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http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_w04_qp_4.pdf
can anyone solve Q4c(i)
can anyone solve Q4c(i)
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Hi, the formula for gradient is : (y2-y1)/(x2-x1).....take two coordinates but keep in mind that the x-axis is in SQUARE form, so u need to square any 2 points. substitute and ull get the answer.
For b the answer is 1, cause this is the standard value
ummm link??Hi guys, could somebody help with questions, 6 , 13 , 14 , 17!! , 21, 24, 27, 28 , 32,
thanks would be very much appreciated
sorry hahaha!ummm link??
link please ?!physics paper4 mayjune 2009 question 1c(ii),3(i)2,3(ii),6C,11b(iii) i dont understand the answer sheet help please
link please ?!
link please ?!
Guys, I need some help in this paper:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w08_qp_4.pdf
Q 1 b)i)
judging from the fact that you asnwered part A, you know its a sinusoidal stationary wave and X has same amplitude as Y. But at any instant they are moving in opposite direction so they're 180 degrees out of phase.
pd is 2.7 because V = V(total) x resistance (thermistor) / resistance (total) = 4.5 x 1800 / (1200+1800) = 2.7 this is a potential divider.9702_w05_qp_2.pdf Can anyone help me with 7 (b) (ii) 1. Why is the pd 2.70? galvanometer reading would be zero not voltmeter?
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