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    (Original post by Anon-)
    ay what did you guys put for what the highest dot is in the oscilloscope reading? i put emf. teacher thinks it's terminal PD.
    Emf because there's no current flowing around the whole circuit
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    (Original post by JamieP123)
    Last question was 3.4 ohms yup. V=IR 1=0.277777r so 3.4 ohms (I think if you used the rounded value of 0.28 then you get 3.6?)
    Yup. Sounds about right! What do you guys think the A grade would be at then? I reckon it'll be at around 56+-3.
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    (Original post by Anon-)
    ay what did you guys put for what the highest dot is in the oscilloscope reading? i put emf. teacher thinks it's terminal PD.
    Emf is terminal pd when there's no current
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    (Original post by Jonah1958)
    For the rms voltage over the heating element i ended getting 229.5V as 0.5v lost to resistance of wires, got a current of about 19A.

    Pretty sure i messed up somewhere
    I got the same
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    (Original post by Wonkylou)
    Did the particle question explicitly say that it was via the weak interaction.
    No but I vaguely remember the mark schemes of some past papers implying that strange particles are only affected by weak interactions? I could be completely wrong here. But if I am right, then I am most disappointed in myself for forgetting about that in the exam.
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    (Original post by JamieP123)
    Don't quote me on this but if you put K^0 then mass is being created from nothing which is impossible.
    wish I knew this before the paper lol
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    (Original post by Anon-)
    ay what did you guys put for what the highest dot is in the oscilloscope reading? i put emf. teacher thinks it's terminal PD.
    I think you're correct. The highest dot was EMF and the one below would have been the terminal PD
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    Name:  image.jpg
Views: 99
Size:  511.5 KB For the internal resistance on last question, i worked out current by doing 1/18 = v/r as when switch two was closed the pd across ocillator reduced from 6 to 5v so 1 volt is lost because of the 18 ohm resistor so it was pd of resistor = 1 divide by resistance of resistor = 18 to get current = 0.056 and so internal resistance of 89 ohms

    Not sure where i went wrong :/
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    (Original post by Matthewg97)
    The emf was 6v (worked out on the previous page) when the circuit was connected the pd accorss the battery dropped to 5v so the external resistor was taking 1v so current across it was 1/18= 0.056
    The pd (5 volts) when the circuit is connected is Emf minus voltage over internal resistance. You're thinking of it the wrong way round.

    Emf is the terminal pd when there's no current. When current flows the terminal pd is Emf - Ir
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    (Original post by RUNSran)
    and current was 0.28 right?
    I dont remember sorry
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    (Original post by ErwinJ)
    I got 48 ohms, but i think i got it wrong. However I definitely think its not 3.6 ohms. The voltage across the 12ohm resistor was 1V, so the internal resistance got the remaining P.D. Surely this means it has a higher resistance?
    I think it was 18 ohm resistor so that might explain it??
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    (Original post by kelvin1338)
    If there was no current, wouldn't there be no dot at all? lol since no power is going through the oscilloscope
    Isn't it because the voltage wasn't alternating? otherwise there would be a dot on the negative y gain too
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    The decay is K^+ = pi^+ + pi^0.

    If you look at their rest energies, K^0 has a higher rest energy than K^+, so it is not possible to create this from the decay of pi^0.

    I agree that it seems like K^0 could have been a possibility due to the wording of the question, and how it's possible to expect strangeness to be conserved, but unfortunately pi^0 is the only possible answer. However 2/3 of the marks should be for working out the charge and the fact it is a meson, not a baryon or lepton.
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    (Original post by TajwarC)
    Isn't it because the voltage wasn't alternating? otherwise there would be a dot on the negative y gain too
    I wrote because it was direct current
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    (Original post by Jonah1958)
    Name:  image.jpg
Views: 99
Size:  511.5 KB For the internal resistance on last question, i worked out current by doing 1/18 = v/r as when switch two was closed the pd across ocillator reduced from 6 to 5v so 1 volt is lost because of the 18 ohm resistor so it was pd of resistor = 1 divide by resistance of resistor = 18 to get current = 0.056 and so internal resistance of 89 ohms

    Not sure where i went wrong :/
    That's awfully high for an internal resistance. I did the emf (6) minus the voltage (5) over the current (calculated from the previous question). 3.4 should be the correct answer, although they could accept 3.6 due to rounding.
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    (Original post by Tayeb)
    Yup. Sounds about right! What do you guys think the A grade would be at then? I reckon it'll be at around 56+-3.
    Apparently they decide upon the grade boundaries different to normal. (Because the sample size wont be large enough they just take a feel for what the boundaries should be)

    IMO the paper was fairly straight forward (nothing too tricky) I think your guess is about right by the looks of things.
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    (Original post by Jonah1958)
    Name:  image.jpg
Views: 99
Size:  511.5 KB For the internal resistance on last question, i worked out current by doing 1/18 = v/r as when switch two was closed the pd across ocillator reduced from 6 to 5v so 1 volt is lost because of the 18 ohm resistor so it was pd of resistor = 1 divide by resistance of resistor = 18 to get current = 0.056 and so internal resistance of 89 ohms

    Not sure where i went wrong :/
    I got the same as you at first, but I thought there was no way IR could be that much higher than resistance, so recalculated it as 5V/18ohms gives 0.28A current.
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    For the last question, The oscilloscope reads 6v when it only involves the internal resistor but when the 18ohm resistor is introduced the osciliscope reading goes from 6 to 5 which means that 1v went to the 18 ohm resistor but when you proceed with the calculations the internal resistor turns out to be 90ohms but thats far to high for internal resistance. Though when you assume that the 5v went to the 18ohm resistor then the internal resistance turns out to be 3.6 ohms
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    For 5.
    People whacking in a variable resistor in there to alter current?
    Also to get readings before the spike in graph would you turn diode to reverse bias
    And would resistance increase for increasing pd for last in 5?
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    Quarks decay through the weak interaction, and they stated that the K^+ meson decays. So it didn't need to be stated that it's a weak interaction
 
 
 
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