# How on earth do you do the graph calculation part of AQA's A2 Empa Watch

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Help,

I can do any other part of the sylabus however the first part of every single Section B in the empa exam is screwing me over, I can never make any sense of it, they seem to re arrange these equations for parts of the graph and deduce them from nothing.

For example Parts 1bi and 1bii here http://filestore.aqa.org.uk/subjects...-QPE-JUN11.PDF

It makes no sense, you have Ln(v/vm) on one axis and Q on the other.

I can get the equation to ln(v) = Q + ln(vm)

and thus V = e^(Q+ln(vm))

However in no way does that get anywhere to their supposed answerhttp://filestore.aqa.org.uk/subjects/AQA-PHYAB6-6X-W-MS-JUN11.PDF

HELP, it makes no sense. I'm pretty much getting in tears over this because no matter how much I look at it it makes no sense. If anyone has any revision material for this that would be brilliant, however when I search for EMPA help it just comes up with definitions of uncertainty which I know back to front.

I can do any other part of the sylabus however the first part of every single Section B in the empa exam is screwing me over, I can never make any sense of it, they seem to re arrange these equations for parts of the graph and deduce them from nothing.

For example Parts 1bi and 1bii here http://filestore.aqa.org.uk/subjects...-QPE-JUN11.PDF

It makes no sense, you have Ln(v/vm) on one axis and Q on the other.

I can get the equation to ln(v) = Q + ln(vm)

and thus V = e^(Q+ln(vm))

However in no way does that get anywhere to their supposed answerhttp://filestore.aqa.org.uk/subjects/AQA-PHYAB6-6X-W-MS-JUN11.PDF

HELP, it makes no sense. I'm pretty much getting in tears over this because no matter how much I look at it it makes no sense. If anyone has any revision material for this that would be brilliant, however when I search for EMPA help it just comes up with definitions of uncertainty which I know back to front.

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#2

V = Pe^–λQ

Taking lns on both side (slowly)...

ln V = ln (Pe^–λQ)

ln V = ln P + ln e^–λQ

ln V = ln P –λQ ln e if you would like me to explain any of these steps in detail...let me know

ln V = ln P –λQ

y = c + m x

So, if a graph of ln V is on the y axis, and Q is on the x axis....the y-intercept will give you ln P and the gradient will be –λ

Taking lns on both side (slowly)...

ln V = ln (Pe^–λQ)

ln V = ln P + ln e^–λQ

ln V = ln P –λQ ln e if you would like me to explain any of these steps in detail...let me know

ln V = ln P –λQ

y = c + m x

So, if a graph of ln V is on the y axis, and Q is on the x axis....the y-intercept will give you ln P and the gradient will be –λ

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#3

So, if the student's analogy is correct and that mathematical relationship is true, plotting a graph of ln V vs Q should give a straight line, of negative gradient with a positive intercept on the y-axis

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(Original post by

V = Pe^–λQ

Taking lns on both side (slowly)...

ln V = ln (Pe^–λQ)

ln V = ln P + ln e^–λQ

ln V = ln P –λQ ln e if you would like me to explain any of these steps in detail...let me know

ln V = ln P –λQ

y = c + m x

So, if a graph of ln V is on the y axis, and Q is on the x axis....the y-intercept will give you ln P and the gradient will be –λ

**physics helper**)V = Pe^–λQ

Taking lns on both side (slowly)...

ln V = ln (Pe^–λQ)

ln V = ln P + ln e^–λQ

ln V = ln P –λQ ln e if you would like me to explain any of these steps in detail...let me know

ln V = ln P –λQ

y = c + m x

So, if a graph of ln V is on the y axis, and Q is on the x axis....the y-intercept will give you ln P and the gradient will be –λ

So would you suggest with questions like this working their equation towards Y = mx+c rather than the other way (which I what I was trying to do)?

Could you also if possible (because your such an amazing help ) have a quick look at question 3a. I thought for the EMPA you didn't need to know/use specific equations from the other exams without them giving you them specifically, however they expect you to use N=N0*e^(-(lambda)Q) which is lovely of them because from what I have read you don't get the formula book to even look it up from to double check it's correct.

Thanks for your help

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#5

Yes - definitely work the equation towards y = mx + c. Always show clearly the steps you are taking, and at the end you always have to finish it off by saying the gradient = -λ and the intercept = ln P. You have to finish it off clearly. I am just looking at question 3a....

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#6

This one is quite applied - I would consider it one of the most difficult on the paper.

If we start with half-value thickness - it explains in the question that this is thickness of glass that would reduce the output voltage by half. This is like half life...half life being the TIME (rather than thickness) needed to reduce the voltage by half. So time and thickness in this context are sort of the same thing.

The graph in the question is clearly an exponential relationship, as it is a straight line. However, it is different to the earlier question, as there is now N - so the equation from before doesn't work. But, all exponential decay is of the same form...

V= Voe^-Lambda t

So, it becomes

V= Voe^-Lambda an

Now, from the graph you could find Vo - read off the y intercept, and antilog. You could then find half of Vo (divide by two).

You you could also find -lambda from the gradient (same as you did in the earlier question).

Then, substitute all that in the equation to find N - this will give you the number of slides needed to half the voltage.

Really sorry if the explanation isn't very clear - difficult to type accurately with subscripts and superscripts!

As far as the formulae are concerned, you do not get the formula sheet, but are usually given a bit of help - in this case they gave the exponential one earlier. I would definitely have all of those in your head anyway.

Not sure if I have explained your question clearly!

If we start with half-value thickness - it explains in the question that this is thickness of glass that would reduce the output voltage by half. This is like half life...half life being the TIME (rather than thickness) needed to reduce the voltage by half. So time and thickness in this context are sort of the same thing.

The graph in the question is clearly an exponential relationship, as it is a straight line. However, it is different to the earlier question, as there is now N - so the equation from before doesn't work. But, all exponential decay is of the same form...

V= Voe^-Lambda t

So, it becomes

V= Voe^-Lambda an

Now, from the graph you could find Vo - read off the y intercept, and antilog. You could then find half of Vo (divide by two).

You you could also find -lambda from the gradient (same as you did in the earlier question).

Then, substitute all that in the equation to find N - this will give you the number of slides needed to half the voltage.

Really sorry if the explanation isn't very clear - difficult to type accurately with subscripts and superscripts!

As far as the formulae are concerned, you do not get the formula sheet, but are usually given a bit of help - in this case they gave the exponential one earlier. I would definitely have all of those in your head anyway.

Not sure if I have explained your question clearly!

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