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ANY1??http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w06_qp_1.pdf
ok i need help wd Q no 18... y do we use da last step as (160/80)*100 ?
20...umm shoulnt it be C?
21... me dont get it
27....
30...umm i got B :S
plz help
plz any one
yep! thanks!is the answer c? if it is this is because the least number of significant figures given in the data is 0.1 (1 sf) so the uncertinty must also be to 1 sf, so we round 0.36 to 0.4. same with the speed given to 3 sf
HELP!!I NEED HELP IN QS. :
14
15
27
AND
34!!!!!!!!!!!!!!!!!!!
PLSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS ((
Questions: 10, 11 , 21, 24, 26, 27, 31, 33 ,36
PLEASE HELP URGENTLY!!!! AND EXPLAIN AS WELL, I CAN GET THE ANSWERS FROM A MARK SCHEME I NEED EXPLANATIONS!! THANK YOU GOD BLESS
OK please post the questions again or link me to your old post, I don't want to dig deep down to find it. :\leosco1995 pls help me out brother then i can post my last doubts pls waited whole day for ur reply pls mate
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the answers B. But how? any description will be appreciated.thank you.
and can anyone tell the effects of changing any one one value will have on other values in d.sin(teta)=n.lamda
I told u I promise..ok, now...I NEED HELP IN QS. :
14
15
27
AND
34!!!!!!!!!!!!!!!!!!!
PLSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS ((
I told u I promise..ok, now...
Q15: first u need to get the angle theta of the triangle using...opp/hyp=sin ( theta) so 3/7 = sin ( theta) so sin^-1 ( 3/7) = 25.4...then when resolving forces..the force down the inclined plane is mg sin ( theta) so 2 x 9.81 sin ( 25.4) = 8.42...then 8.42 - 5n ( 5 N is the friction) so 3.42N then...F=ma so 3.4 /2 = a = 1.7...finally V^2= 2as...so V= square root of 2 x 1.7 x 7= 4.9 so A
27: 500 lines per metre..so to get d..d = (1/500) x 10^-3...( i multiplied by 10^-3 cause it is mm) so ull get 2 x 10^-6 then d x sin( theta) = n ( lambda) so 2 x 10^-6 sin ( 90) = n x ( 600 x 10^-9)..note( sin 90 means to observe the whole grating ..and 600 x 10^-9 because it wqas nano metre)..so n=3.3 so 3....then he asked for the images..here we got n which is the number of the maximum level..so we have the normal light pasin straight so this is 1..and we have 3 above the normal and 3 down the normal so the total is 7.....note( in the future..if he asked for the images or the beams the observer will see, and u got n from the calculation..use the equation 2n +1) if we used it here we will get 7...as (2 x 3) + 1 = 7
34: resistance is equal to = pl/A and the volume is Ah...we have and h which is the l...the area of a cube is L1 x L2 so pl/L1 x L2 and the volume of a cube is L1 x L2 x L3 so it is, when substituting.....p/V^1/3 so C
please do mine too leosco, your like a genius!! <3OK please post the questions again or link me to your old post, I don't want to dig deep down to find it. :\
Please answer the questions above. Thank you.http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w10_qp_12.pdf
Please help me with these questions. 8, 9, 12, 14, 15, 22 and 25. Thank you very much.
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