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    (Original post by sceezy)
    I have it downloaded, ill attach it below...
    Thanks very much!!
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    I just need 88 UMS in this resit. Last time I got 78, and I forgot a lot since then!
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    Is the only way to know whether something is weak or strong interaction to look at the strangeness of all particles and whether it has changed? How else do you know if it is strong or weak interaction? Cause I always pretty much have educated guesses but id rather have a rule to go by.
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    You will hopefully get more than 88 ums.
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    (Original post by StalkeR47)
    You will hopefully get more than 88 ums.
    The question is, why the existence of threshold frequency supports the particle nature of electromagnetic waves
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    (Original post by sceezy)
    Can anyone help with question 6b jan 2013 ?????

    PLEASE!!!
    I assumed your stuck with the third part of finding V across CD, so drew you a diagram... Name:  ImageUploadedByStudent Room1368975141.293229.jpg
Views: 114
Size:  97.9 KB

    Hope that helps


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    (Original post by Steroidsman123)
    Is the only way to know whether something is weak or strong interaction to look at the strangeness of all particles and whether it has changed? How else do you know if it is strong or weak interaction? Cause I always pretty much have educated guesses but id rather have a rule to go by.
    Ok... Here is the RULE! EVERY decay happens in weak interaction/weak nuclear force. Strong nuclear force is never responsible for any decay.
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    (Original post by posthumus)
    I assumed your stuck with the third part of finding V across CD, so drew you a diagram... Name:  ImageUploadedByStudent Room1368975141.293229.jpg
Views: 114
Size:  97.9 KB

    Hope that helps


    Posted from TSR Mobile
    Ok.. we have 4 resistors in series. 2 in each series. Total voltage is 12 volts. (Remember voltage in parallel is the same all the way around and in series it adds up to). Since the resistors are in series, they must have voltage adds up to total depending on the amount of resistance. There is an easy way to handle this problem and a harder way to solve. Easy way----- Voltage in parallel----same. So, across ac and ce, they must share same amount of voltage. so it must be 6V for both ac and ce. This is total of 6 Voltage used up in A-E. So you have 6 left. Since the resistance of the resistor is twice the resistance of the thermistor. Resistor BD must get twice the voltage as DF. So, BD must be 4 V and thermistor which is DF or CD must get 2V. so our voltage adds up to total. 6+4+2=12V. Harder way... calculate the current through AE by using V=IR. I=12/4xx10^3 = 3x10^-4A... Use V=IR again to calculate voltage across AC. V=3x10^-4 times 20x10^3 = 6V. The rest can be solved by using V=IR. Hope this helps... Haven't I just answered the 6marker?:cool::cool::cool::eek:
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    (Original post by StalkeR47)
    Ok... Here is the RULE! EVERY decay happens in weak interaction/weak nuclear force. Strong nuclear force is never responsible for any decay.
    Well that was helpful...thanks!
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    (Original post by Qari)
    The question is, why the existence of threshold frequency supports the particle nature of electromagnetic waves
    Give a definition-- minimum energy required by photon to eject an electron from the atom. Electrons can be emitted from a metal surface if given energy greater than or equal to the work function of the metal ( h * f(threshold) ). This energy can be supplied by incident light photons of energy h * f(threshold) or greater - the threshold frequency therefore being the minimum frequency that the incident light must have in order to liberate electrons. This is shown to be the case as if f < f(threshold), no electrons are emitted, regardless of the intensity of the incident light. The light must therefore act as particles (photons) whose energy = h * f. Increasing the intensity if f is greater than or equal to f(threshold) will increase the rate of emission of electrons, indicating that the number of incident photons per second has increased. Each electron can absorb only one photon.
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    (Original post by posthumus)
    I assumed your stuck with the third part of finding V across CD, so drew you a diagram... Name:  ImageUploadedByStudent Room1368975141.293229.jpg
Views: 114
Size:  97.9 KB

    Hope that helps


    Posted from TSR Mobile
    Woops! Made a huge mistake

    Bottom set adds up to 10V


    The different seems to be 4V but because the current splits up on both ends you half this value (since...V=IR)


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    (Original post by posthumus)
    I assumed your stuck with the third part of finding V across CD, so drew you a diagram... Name:  ImageUploadedByStudent Room1368975141.293229.jpg
Views: 114
Size:  97.9 KB

    Hope that helps


    Posted from TSR Mobile
    I'm also not seeing how you figured out to do 14-12 why not 12-14?
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    (Original post by usycool1)
    As I said a few pages back, you will only lose significant figures if the question explicitly state that you need to give your answer to an appropriate number of sig figs. So no, you don't lose marks on those.
    Thanks for the clarification. Makes sense but the mark scheme really confused me.
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    (Original post by BenChard)
    I'm also not seeing how you figured out to do 14-12 why not 12-14?
    Either way is correct: Potential difference means exactly that, the difference between 14 and 12 is 2 and the difference between 12 and 14 is still 2. The signs doesn't matter.
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    (Original post by StalkeR47)
    Give a definition-- minimum energy required by photon to eject an electron from the atom. Electrons can be emitted from a metal surface if given energy greater than or equal to the work function of the metal ( h * f(threshold) ). This energy can be supplied by incident light photons of energy h * f(threshold) or greater - the threshold frequency therefore being the minimum frequency that the incident light must have in order to liberate electrons. This is shown to be the case as if f < f(threshold), no electrons are emitted, regardless of the intensity of the incident light. The light must therefore act as particles (photons) whose energy = h * f. Increasing the intensity if f is greater than or equal to f(threshold) will increase the rate of emission of electrons, indicating that the number of incident photons per second has increased. Each electron can absorb only one photon.
    Oh thanks I understand it now
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    (Original post by NedStark)
    Some one else might explain it better but here goes:

    The current in the diagram is going clockwise, so diode X is forward biased and diode Y is reverse bias.
    Diode X lets only 0.6v through and since it's in parralel with Q (10k resistor), then Q also has 0.6v.
    Therefore Y(diode) must have 2.4v, and once again it's in parralel so P has 2.4v.

    That's the pd's calculated for all 4. Next is the currents:

    Since Y is reverse bias, it doesn't let current through so for that parralel combination, all current passes P. Therefore current through Y is 0A.
    Use V=IR to calculate the current through P. This gets 0.48mA (i think)

    0.48mA must also go through the combination of Q+X. Use V=IR to calculate the current through Q which is 0.42mA, the remaining 0.06mA goes through diode X.

    Sorry if that isn't clear, it's basically me typing my thought process.
    Thanks a lot!
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    (Original post by Qari)
    Oh thanks I understand it now
    Your most welcome
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    (Original post by dada55)
    Either way is correct: Potential difference means exactly that, the difference between 14 and 12 is 2 and the difference between 12 and 14 is still 2. The signs doesn't matter.
    fantastic, thanks
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    (Original post by SpartASH)
    Does anyone know how to do question 4 on page 69 of the official AQA Physics book?
    just did this question yesterday. HERE.... sorry for the poor quality...
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    (Original post by posthumus)
    It's the work done per coulomb charge when there's no resistance including internal resistance.

    The voltage equals the EMF when current equals zero

    V= EMF - Ir

    I think that's about all you need to know
    Its a bit easier if you understand it in relation to potential difference.

    The potential difference is the energy transferred per coulomb through any two points in a circuit whereas the electromotive force is the electrical energy per unit charge PRODUCED by the source.

    Also the mark scheme definition is “work done per unit charge".
 
 
 
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