy was 25% of the kinetic energy ( thats what i understood :/)
P=E/T
P=0.25(o.5*m*v^2)/1 (it said mass per second i.e. one second)
P=0.125*m*(15.49)^2
11000/0.125*240 =m
m=1100/3 Kg s^-1
m not sure btw :/
usama321 ??

The method is correct. However, you get 15.344m/s in the first part, and if you use that whole value in your second part, you will get the correct answer. The concept here is that 110000W means 110000J is being produced per second. Thus, we equate it to energy, and we will get the mass flowing in one second

 

[kitkat <3 :P]

The method is correct. However, you get 15.344m/s in the first part, and if you use that whole value in your second part, you will get the correct answer. The concept here is that 110000W means 110000J is being produced per second. Thus, we equate it to energy, and we will get the mass flowing in one second

thank u

 

[usama321]

Question! In this paper (http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s12_qp_22.pdf) Q6c (ii), the answer in the marking scheme is 490 - 500 Hz. This is as a result of multiplying the distance between two consecutive antinodes or alternate nodes (which is 0.34 m) by 2 (giving 0.68 m) and taking that value as the wavelength. My question is, why do they multiply it by 2? The stationary wave produced isn't the first harmonic, so why do we multiply it by 2?

I don't know what harmonic is However, distance between two nodes or antinodes always equals to half of a wavelength from what i have read so far. That is why it is multiplied by 2.

 

[Jigar]

I was viewing the Xtremepapers, Physics Revision page!!! I was wondering, is ' LASER AND SEMICONDUCTORS' for As level or A2 Level!!!!! Please, help

 

[DeViL gURl B)]

papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s08_qp_2.pdf

Question 2 (b)
Part one as well as 2 ?
How ??
usama321 please help

 

[DeViL gURl B)]

DeViL gURl B)
For the b part, while the velocity is positive, the ball is going upwards. When it is negative, it is coming downwards. Use area under the graph 1/2 (2.4) * 9 - 1/2 1.6 * 6

For c part, the safest to do this question would be to form an equation of the line and find the co-ordinates at time 3.5s. However, in the mark scheme, they have written (allow 4.2 +- 0.2) thus 4 would also be accepted

Could u please explain y at 3.5 seconds? I mean any specific reason?
And thank u for the first part

 

[DeViL gURl B)]

usama321 okhay okhay I got it why 3.5 cuz they've mentioned in the question .. -___-
Sorry didn't read the question properly ..
Thank u

 

[usama321]

papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s08_qp_2.pdf

Question 2 (b)
Part one as well as 2 ?
How ??
usama321 please help

Total momentum before the spring is released is zero. Thus total momentum after releasing would also be zero. Conservation of momentum
.8 *v1 = 2.4* v2
v1/v2 = 3/1

part 2.
1/2 * .8 *v^2 + 1/2 * 2.4 * (1/3v)^2 = .49
Solve it and you will get the answer

 

[DeViL gURl B)]

DeViL gURl B)
For the b part, while the velocity is positive, the ball is going upwards. When it is negative, it is coming downwards. Use area under the graph 1/2 (2.4) * 9 - 1/2 1.6 * 6

For c part, the safest to do this question would be to form an equation of the line and find the co-ordinates at time 3.5s. However, in the mark scheme, they have written (allow 4.2 +- 0.2) thus 4 would also be accepted

I'm sorry but I tried doing the first part ..
The thing I don't get is .. Why don't u take the ENTIRE area under the graph ... As in we took .5 *9*2.4 okhay understood ..
Then why not L *B ?
Y do we take 1/2 *6*1.6 only?
And lastly, why do we subtract them ? :$

 

[Suchal Riaz]

at maximum point, the crest and crest meet: so displacement = 1.4+2=3.4 units
at minumum point crest and trough meet so displacement = 2-1.4=0.6
I is directly proportional to a²
so ratio of I = 3.4²:0.6² = 32.1
Snow Angel

 

[DeViL gURl B)]

Guys can anyone explain .. In what happens to the fringes when the intensity is increased or decreased ... Explanation needed! Detailed !
Please..
Thank u

 

[Mohammed salik]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_22.pdf
Q2 ) b) Its related to graph and M.S Only Says Resonable shape ! Could anyone Help me Out !
Tnx in Advance!

 

[Wolfgangs]

http://papers.xtremepapers.com/CIE
/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w12_qp_23.pdf

Can someone explain the first part of Q2 please?

 

[princeali97]

[URL='https://www.xtremepapers.com/community/members/mohammed-salik.36490/']Mohammed salik[/URL]
Dont wanna explain.Yahin samjh jaogay

 

[Mohammed salik]

Gr8..! I actually Only needed this !
Thnx Man

Mohammed salik
Dont wanna explain.Yahin samjh jaogay

 

[usama321]

I'm sorry but I tried doing the first part ..
The thing I don't get is .. Why don't u take the ENTIRE area under the graph ... As in we took .5 *9*2.4 okhay understood ..
Then why not L *B ?
Y do we take 1/2 *6*1.6 only?
And lastly, why do we subtract them ? :$

It is a triangle not a square or a rectangle xD Area of triangle is 1/2 base * height. Once the velocity is in negative, the ball is coming back downwards. Now, the question asks the distance from the ground to the hill, where the ball is now resting. The ball reached its highest point (v = 0 here) then started coming back downwards (v in negative here) and stopped at a hill. Thus the areas are subtracted

 

[_Ahmad]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_23.pdf

can anyone PLEASE explain Q5 b iii)

 

[usama321]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s13_qp_23.pdf

can anyone PLEASE explain Q5 b iii)

-__- Zepudee solved it for you on the previous page

 

[robinhoodmustafa]

Han kafi gand tha Kafi der lag gai thi solve kerte hue

Do u have the David Sang wali book?
I have to ask a small same question from it.

 

[usama321]

Do u have the David Sang wali book?
I have to ask a small same question from it.

Yup i have it