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Unit 2 Physics notes!!

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im worried about physics unit 2 X_x didnt start yet

I'm worried too :S But they can't really twist questions in unit 2 like they do in Mechanics yeah? :/ Uhf I don't even know what kinda help to get from this forum :S
 
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I'm worried too :S But they can't really twist questions in unit 2 like they do in Mechanics yeah? :/ Uhf I don't even know what kinda help to get from this forum :S

yaa same unit 2 can have some twisty question but thanks to its many calculation question they are definitly less than unit 1 twists
 
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In that question it says that the light reflected is 180 degrees out of phase with the light reflected from B. Therefore it's path difference will be 1/2 lambda, since it's destructive interference. So delta path difference=1/2 lambda. U can say that delta path difference is say 2h (because the light ray comes and reflects back-so distance wil be twice) So 2h=1/2 lambda. U already have lambda from the 1st part. And then by equating it, u can get the value of h i.e. the minimum distance...

thnx dude;)
 
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Can anyone tell how to get the frequency from the question of Jan 2010, Q-5. Plsssssssssss
 
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Can anyone tell how to get the frequency from the question of Jan 2010, Q-5. Plsssssssssss

the equation is F=1/T
so in one wave, you have 4 squares across it, and the 1 square is 5ms, 4*5 = 20
1/20*10^-3 = 50
 
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Here's a question..in a circuit, if the resistance of a component increases, we say that the voltage dropped across it increases...why is that??
 
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Here's a question..in a circuit, if the resistance of a component increases, we say that the voltage dropped across it increases...why is that??

This is how we learnt it:
Since the resistance of one component decreases, the overall resistance decreases (it's in series)
When the overall resistance decreases, and EMF is constant, the current 'I' increases (V= IR)
Since the current now increases, and the voltage drop 'V' = IR, the voltage drop increases.

Hope that helps...? :)
 
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This is how we learnt it:
Since the resistance of one component decreases, the overall resistance decreases (it's in series)
When the overall resistance decreases, and EMF is constant, the current 'I' increases (V= IR)
Since the current now increases, and the voltage drop 'V' = IR, the voltage drop increases.

Hope that helps...? :)

I'm saying for 2 components in parallel...is the same concept applied or is it different? it's this question in june 2011 paper Question no. 15 last part
 
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I have 2 more questions..jan 2011 MCQs 4 and 5..any explanations for that???
 
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This is how we learnt it:
Since the resistance of one component decreases, the overall resistance decreases (it's in series)
When the overall resistance decreases, and EMF is constant, the current 'I' increases (V= IR)
Since the current now increases, and the voltage drop 'V' = IR, the voltage drop increases.

Hope that helps...? :)

what is the voltage drop? :S
 
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I'm saying for 2 components in parallel...is the same concept applied or is it different? it's this question in june 2011 paper Question no. 15 last part

Lol no two resistors in parallel will have the same voltage drop across them, it doesn't vary.

Remember: In series, voltage drop across resistors vary. Current will be the same throughout the circuit. In parallel, the current in the resistors is different and varies. Voltage drop will always equal to the emf of the battery.
 
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what is the voltage drop? :S

the voltage drop is just the amount of voltage dropped across the resistor lol. sorry that probably doesnt make any sense.

See when you apply a voltmeter across any resistor, you get a reading. that's the value of your voltage drop.
 
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I have 2 more questions..jan 2011 MCQs 4 and 5..any explanations for that???

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... ?
 
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I have 2 more questions..jan 2011 MCQs 4 and 5..any explanations for that???

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
 
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the voltage drop is just the amount of voltage dropped across the resistor lol. sorry that probably doesnt make any sense.

See when you apply a voltmeter across any resistor, you get a reading. that's the value of your voltage drop.

kkkkkkkkk thx
 
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