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find da total currnt in da circiut..V/R=I..so 9/60= 0.15 Acan u plz help me in q 34 of the same paper?
using ths currnt find da pd across P n Q... V=IR = 0.15* 50= 7.5v.. so and B
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find da total currnt in da circiut..V/R=I..so 9/60= 0.15 Acan u plz help me in q 34 of the same paper?
can u help me out too pls matefind da total currnt in da circiut..V/R=I..so 9/60= 0.15 A
using ths currnt find da pd across P n Q... V=IR = 0.15* 50= 7.5v.. so and B
sure..bt do u wnt hlp in ths qustion...can u help me out too pls mate![]()
its posted on previous page in big yellow fontssure..bt do u wnt hlp in ths qustion...
gve me 10 mis ..il solve them fr u!!its posted on previous page in big yellow fonts![]()
gve me 10 mis ..il solve them fr u!!
any one pls help needed queston no 5
gve me 10 mis ..il solve them fr u!!
oh right. sorry, my bad. I was looking at biology.
this question seems hard, can anyone explain?
Q 13..in ths question v consider KE to b eqaul to the components... so KE= the horizontal component, i,e 1) KE=1/2 m (v cos 45)^2 or 2) KE= 1/2 m V^2 cos 45^2....Anyone pls plss q13 25 and 26 pls pls do reply![]()
this question seems hard, can anyone explain?
tat wont hlp..knw ur theory n do all da possibl pst prps..u will b abl to do it..aftr all u cn post ur doubts or any kind of hlp u needf********ck!!!! after seeing these questions and realize that i m not able to solve thme so i felt that i will fail!!!! should i just stop studying?
sum1 do clear ma doubts!!!time for ma doubts nw... pls do solve them.....plsssssssssss!!!!!!!!!!
Q- 12, 25, 28, 31, 33, 34
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_qp_1.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s06_ms_1.pdf
yeah but it didnt make sense... the numbr of neutrons on both sides is the same.. so how can u say they are not conservedI already did..
nothing clear !Q 13..in ths question v consider KE to b eqaul to the components... so KE= the horizontal component, i,e 1) KE=1/2 m (v cos 45)^2 or 2) KE= 1/2 m V^2 cos 45^2....
.. at the highest point velocity = 0, tats da reason ve use the horizontal component..n not the verticle...here we will use da 2nd equations...
1/2mv^2= E.. ..therfr substitue ths in da equation where u will gt KE= E cos 45^2...
= cos 45^2= o.5. =o.5E------- ans A
fr qustn 25,,i keep gtn da ans as A..bt ms sys its D...il try figurin out n thn let u knw...
hav i made ma slf clear???![]()
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