Physics: Post your doubts here!

[Hassan Ali Abid]

actually is nothin to do with a graph....its like a situation is given....weve to use rms...

ill upload a sample then help me with it,can u?

rms is used when ever the AC voltage or current is given,...
yes sure ...i'll see in to it ..

 

[RoOkaYya G]

rms is used when ever the AC voltage or current is given,...
yes sure ...i'll see in to it ..

ok
ty...ill upload the word problem soon

 

[Eye catcher]

View attachment 36357
question b: the time the ball falls before the photograph is taken??

If you read the question it says :
The scale shows the distance fallen from rest by the ball. At time t = 0, the top of the ball
is level with the zero mark on the scale. Air resistance is negligible.

On the diagram above, it tells the initial position of the ball in photograph. Therefore, the distance, the ball fell before the photo was taken was 79 cm ( 0.79m)

therefore, s= 0.5*a*t^2
0.79=0.5*10*t^2
t = 0.4 s


question c: the time interval during which the photograph is taken???
therefore,
s= 0.5*a*t^2
0.90=0.5*10*t^2
t = 0.03s

ty

 

[Igcse stuff]

Just focus on P4 as it is having the max weightage and try to score maxx marks ..like around 80-85 in this paper and you can easily get an A* or if you are having less time then you can only prepare the application part well along with other two topics like magnetism and quantum or heat ch ...and then give you max on P2 so this will easily let you to score atleast an A.
For paper 5 , you only have to memories the mark schemes or 3-4 specific papers and copy pasting that things in any paper can make an A in that component too ..

Thanks a lot Hassan!...for P4 does it help to memories the mark schemes as well?....and in P5 are you supposed to answer in point form or paragraphs?

 

[Faithix MFSH]

Answer is D.
Please someone do provide a concise explanation without skipping of any points.
Quick help would be appreciated.

 

[ZaqZainab]

Answer is D.
Please someone do provide a concise explanation without skipping of any points.
Quick help would be appreciated.

calculate the change in momentum every second.
I am not getting the exact answer but i am pretty sure my method is right
F = (change in momentum)/time
F=(mass*velocity)/time.
the mass is obtained by multiplying the area (through which the wind is flowing) x density of air x velocity of wind
you will get F= 475.2*33=15681.6
round it off and you get 16000N

 

[Suchal Riaz]

Answer is D.
Please someone do provide a concise explanation without skipping of any points.
Quick help would be appreciated.

I created this picture in my mind and the answer came true. It is obvious that in on second all particles of air 33m away from wall hit the wall as speed of air is 33m/s. See attachment i hope it clears your concept.

 

[Faithix MFSH]

I created this picture in my mind and the answer came true. It is obvious that in on second all particles of air 33m away from wall hit the wall as speed of air is 33m/s. See attachment i hope it clears your concept.

Yeah thanks your answer sort of makes sense after the diagram. Sadly <ZaqZainab> just quoted what the examiner report said which was not so clear.
Its actually as you stated, area x distance traveled in one second x density. And then into velocity to achieve rate of change of momentum.

Thanks alot for your fast response!

 

[ZaqZainab]

Yeah thanks your answer sort of makes sense after the diagram. Sadly <ZaqZainab> just quoted what the examiner report said which was not so clear.
Its actually as you stated, area x distance traveled in one second x density. And then into velocity to achieve rate of change of momentum.

Thanks alot for your fast response!

Oh I am sorry though i don't even know which paper its from
I thought you were having trouble in finding the values and not in understanding the question

 

[ahmed abdulla]

Application part
can anyone solve it?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf
q. 11 b(iii)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s10_qp_41.pdf
q. 10 b

 

[sitooon]

question 11 b

 

[Suchal Riaz]

Yeah thanks your answer sort of makes sense after the diagram. Sadly <ZaqZainab> just quoted what the examiner report said which was not so clear.
Its actually as you stated, area x distance traveled in one second x density. And then into velocity to achieve rate of change of momentum.

Thanks alot for your fast response!

My pleasure. I am flattered to hear that somebody benefitted from me. Always ready to help.

 

[Snow Angel]

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

question 6, 2, 10.

please can anyone explain it.

please explain as soon as possible.

 

[usama321]

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

question 6, 2, 10.

please can anyone explain it.

please explain as soon as possible.

Here is how i applied logic to these questions :

2: As Q is negative, its direction would be opposite to its actual direction. Now, just use the head to tail rule of addition of vectors and you can see that C shows the required direction

6: (I am not quite sure about this one )
y = mx + c
As both values depend on each other, we can say that an increase in uncertainty of y will increase uncertainty in x.

Now, the percentage uncertainty of y remains same across all temperatures. Seeing the formula of percentage uncertainty,

uncertainty/actual value * 100...

Now, as the values of y change, so do our actual values in our formula. But as the percentage always remains same at all values, it is evident that uncertainty must also increase with the actual value to accommodate the always constant value of 1 % uncertainty. Therefore, it can be seen that uncertainty must be smallest at the lower values of y to accommodate the constant value of percentage uncertainty.

As the uncertainty of x depends on y, therefore we can see that the same must apply to x too.

10: not sure why A can not be considered the answer here

 

[goodluckayesha]

i have a few doubts plz someone kindly explain (A2 questions)
http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s09_qp_4.pdf Q10 part cii mark scheme says 9v how does the answer come?
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s10_qp_42.pdf Q10b
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_41.pdf Q9biii the V- decreases as temp is increasing but y is vout -9 or +9 y not taking +5v into account as in part Q9bii?

 

[goodluckayesha]

Salam.

Can anyone help me with this A2 Paper 4 Physics question on electromagnetism and nuclear physics?

Please provide with as much explanation as possible...

Thank you.

May Allah repay your kindness.

Amin.


Electromagnetism:

Question 6. (c)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf


Nuclear Physics:

Question 9 (b)

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_4.pdf

answer to part 9b is :

possible error is 10% and uncertainity is 2% so it can be 10+/- 2 which is 8 and 12 but maximum error is 10 so value cant exceed 10 which means it has to be 8 and not 12. 8% is error so we can measure (100-8)=92% accurately. now find lamda by the formula lamda=ln2/t = ln2/5.27 x 365 x 24 x 3600 = 4.17 x 10^-9
now use the formula A=Aoe^-lamda x t where A=92 and Ao=100 lamda we found out n now find t by substituting the rest of the values. you will get the answer in seconds which is 2 x 10^7 n convert it to days which is 231.
sorry i cant properly understand Q6c myself if u get it plz do help me also n i hope u understand Q9b. Above r my doubts if u know plz explain

 

[ijlalm]

Q A neutron (relative atomic mass 1) undergoes a head on elastic collision with a stationary nitrogen nucleus (relative atomic mass 14). What is the final velocity of the neutron?
A) It is zero
B) It is less in magnitude than the final velocity of the nitrogen atom
C) It is equal to the initial velocity but in opposite direction
D) It is less in magnitude than its initial velocity
E) It is greater in magnitude than its initial velocity

 

[Suchal Riaz]

Q A neutron (relative atomic mass 1) undergoes a head on elastic collision with a stationary nitrogen nucleus (relative atomic mass 14). What is the final velocity of the neutron?
A) It is zero
B) It is less in magnitude than the final velocity of the nitrogen atom
C) It is equal to the initial velocity but in opposite direction
D) It is less in magnitude than its initial velocity
E) It is greater in magnitude than its initial velocity

D
It cant be zero. Its velocity can not increase. It's speed can not be same as some of the momentum is transferred to nitrogen. The speed of it would still be more than the speed of nitrogen as the nitrogen has more mass so it's velocity would be less in magnitude than neutron, and speed of it would decrease obviously. Therefore the answer is C.
I have considered these two principles to make these assumptions: relative speed of approach = relative speed of separation. And total momentum before collision=total momentum after collision. Even if you have slight confusion in anything i mentioned above, tell me. As it is very important that you don't have any confusions.

 

[ijlalm]

http://papers.xtremepapers.com/IB/Physics/Higher/2008 May/Physics HL paper 1 TZ2.pdf
Can someone explain Q7.

 

[Suchal Riaz]

ijlalm i guess you deleted your response. Actually very large means that the m2/m1-->infinity. That means that this will never happen unless the bigger mass is much much bigger. Such as a ball and the earth. But the neutron and nitrogen are not the same way. M2/m1=14 which is not as big as infinity obviously. Was my answer wrong by the way?