Physics: Post your doubts here!

[Thought blocker]

But we also need to know the battery's EMF considering this formula!

Which year paper ?

 

[Mohammad Farzanullah]

Consider formula : V = E - Ir, So we need quantity is Ir that is I(current) and r(internal resistance) so option C

Oh i got it. thank u very much

 

[Mohammad Farzanullah]

Which year paper ?

May june 04

 

[Thought blocker]

Oh i got it. thank u very much

May june 04

Ok

 

[_Ahmad]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_13.pdf

Can anyone please explain Q22 and 31

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_13.pdf

Can anyone please explain Q22 and 31

Gimme time of 3 hours, I am busy doing chemistry, if anyone help you, thats good, but if anyone don't just remind me at 12AM

 

[_Ahmad]

Gimme time of 3 hours, I am busy doing chemistry, if anyone help you, thats good, but if anyone don't just remind me at 12AM

ok

 

[mehria]

Well Guys, kitkat <3 :P mehria
Mohammed salik


I am sorry for yesterday night.. I got it, it was v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v. SIMPLE I made it complicated, Abb shamajh aya hadu ne a ko 1 q liya tha...
I was so stupid, sorry for wasting your imp hours on this topic.

its ok... m happy that finally u got it... n see i told u that u need rest

 

[mehria]

I M Fifteen ! U Cant Beat ME !
20/10/1998 ! I m not Joking!

i still cnt believe that u r 15

 

[_Ahmad]

Gimme time of 3 hours, I am busy doing chemistry, if anyone help you, thats good, but if anyone don't just remind me at 12AM

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_13.pdf

Can anyone please explain Q22 and 31

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_13.pdf

Can anyone please explain Q22 and 31

Sure, now lemme type it

 

[Thought blocker]

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_13.pdf

Can anyone please explain Q22 and 31

Well, I got the answers, but m not sure whether my method is correct or not, lemme conform it... It would take some time, but as soon as I get the correct way I would post the answers

 

[Thought blocker]

Well I am sure of Q31 - greater the charge the more the electric field. since both Experiences a same force F it should be F/q. or u could say like F=E*charge now since F is constant if u increase the electric field strength u decrease the charge

 

[_Ahmad]

Well I am sure of Q31 - greater the charge the more the electric field. since both Experiences a same force F it should be F/q. or u could say like F=E*charge now since F is constant if u increase the electric field strength u decrease the charge

Can you please elaborate it a bit more,
also can you please tell why ans is not B

 

[Thought blocker]

Can you please elaborate it a bit more,
also can you please tell why ans is not B

Its simple dude, Because it is itself the formula of electric field strength---->E=v/d
I cannot elaborate more ._.
If mehria kitkat <3 :P sagar65265 Suchal Riaz may explain in detail

 

[Mohammed salik]

i still cnt believe that u r 15

Wow..! How Should I Prove it ? :/!

 

[Thought blocker]

Wow..! How Should I Prove it ? :/!

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_13.pdf
Solve Q22 fast I wanna check my method ._.

 

[Mohammed salik]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_13.pdf
Solve Q22 fast I wanna check my method ._.

Give Me Few Minutes ..! Busy Hoon !

 

[Thought blocker]

Give Me Few Minutes ..! Busy Hoon !

Jaldi please.

 

[mehria]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w13_qp_13.pdf
Solve Q22 fast I wanna check my method ._.

http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w13_qp_13.pdf

Can anyone please explain Q22 and 31

yound modulus=stress/strain
E=Fl/Ae
(2.1x10^11)= (70 x 9.81) x (20) / ((3.2 x 10^6)x(200) x e)
e= 13734 / (( 2.1 x 10^11) x ( 6.4 x 10^-4))
e= 1.02 x 10^-4 m
e= 0.1 mm