Physics: Post your doubts here!

[rideronthestorm]

 

[rideronthestorm]

D,A,D,B,C,B,B,C,B in the order uploaded .

 

[guigao1994]

Hey everyone!

this question is really anoying me. The question asks us to calculate the potential difference at B when the temperature is 15 C The marking scheme says that the potential difference at B is 2.4 V but I have calculated it to be 6.6 V, using 2200*9/(2200+800).
Consequently it asks for the output power which would be +9.0 V in case I am correct. And following which LED would be bright, if I am correct the first one (green).
All the answers in the marking scheme are the opposite of mine because of the first difference in the answer. 2.4 vs 6.6.

is the marking scheme wrong?

Thank you very much!

 

[omg]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_21.pdf
Q4 C (ii) plss helpp

 

[darknessinme]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_21.pdf
Q4 C (ii) plss helpp

Efficiency=(Useful Energy Output/Total Energy Input)x100%
It says the machine is only 25% efficient. So only a quarter of the kinetic energy of the water passing through the wheel can be used to produce a useful 110kJ output (this is every second).
So you get : 0.25 x (Kinetic Energy of water flowing every second through wheel)=110kJ
0.25 x 0.5 x (mass of water) x (15.3440...)^2=110
mass of water= 3740kg
Final answer= 3740kg/s

 

[unique840]

http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w10_qp_22.pdf
how to solve q4 part b

 

[Scafalon40]

hahaha Bombastic! Loved it i hope u don't get a warning by mods caz of having fun here

yes man the matter of fact is dat this dignified, honourable, disgusting, money making machines, world wide status uni, disrespected, renowned and so on examining body can make MISTAKES!!

And did it ever occur to you that on one of the front pages of their syllabus for any subjects they include:
And I quote:

Not-for-profit, part of the University of Cambridge
CIE is part of Cambridge Assessment, a not-for-profit organisation and part of the University of Cambridge.
The needs of teachers and learners are at the core of what we do. CIE invests constantly in improving its
qualifications and services. We draw upon educational research in developing our qualifications

BLOODY HELL IF IT'S A 'NOT FOR PROFIT' ORGANIZATION THEN WHY DID I HAVE TO PAY ABOUT Rs 80 000 FOR MY O'LEVEL EXAMS AND ANOTHER Rs 50 000 FOR MY A LEVELS
Would somebody tell me why please?
Hell I'd hate to see what I would have to pay if CIE were a 'for profit' organization!

 

[confused123]

And did it ever occur to you that on one of the front pages of their syllabus for any subjects they include:
And I quote:


BLOODY HELL IF IT'S A 'NOT FOR PROFIT' ORGANIZATION THEN WHY DID I HAVE TO PAY ABOUT Rs 80 000 FOR MY O'LEVEL EXAMS AND ANOTHER Rs 50 000 FOR MY A LEVELS
Would somebody tell me why please?
Hell I'd hate to see what I would have to pay if CIE were a 'for profit' organization!

hahaha yeah they are selling education man. i wish our own education system improves so that we don't need to buy these expensive forms of exams n all. nothing is not for profit

 

[confused123]

View attachment 6428

the only force acting on the stone is gravity which is downwards so XV is the direction.

 

[omg]

Efficiency=(Useful Energy Output/Total Energy Input)x100%
It says the machine is only 25% efficient. So only a quarter of the kinetic energy of the water passing through the wheel can be used to produce a useful 110kJ output (this is every second).
So you get : 0.25 x (Kinetic Energy of water flowing every second through wheel)=110kJ
0.25 x 0.5 x (mass of water) x (15.3440...)^2=110
mass of water= 3740kg
Final answer= 3740kg/s

thank you

 

[omg]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_s07_qp_2.pdf
q1. i didnt understand the calibration :/

 

[confused123]

View attachment 6453

i quoted all the answers and the god damn light went away when i was about to finish...
anyways for this one => Intensity is directly proportional to Amplitude square.
so 2I directly propor to A ^2

put A^2 instead of I, it becomes 2A^2 = A^2
the new amplitude is under root 2A , it comes 1.4A , frequency halved means number of wavelength also halved as frequency and wavelength r directly proportional

 

[rideronthestorm]

thanks a million ... plz do the rest tooo

 

[A.ELWY 7]

please help in june 2008 paper 1 question 27...thanx

 

[rideronthestorm]

wave speed = frequency x wavelength ..... they mentioned the antinode in question ... distance between adjacent antinodes or adjacent nodes is always half the wavelength , learn this thing .... so from the wall to the node the distance will be 1/4 of the wavelength ...thus 4 into x wud give us wavelength /// in the formula frequency = c/wavelength ... f=c/4x

 

[A.ELWY 7]

thanx very much...i got confused, i thought the ditance between 2 adjacent nodes is one wavelength so i choosed 2x not 4x

 

[KurayamiKimmi]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w02_qp_1.pdf
can someone explain question no.9? pls ^^

 

[hmlahori]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w02_qp_1.pdf
can someone explain question no.9? pls ^^

7) s= ut + 1/2 gt^2 . ut is zero as initial velocity is xero. g = 2s / t^2 right so now we need to subtract both times in order to get one expression giving acceleration : 2s / t2^2- t1^2 D option

 

[KurayamiKimmi]

7) s= ut + 1/2 gt^2 . ut is zero as initial velocity is xero. g = 2s / t^2 right so now we need to subtract both times in order to get one expression giving acceleration : 2s / t2^2- t1^2 D option

okay -but why is it t2^2- t1^2 and not (t2-t1)^2

 

[hmlahori]

okay -but why is it t2^2- t1^2 and not (t2-t1)^2

(t2-t1)^2
=(t2-t1)(t2-t1)
= t2^2 - 2(t1)(t2) + t1^2
which will give a different value but
(t2^2 - t1^2)
= t

Hope this helps.