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ALL CIE PHYSICS DOUBTS HERE !!!

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Deadly_king:

--- Quote from: $!$RatJumper$!$ on November 10, 2010, 04:16:02 pm ---Hmm im still confused. What i thought was that since X was 1/8th of the wavelength away from the start and Y was 1/8th of the wavelength away from the end, they would each be 45 degrees away from the start and end. Thus 360-(45*2) = 270 degrees.

So what if point X was at the end to the left (by the oscillator) and point Y was at the end to the right (by P), what would the phase diff be then?

:(

--- End quote ---

Ooh..........i understand what you did. But we don't measure phase difference like this. :-[

If the wave is placed on a graph paper, then the x-axis would be length and not angle. This is why you can't use length as angle.

Phase difference is described by the positions of the wave at the respective points. Whether at X or at Y, the waves undergo similar but opposite motion. This is why they are said to be in anti-phase.

For your question, The phase angle would still be 180o since even then both points will undergo similar motions but will be in phase as they are not opposite.

$!$RatJumper$!$:

--- Quote from: Deadly_king on November 10, 2010, 04:47:01 pm ---Ooh..........i understand what you did. But we don't measure phase difference like this. :-[

If the wave is placed on a graph paper, then the x-axis would be length and not angle. This is why you can't use length as angle.

Phase difference is described by the positions of the wave at the respective points. Whether at X or at Y, the waves undergo similar but opposite motion. This is why they are said to be in anti-phase.

For your question, The phase angle would still be 180o since even then both points will undergo similar motions but will be in phase as they are not opposite.

--- End quote ---

Alright i get it now thank you :) +rep

Can you please do S09 P21 Q5b

Thanks

Deadly_king:

--- Quote from: TJ-56 on November 10, 2010, 04:12:32 pm ---Can sumone pls explain s10 P2 Q7 (b) (ii)
thank you

--- End quote ---

7.(b)(ii)

1. 1.1 MeV is supplied in the form of kinetic energy of the alpha-particle.

2. First you should convert 1.1 MeV into joules. => K.E = (1.1 x 106) x (1.6 x 10-19) = 1.76 x 10-13 J

Now you equate K.E = 0.5mv2 = 1.76 x 10-13

Mass of an alpha-particle is 4u = 4(1.67 x 10-27)kg

Hence speed of alpha-particle is found to be the square root of 2(1.76 x 10-13)/4(1.67 x 10-27)

v = 7.3 x 106 ms-1

Hope it helps :)

Dania:
Help! Please. :)
The whole question.

$!$RatJumper$!$:

--- Quote from: Dania on November 10, 2010, 05:03:57 pm ---Help! Please. :)
The whole question.

--- End quote ---

Haha looks like the same question is getting to all of us :P I also need help on that :P

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