Physics: Post your doubts here!

[Kumkum]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_21.pdf
Please help me with question number 7(b)(i) and (ii), and also (c), and question number 5(a) and (b).

 

[Asheeta]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w06_qp_2.pdf

Please help me for no. 2(a) on how to get the gradient !
and for no. 7(b)
Thanks

So for the gradient just draw a tangent at t=0.8 s and use 2 points.

7(b) (i) Kirchoff's first law states that the sum of currents arriving a junction = the sum of currents leaving the junction
therefore, I1+I2=I3

(ii) 1. 2nd law is that the sum of emfs around any closed circuit = the sum of pds . Therefore,
E2 = I2R2 + I3R3
2. E1 - E2 = I1R1 - I2R2

 

[Asheeta]

Hey ppl in paper 2 sometimes there is a number involving estimates of certain things for e.g the mass of a protractor- 10 g
could someone just list some quantities whose value must be learnt

 

[A_K_D]

Can someone Please Please Please answer my Question's (it got ignored 3 times)

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_23.pdf
7bii) I don't get why charged particle is not an isolated system

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w11_qp_13.pdf
29) I just don't understand the question, I looked at examiners report and it said that the lines R to S has a constant phase difference


I will thank you a lot

 

[zephyr86]

Hi!can anyone solve o/n 09 paper 22 question 3b?
Thanks!

 

[Asheeta]

Hi!can anyone solve o/n 09 paper 22 question 3b?
Thanks!

so your diagram is something like this, you have to complete the diag to show the resultant velocity. Just use the parallelogram law and it should be ok.
----------------------->
|
|
|
|
\/

You have been told that the vertical component of the velocity is 6.2m/s and the horizontal is 4.0m/s
resultant velocity = (squareroot)( (6.2^2) + (4^2)) giving 7.4 m/s
then find the angle using tan A = 4/6.2 , therefore A (i.e the angle it makes with the vertical) = 32.8

Now can you please help me with a question? Its N10 p2 no. 4
What is the region Es and how to find it?
Thanks in advance.

 

[bamteck]

So for the gradient just draw a tangent at t=0.8 s and use 2 points.

7(b) (i) Kirchoff's first law states that the sum of currents arriving a junction = the sum of currents leaving the junction
therefore, I1+I2=I3

(ii) 1. 2nd law is that the sum of emfs around any closed circuit = the sum of pds . Therefore,
E2 = I2R2 + I3R3
2. E1 - E2 = I1R1 - I2R2

Thank you ! That's so humble from you

 

[Zaid Beiruty]

Hi, Can any one explain p1 may 2004 q24

 

[queen of the legend]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s11_qp_21.pdf
Please help me with question number 7(b)(i) and (ii), and also (c), and question number 5(a) and (b).

for question 5 a: since I=V/R , and we know that v=12 and R is total resistance in the circuit (6 for resistor and 12 for max. variable resistor's resistance) so substitute and u get 0.67A. just to make you understand the concept better , Vin is constant, so increase in amount of R decreses the current and the question asks for minimum current so you should include the total resistance.
5ii: you got two values of current for two values of variable resistor ...just plot it and draw a descending curve.

for part b of question 5 ....i am getting wrong answers myself , so better ask a physics teacher

now lets go to question 7bi: if yo have your physics AS textbook you can refer to the expeiment of two dippers in water tank connected to same vibrating source an they produce circular wavefronts in water. wavefronts produced by each dipper interfere producing constructive interference at some places and destructive on other places.
7bii: the ripple tank interference can be seenby placing a viewing screen under the water tank and placin a lamp on top ofthe tank so that shadows made on the viewing screen show movement and shape of waves produced.

c: is a bit difficult t explan coz it includes drawing but anyhow at places where you see lines exactly crossing for ex: exactly in the middle its constructive (maxima) , where as which do nt cross exactly its minima .

i hope you understood !

 

[LeQuavina]

Can someone please explain to me the answers of questions number 4,5 and 6 from paper 11 2010 June?

 

[champ-student]

hi everyone........ i m a new member here........i m doing ma first part of o level ..............i need notes about problems KE & PE

 

[champ-student]

anyone gonna help......plzzzzzzzzzzzz

 

[Kumkum]

for question 5 a: since I=V/R , and we know that v=12 and R is total resistance in the circuit (6 for resistor and 12 for max. variable resistor's resistance) so substiture and u get 0.67A. just to make you understand the concept better , v in is constant, so increase in amount of R decreses the current and he question asks for minimum current so you should include the total resistance.
5ii: you got two values of current for two values of variable resistor ...just plot it and draw a descending curve.

sorry i am busy i will explain the other questions few hours later

ok no problem, thanx so much!

 

[Asheeta]

Hi, Can any one explain p1 may 2004 q24

Hi, the qu: Which Observation indicates that sound waves are longitudinal?
the answer is B. Well there is no such explanation involved it is just a fact that sound waves cannot be polarised and this should be learnt. That's it. Anyways let me know if something is still not clear.

 

[bamteck]

Hi, the qu: Which Observation indicates that sound waves are longitudinal?
the answer is B. Well there is no such explanation involved it is just a fact that sound waves cannot be polarised and this should be learnt. That's it. Anyways let me know if something is still not clear.

Hey you do M1 as well ?
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w09_qp_41.pdf
would you mind help me with the no. 4 ?

 

[Asheeta]

Can someone please explain to me the answers of questions number 4,5 and 6 from paper 11 2010 June?

so for no.5
X represents XCos(theta) and Y represents Ysin(theta)
Consider a sine curve, as theta increases sin(theta) reaches a max value of 1, therefore magnitude of Y increases
Consider a cos curve, as theta increases cos(theta) decreases from a max value of 1 to zero at theta=90, therefore magnitude of X decreases
so the ans is C

 

[Kumkum]

for question 5 a: since I=V/R , and we know that v=12 and R is total resistance in the circuit (6 for resistor and 12 for max. variable resistor's resistance) so substitute and u get 0.67A. just to make you understand the concept better , Vin is constant, so increase in amount of R decreses the current and the question asks for minimum current so you should include the total resistance.
5ii: you got two values of current for two values of variable resistor ...just plot it and draw a descending curve.

for part b of question 5 ....i am getting wrong answers myself , so better ask a physics teacher

now lets go to question 7bi: if yo have your physics AS textbook you can refer to the expeiment of two dippers in water tank connected to same vibrating source an they produce circular wavefronts in water. wavefronts produced by each dipper interfere producing constructive interference at some places and destructive on other places.
7bii: the ripple tank interference can be seenby placing a viewing screen under the water tank and placin a lamp on top ofthe tank so that shadows made on the viewing screen show movement and shape of waves produced.

c: is a bit difficult t explan coz it includes drawing but anyhow at places where you see lines exactly crossing for ex: exactly in the middle its constructive (maxima) , where as which do nt cross exactly its minima .

i hope you understood !

thnx for number 7...understood it!

 

[Asheeta]

Hey you do M1 as well ?
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w09_qp_41.pdf
would you mind help me with the no. 4 ?

From fig 2 after resolving forces at S
R=T2 sin(theta) + W , R= (3T2/5) + W
F=T2 cos(theta) , F = 4t2/5

From fig 3 after resolving forces at P
T1cos(alpha)=T2cos(theta)
3T1/5 = 4T2/5 therefore, 3T1=4T2
T1 = 4T2/3

T1sin(alpha)+T2sin(theta) = 5 therefore (4T1/5) + (3T2/5) = 5
Therefore, 4T1 + 3T2 = 25

So replacing T1 by 4T2/3
16T2/3 + 3T2 =25
25T2 = 75
so T2 = 3 N (Tension in longer string)

F = T2cos(theta)
therefore F = 3(4/5) = 12/5 = 2.4 N

Let coefficient of friction = u
R = (3T2/5) + W
F=(u)R
2.4 = 0.75(0.6 T2) + 0.75W
0.75 W = 2.4 - (0.75*0.6*3)
Therefore W = 1.4 N

*there are some pics that i am not able to upload* sorry for that

 

[xyz!]

hiee..dis is a doubt regarding paper 5
here's the link..
http://www.xtremepapers.com/papers/...nd AS Level/Physics (9702)/9702_w11_qp_51.pdf
can sum1 pls the diagram(a labelled one pls) fr the first question and attach it..?? i read the ms but m not able to undrstnd!

 

[aiskw1]

Hi Everyone I'm new here,

Can Someone Explain October 2011 QP 13 Question 29?