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    (Original post by simon105)
    My default expression is 'An isotope of an element is made up from...'
    That's a good one too, I just found this past paper that had "an isotope is a variation of an element with..." so I think I'm good for that now


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    (Original post by Cheesegrater4)
    i thought it varies the resistance so changes voltage. ??
    Ah great. But surely that would vary the current too?

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    Its probably a stupid question, but how do you know what type of interaction a reaction is? It always comes up and i always tend guess and put weak which tends to be right most of the time but is their an actual way of telling which one it is!

    Thankss
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    (Original post by Surf)
    Well, see ya guys, good luck.... I've got to leave now...
    Remember to relax - take it one question at a time!
    :cool:
    Best of luck! xD

    You're leaving early!

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    Can some one go through jan 2013 question 7biii) please . And also is it only hadrons that experience strong interaction and leptons don't?
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    (Original post by x-Sophie-x)
    Did they use 2sf in the question?
    That's why they would have rounded to 120.

    But don't worry, unless they ask you for a certain number of Sig figs, you won't lose marks if you give your answer to something appropriate

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    Ah, they probably had. Cheers anyway


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    hows everyone feeling bout the exam today?
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    any suggestions on what the 6 mark might be on?
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    (Original post by x-Sophie-x)
    Does a variable resistor vary the current initially? Or resistance? Or both?

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    I'm pretty sure a variable resistor alters the current that flows through the circuit, and as V is constant and V=IR then that affects the resistance.
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    Can someone confirm that the W+ boson mediates beta+ decay and that the W- boson mediates beta- decay. Cheers
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    (Original post by Micheal123456)
    Can some one go through jan 2013 question 7biii) please . And also is it only hadrons that experience strong interaction and leptons don't?
    The emf is the total voltage in the circuit.
    So, 12v + (2x2.04v) as there are two wires which equals 16.1V.

    Ans yep, that's correct

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    (Original post by FrankB3)
    Guys can someone explain to me in Jun 12, Q7.a.ii, it's asked us to calculate the resistance, and the correct equation is: R=(8/0.067) it even backs this up in the mark scheme. The answer to that is 119.4, so 119 right? But in the mark scheme it has rounded up to 120. I don't get why they've done this and it's really annoying me because I'll lose marks this way. Can someone please explain when I'm supposed to round up all of a sudden?


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    They like to put their answer to 2 sig figs I've noticed, don't worry as long as you've done it right you'll get the marks. It's only on the question where it specifically asks you to put your answer to the appropriate sig figs, then you put your answer to the amount of sig figs indicated in the question.
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    In this interaction, can someone please explain why strangeness is not conserved?

    Sigma plus--> pi plus + n

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    I'm okay with Particles and Quantum Phenomena but Electricity is going to be the end of me. Good luck guys!
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    (Original post by x-Sophie-x)
    Thank you, that was a good explanation.

    I understand why for excitation, although one thing that still confuses me is that in ionisation the electron completely leaves the atom in one go, so doesn't need exact energy values incident on it.. and the 'extra' energy is the KE on the electron.

    Or have I completely missed the point here?
    This one is a bit weird and perhaps a bit tricky to explain, and I'd just like to point out that you are right in terms of Ek of the electron there - but this is a longer explanation as to why that happens:

    Using the stair analogy again (I just made that analogy up on the spot btw :cool:), this time think of the "ionisation shell" as the infinite shell, n= - you will most likely have seen this on those energy level diagrams in past papers, maybe without realising. Also on these diagrams, have you ever noticed that the energy levels are labelled as negative numbers? This is also important.
    The ground state, usually n=1, will be labelled with a negative number - let's say that an atom has an n=1 of -11eV. This means that in order to leave the atom entirely, the atomic electron has to take at least 11eV from a bombarding electron in order to "get back to 0eV" (shoddy terminology there). From there, any extra energy will be converted to Ek of this electron - just as you said at the beginning! So in a sense, that discrete 11eV is still required, but this time excess energy can still be transferred.
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    (Original post by Micheal123456)
    Can some one go through jan 2013 question 7biii) please . And also is it only hadrons that experience strong interaction and leptons don't?
    Hadrons INTERACT through the strong interaction, but charged hadrons DECAY through the WEAK interaction. So whenever there is an equation with a hadron becoming something else, it is through the WEAK interaction. The only charged hadron that doesn't decay (afaik) is the the proton, because it is stable.
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    (Original post by Micheal123456)
    Can some one go through jan 2013 question 7biii) please . And also is it only hadrons that experience strong interaction and leptons don't?
    7biii it tells you in the question that the bulb is 12v and it tells you there are two wires (which you previously calculated the p.d. across). Hence 12 + 2.1 + 2.1 = 16.2v. The strong nuclear force only acts on quarks where as the weak acts on quarks and leptons
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    (Original post by NtainS)
    Can someone confirm that the W+ boson mediates beta+ decay and that the W- boson mediates beta- decay. Cheers
    Confirmed!
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    (Original post by jazzynutter)
    Confirmed!
    Excellent - thanks
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    (Original post by BayHarborButcher)
    I'm pretty sure a variable resistor alters the current that flows through the circuit, and as V is constant and V=IR then that affects the resistance.
    Okay thank you surely the voltage would change as well then?

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