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[efficiency=useful energy/ total energy used x100]
for Q4
resistance of C is what that u might not have figured if u draw tangents at 1V 2V 3V u will notice that the gradient for R is greater than C u know that resistance does not equal I/V it equals V/I so u have to find the inverse of that gradient in short greater gradient means less resistance. the rest u know that resistance is directly proportional to p.d so C has more V and I is constant as it is connected is series so according to P=IV C will dissipate more heat hope u get it
waisay lite janay wali h bute lets see what i could
for 4c i memorised this part and have problem in part 2
i will solve it later maybe 2-3hours later i do know how to solve but loadshedding is a hindrance
... and part 2 would be solved via Stress/Strain..
lol na na i knw that but i dnt know how to solve 1.they find the change in temp then multiply the original length*change in length for the decrease in temperaturehaha memorizing is my weak point
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_2.pdf
Q6 b) all parts.
f=qe(1.2*10^-15*2*10^3)haha memorizing is my weak point
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_2.pdf
Q6 b) all parts.
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