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OCR Physics B G495 Field and Particle Pictures June 21st 2011 Exam Thread

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Hmm it probably is :tongue:

What I was trying to say though, is that only low-risk patients will get a scan. So low-risk patients, getting a very small amount of radiation.

Anywho, I think we've had enough confirmation on here that it's 4.35. If that's too low then blame OCR :wink:

OCR doesn't wanna scare students so kindly said that it's 4.35 that's wt i thought.. anyway i should hv put 4/5 instead ?

Reply 701

arianex

Original post by wki

got ignore :frown:

anyone know the phasors one in section c? n also how to find the mass of uranium per year or anything?

Mass of U-235 per year:

Multiply Watts/power given by number of seconds in a year, get total energy used per year.

Divide total energy by energy released per fission to get no of nuclei that decay.

Get actual mass of one U-235 nucleus by multiplying 235 by u to get mass in kg

Multiply no of nuclei by the mass...result?

Reply 702

Summerdays

Original post by Revolution is my Name

Well, it asked to comment on the amplitudes, so I said that of OX (or whichever it was where they were removing the air) would increase because less was being refracted and so more reflected, and there'd be no change for OY. Then it said to mention the detector signal, and I think I guessed at their being more minima.

Hmmm... is this similar to my answer :confused:

Reply 703

Geekchic17

Original post by jimmeh

Nah, definitely meant to get a positive root :/

What equation did you end up with after you rearranged to find V?

I think (can someone confirm this?) it was meant to be

v = \sqrt{c^{2}(1-\gamma ^{2})}

Damn yeah i get that, must have messed something up in the exam
I got pretty stressed on that question

Reply 704

Revolution is my Name

Original post by wki

same here

how to find the mass of uranium per year or anything?

Divide the wattage given by the number of J per fission, to give the fissions per second. Mulitply that by seconds in a year, to give the number of fissions per year. Then multiply that by the amount of uranium per fission (mass number x number of u per kg)

Reply 705

Summerdays

Original post by Revolution is my Name

Same

I got 980Kg

Reply 706

Geekchic17

Oh another thing,
Radioactive question:
something like find N to be 9x10^something (can't remember))
I got 9.89 .. and so did like alllll my friends .. is 9.89 supposed to be ~ 9? :s

Reply 707

Swindan

Original post by arianex

I can't remember my exact method, but I think I got just over 1000kg for this question. Anyone else get something similar?

Edit: Different answer to Summerdays, hence wrong :wink:

Reply 708

ejw93

Original post by Summerdays

They were, before the air was put in front of one of the mirrors, I think?

air was being removed not put in front of the mirror? I dunno what the answer would be but the wavelength would be refracted less as the refractive index would be less.

Reply 709

Summerdays

Original post by ejw93

air was being removed not put in front of the mirror? I dunno what the answer would be but the wavelength would be refracted less as the refractive index would be less.

I put that. I am not even sure about my answer, I just guessed.

Reply 710

Revolution is my Name

Original post by Swindan

Edit: Different answer to Summerdays, hence wrong :wink:

I imagine there's different potential answers, depending on whether you used 3x10-11 from the previous question, or the exact answer that you found.

Reply 711

arianex

Original post by Swindan

I can't remember my exact method, but I think I got just over 1000kg for this question. Anyone else get something similar?

Edit: Different answer to Summerdays, hence wrong :wink:

Original post by Summerdays

Same

I got 980Kg

Feckload of uranium, that is. I might've gone wrong, divvied it by 1000 because I thought I got it in g. Oops. XD

EDIT// I did get an answer that was over 1000 though. :smile:

Before the divvying. XD

Reply 712

Summerdays

Original post by Revolution is my Name

I imagine there's different potential answers, depending on whether you used 3x10-11 from the previous question, or the exact answer that you found.

Yeah. I actually foun the average binding energy and then went from there, like it doe in the CPG book :tongue:

I was talking bout the air and mirror question :colone:

Reply 713

ejw93

Original post by arianex

i did that, it seems correct.

Reply 714

41jms

Original post by Swindan

I can't remember my exact method, but I think I got just over 1000kg for this question. Anyone else get something similar?

Edit: Different answer to Summerdays, hence wrong :wink:

i got this!

Reply 715

Summerdays

Original post by arianex

Feckload of uranium, that is. I might've gone wrong, divvied it by 1000 because I thought I got it in g. Oops. XD

EDIT// I did get an answer that was over 1000 though. :smile:

Before the divvying. XD

I used MY answer that's why, which was 3.410^-11J :smile:

Reply 716

arianex

Original post by Summerdays

Yeah. I actually foun the average binding energy and then went from there, like it doe in the CPG book :tongue:

I was talking bout the air and mirror question :colone:

Well since it gets refracted less, wouldn't the speed (and hence the wavelength) of said beam change, thus rendering the phasor longer/shorter and thus unable to totally cancel out the other one? :tongue:

I totally blagged that Q xD