Physics: Post your doubts here!

[asenfire]

Link please?

Its May/June 2008 Paper 4 (Q No. 8)

 

[SomeStudent]

Guys, do we have to give our calculations correct to 2 significant figures or 3?

equal to or one greater than the least sf of the data given in questions.

 

[abruzzi]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s05_qp_4.pdf
I desperately need your help in qn no. 7(c), pleaseee help me understand the solution

 

[Yousif Mukkhtar]

Can some one help me in
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_43.pdf
May June 13 Q 7c) Paper 43
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_43.pdf
Can some one explain Q 7c)

 

[unique111]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s07_qp_4.pdf

Q 7b? Please.

 

[unique111]

Can some one help me in
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_43.pdf
May June 13 Q 7c) Paper 43
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_43.pdf
Can some one explain Q 7c)

For m/j 7c)higher frequency means the total ke of electrons will rise.
But since same intensity, the no. Of electrons emitted will be same or less per unit time.

There is no 7c for oct/nov :/

 

[princeali97]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s11_qp_11.pdf.
Q25.

 

[AmeeraKhan]

Factors that determine the frequency response and sharpness of resonance?

 

[sitooon]

Can some one help me with W13 graph part of Vout
in section B plz?

 

[Sean002]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_qp_4.pdf
ANYONE Q2(b) part two

P1v1=P2V2
(2.5 x 10^7)( 4 x 10^4 x 10^-6 ) = ( 1.85 x 10^5 ) V2
V2 = 5.41 m^3
( volume of a super large balloon when all the air in cylinder is completely transferred to which is impossible. There must be some air left in the cylinder at the end )

the actual volume of super large ballon = 5.41 - 4 x 10^4 x 10 ^-6 = 5.365 m^3

# of small balloons = 5.365 / ( 7.24 x 10^3 x 10^-6 ) = 741

 

[Sean002]

View attachment 43076 can anyone help me in bi)

Energy for TWO gamma ray photons
E=mc^2
E=(2.4 x 10 ^-28 ) ( 3 x 10^8)^2
E= 2.16 x 10^-11
Energy for ONE gamma ray photon
E/2 = 1.08 x 10^-11 J

 

[abruzzi]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_4.pdf
I desperately need your help in qn no. 7(c), pleaseee help me understand the solution

Anyone?

 

[waleed302]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_11.pdf
q18 why is it B?

 

[sara kamal]

Anyone?

Anyone?

in such questions first what u need to do is try to identify the mass of iron and Mn left.....so if u getting 9 as ratio so they must be 0.9 for iron and 0.1 for Mn....
hence ratio 0.9/0.1=9
now sine we have half life of reaction ,we can find decay constant of it=ln2/half life in sec=.693/2.6x 3600=7.4 x 10^-5/s
hence using exponential decay eq
0.1=1 x e^(-7.4 x 10^-5 ) x time
proceed further to find time which will be in sec so convert to hours
further procedure...
ln 0.1=-7.4 x 10^-5 x time

 

[salvatore]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s08_qp_4.pdf
qn no. 6(b).. for part i, why is the distance 2.8cm? Shouldn't it be 4.5cm?
I'll be really grateful for a prompt response

 

[LeQuavina]

what's the number they write after the elements e.g Two Helium4 ? and we find the mass of the 2 atoms by multiplying u into 4 or 2?

 

[kitkat <3 :P]

what's the number they write after the elements e.g Two Helium4 ? and we find the mass of the 2 atoms by multiplying u into 4 or 2?

we multiply it by 4

 

[scouserlfc]

Why in satellite transmission the uplinks power is greater than downlinks power ???

 

[abruzzi]

in such questions first what u need to do is try to identify the mass of iron and Mn left.....so if u getting 9 as ratio so they must be 0.9 for iron and 0.1 for Mn....
hence ratio 0.9/0.1=9
now sine we have half life of reaction ,we can find decay constant of it=ln2/half life in sec=.693/2.6x 3600=7.4 x 10^-5/s
hence using exponential decay eq
0.1=1 x e^(-7.4 x 10^-5 ) x time
proceed further to find time which will be in sec so convert to hours
further procedure...
ln 0.1=-7.4 x 10^-5 x time

Thanks a lot!

 

[Rutzaba]

which university offers minors in Pakistan. i also wanted to do a minor :/
naa its a good choice not saying its a bad one just asking .
p.s minor in STATS . i hate it for some damn reason make stupid errors :/

Other options were economics n chemistry