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Can someone explain june 2009 last qustn (c) part 2???
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these are my guess! if im not correct, please correct me!
answer for Q4 is B
UV ray freq is higher(10^15) than that of visible light(10^14)!
according to equation E=hf, energy of UV is high!
so it is from energy level 4 to 1!
the answer for
Q5 is D!
the current in (L+M) will be = to N!
because (L+M) is parallel and N is in series!
so when M is removed,
some of the current will move from N to L!
for example, consider:
L = 2A
M = 2A
N = 4A
when M breaks,
L will gain 1A from N,
making L = 3A & N = 3A
so, this increases brightness of Lamp L and decreases brightness of Lamp N
err I don't have the paper and I can't find it on the papers section, so if you post the question or give me a link I can try... ?
Can someone explain june 2009 last qustn (c) part 2???
for that the answer is Chow to get frequency in question no 6 in jan 2013?
for that question, look carefully!In jan 2013,question no 15 ,two resistors are parallel thn y do we hve to add all resistance to get totall resistance?
welcomeThe answers are correct and yeah I think the explanations makes sense now..okay thanks a lot for the explanation!
for that the answer is C
first convert 5ms to seconds by dividing it over 1000!
that is 0.005!
then multiply it by 4, because it have four squares for 1 wavelength!
that is 0.02!
then use F=1/T
so 1/0.02 is 50Hz
thats it!
F physics its an impossible subject with alot of stupid questions that Einstein cant do
welcome
btw are you really from maldives?
since they said R is directly proportional to l^2
then R=((2.51^2/2.5^2)*0.22)
then the answer you get - 0.22 because they want the increase in resistance...
so change in R is 1.76*10^-3
hey guys from where do u get the mrk schemes from??? i want for chem phy and bio u1 and u3s for june 13
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