Physics: Post your doubts here!

[ahmed abdulla]

Question 2 (c) attached!

For the 3 (i), there's only one way. Split the graph into 3 parts minimum and calculate the area under graph. I believe cambridge have a range of values of the distance moved.

bro .. can u help me again plz with ... eg when they say second wave has same frequency with phase difference of 90 degree .....q 5(ii) ..same paper may/june 2010 23 ..... and with 2 _ when they say a wave has 0.25T where T is time period ... same thing ... sketching ?? for both ... i find a lot of difficult in this :::::::::and can u tell me the way of finding the distance ,, i tried but never got it @gary221 salvatore

 

[gary221]

g = 0 at least ⅔ distance to Moon
gE and gM in opposite directions
correct curvature (by eye) and gE > gM at surface
1357913579
hope i helped!

 

[Minato112]

View attachment 23151
g = 0 at least ⅔ distance to Moon
gE and gM in opposite directions
correct curvature (by eye) and gE > gM at surface
1357913579
hope i helped!

I thought the curvature after it touches the x-axis becomes negative! That is, the same curvature but downwards

 

[salvatore]

does any one know the explanation for the gap in the stress-strain graph of a polythene?

A polymeric material, in this case polythene, is capable of supporting a much higher stress. In other words, the ultimate tensile stress of the material is very large.
The graph can not accommodate all the values of the stress and therefore, a gap is added added in between to differentiate the values. ( The gap is similar to a line break in a histogram)

Hope that helped

 

[gary221]

I thought the curvature after it touches the x-axis becomes negative!

yup...thts d part i ws confused abt...but i thought tht frst d pull is in 1 direction, n thn the nxt...so thts how the overall force wud be...unless u mean like this
but m pretty sure this is wrong!!

 

[gary221]

yup...thts d part i ws confused abt...but i thought tht frst d pull is in 1 direction, n thn the nxt...so thts how the overall force wud be...unless u mean like thisView attachment 23162
but m pretty sure this is wrong!!

Minato112 ??

 

[gary221]

I thought the curvature after it touches the x-axis becomes negative! That is, the same curvature but downwards

uh huh...not to be a bother, bt if u cud draw tht n show me...??
i just cant seem 2 figure tht out...

 

[xxxt]

Two springs X and Y have spring constants k and 2k respectively.
Spring X is stretched by a force F and spring Y is stretched by a force 2F. Each spring obeys
Hooke’s law during the extension.
The work done in stretching spring X is WX and the work done in stretching spring Y is WY.
What is the relationship between WX and WY? [1]
A WY =0.5 WX
B WY = WX
C WY = 2WX
D WY = 4WX why is the answer c and not A

 

[salvatore]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w08_qp_2.pdf
Could someone please help me with no. 7 (b) (ii).. I'm totally confused!

 

[Minato112]

yup...thts d part i ws confused abt...but i thought tht frst d pull is in 1 direction, n thn the nxt...so thts how the overall force wud be...unless u mean like thisView attachment 23162
but m pretty sure this is wrong!!

Here it is :

 

[gary221]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_2.pdf
Could someone please help me with no. 7 (b) (ii).. I'm totally confused!

as u can c, the 5kohm, n the thermistor are in a parallel combo... so the total resistance R ------> 1/R = 1/5 + 1/T
since the total resistance is R, substituting it into the formula... V = P/(P + Q) * E
3.6 = 2 / (2 + R) * 6 -------> so R = 1.33 k ohm
So, substituting in the abv formula, 1/R = 1/5 + 1/T
1/1.33 = 1/5 + 1/T
So, T = 1.82 k ohm

 

[salvatore]

View attachment 23193
as u can c, the 5kohm, n the thermistor are in a parallel combo... so the total resistance R ------> 1/R = 1/5 + 1/T
since the total resistance is R, substituting it into the formula... V = P/(P + Q) * E
3.6 = 2 / (2 + R) * 6 -------> so R = 1.33 k ohm
So, substituting in the abv formula, 1/R = 1/5 + 1/T
1/1.33 = 1/5 + 1/T
So, T = 1.82 k ohm

Thanks a lot for the explanation..
Much appreciated

 

[anmolareeba]

Plz if anybody could help in October 2011 paper 23 in Q.1 part. (D) and Q2 and Q5

 

[Minato112]

Plz if anybody could help in October 2011 paper 23 in Q.1 part. (D) and Q2 and Q5

I'll answer the Qu 1 part (d) for now cuz' im really busy... In case noone answers the rest, I'll answer it

 

[anmolareeba]

Thanks so much ...I would be too greatful if u could help me wid others also BDw do u hav these type of solutions to all papers

 

[Salman Khalid]

can anyone help me with these two questions
with working
thnk u in advance

 

[unseen95]

please help me with question 4 (d)(i) and (ii). http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s07_qp_2.pdf

 

[Tkp]

can anyone help me with these two questions
with working
thnk u in advance

fr question no. 35 the ans is d as the current 5 is divided in 3 parts so the ammeter reading would be 1.66 approx to 1.7(hope its right)

 

[Ariel Robert]

What is the meaning of SHM? and what is Angular Frequency?

 

[taimoor.08]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_11.pdf Q5