Physics AS Paper 22 .... Discussion

[mury456]

hyy ,

post up ur qs here ... only today is left so lets hurry up n get all the past papers done with ...
the reason i made a new thread is because the ones discussing them now have really got cramped n confusing ...

so fire away ...

help me out wd this too

best of luck
cheers

 

[msk]

its so easy....first part extension is 2e...second part it is 0.5e and third part it is 1.5 e.......for value of k 1/e cause
k is directly proportional to 1/e......

 

[mury456]

dude dtz nt much of explainin ...
lyk cn u explain da K part ??
im just gttin confused in dat ??

 

[Abdulrab]

please do this Q 4 (b and c) http://www.xtremepapers.me/CIE/Internat ... 7_qp_2.pdf Hurry

 

[Sully17]

First part the two strings are in series so : e+e = 2e Therefore : Since l=ke , l=k2e thus k=1/2(l/e) and since l/e is k .....it will be 1/2k

second part the two strings are in parallel so : 1/e + 1/e = e/2 therefore since l=ke , l=ke/2 thus k = 2(l/e) and since l/e is k ..... it will be 2k

third part is two string are parallel and serial to one string so : e/2 + e = 3/2e therefore since l=ke , l=3/2(ke) thus k = 2/3(l/e) and since l/e is k ...... it will be 2/3(k)

 

[libra94]

4b is Young Modulus is stress/ strain
stress (from the graph) =9*10^8 Pa
strain=0.013

 

[libra94]

4c....here, u subtract the area of the cylindrical rod by the area when the rod is under tensile stress
---->from the graph, the tensile stress is 9*10^8 Pa...this is when the rod has the smallest cross sectional area (due to necking)
stress= force / area
area= force/ stress
force = 1.9*10^3 N
stress = 9*10^8 Pa
so ur area will be 2.11* 10^-6 m^2

so (3.2*10^-6) - (2.11* 10^-6) = 1.2*10^-6m^2!!

 

[mury456]

hope dat helps

remember to hit the thanks button

 

[utsav156]

please do this Q 4 (b and c) http://www.xtremepapers.me/CIE/Internat ... 7_qp_2.pdf Hurry

for 4 b take out the gradient of the graph and young modulus= the reciprocal of the gradient
for 4 c take out the csa by using the formula stress= force/CSA
now subtract it from the csa of the cylinder!!!
gudlck!!!

 

[Abdulrab]

what about part (c) and i havent really understood part b !! can u explain it further ??

 

[ALevelstudent]

Can anyone help me with june paper nov 2007 ques 5a please ? :'(
http://www.xtremepapers.me/CIE/International A And AS Level/9702 - Physics/9702_w07_qp_2.pdf

 

[mury456]

read part c in my uploaded file ... its quiet self explainaitory

 

[utsav156]

what about part (c) and i havent really understood part b !! can u explain it further ??

ok listen to me
for b the graph is of strain against stress ryt?
the formula of the young modulus is stress/ strain
so if the graph is of stress against strain we can take out young modulus by taking out the gradientof the graph
but as it is of strain against stress graph the young modulus would be reciprocal of the gradient of the graph thus take out the gradient and the reciprocal it its you young modulus!!! get it???

 

[Sweet_angel]

dude dtz nt much of explainin ...
lyk cn u explain da K part ??
im just gttin confused in dat ??

ok so.....just remember to find k u have to do inverse of ur value of e
like if e is 3/2 then value of k would b 2/3

HOPE THIS HELPS U......

 

[libra94]

Can anyone help me with june paper nov 2007 ques 5a please ? :'(
http://www.xtremepapers.me/CIE/International A And AS Level/9702 - Physics/9702_w07_qp_2.pdf

i hope my drawing helps

 

[libra94]

every point of the new wave is lagging by 60' so every point is drawn after 5 boxes(of wave W)

 

[mury456]

Some one help

 

[libra94]

hope this helps!!

 

[libra94]

LOL sorry, i forgot to label nodes and antinodes
im sure u know that part

 

[mury456]

its givn so in the mark scheme ? ?
λ = (330 × 100)/550
λ = 60 cm

cn u explain this ???
thkx