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-_- Continue from where you ended.so my graph is correct ? Oh my, what language are you using? hehe, which line should touch 3T?
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-_- Continue from where you ended.so my graph is correct ? Oh my, what language are you using? hehe, which line should touch 3T?
OHH sorry haha! Thanks dude!-_- Continue from where you ended.
No problem at allOHH sorry haha! Thanks dude!
More doubts?OHH sorry haha! Thanks dude!
My net is damn slow. I need to go.Thought blocker im so sorryyy but mind helping me with this heh heh. this is winter 06 q6 (b) i think i drew it wrongly so yeah
I didn't start my A2 yetMy net is damn slow. I need to go.
Ask ZaqZainab
Q15http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_1.pdf
Question 15 and 25, please. :/
Q29 v^2 = 2ashttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s09_qp_1.pdf
And question 29, I can't seem to understand it.
No problem.I didn't start my A2 yet
No problem.I didn't start my A2 yet
28)
Q15
not only did container X lose half the height even the mass became half
so m now will be m/2
and h will be h/2
g stays the same
we now have mgh/4
Q25 N lines per metre means 1/N and then substituting into the equation it will be (1/N) sin (theta)= n ( lambda)
sin theta/lamda is also = sin theta/((1/N) sin (theta)/n) solving it further gives Nn (if you don't know how to solve further let me know)
Q29 v^2 = 2as
a=F/m
F= EQ means Ee as in the question
so v^2=2*(Ee/m)*s
v^2/2=(Ee/m)*x
making x the subject
(v^2/2)/(Ee/m)=x
which equates to x=v^2*m/2Ee
28)
F= Eq
F = 3 x 10^7 x 1.6 x 10 ^(-19)
F = A
No problem.Thanks again, I keep overlooking small details and bothering you.
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