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    (Original post by Zakee)
    What about a neutron-neutrino interaction where the W- boson causes the neutrino to change into an electron? Or a proton - antineutrino interaction where the W+ boson causes the antineutrino to form a postiron?
    Interaction is always either w+ or w - in decay. Proton ---- turns into a neutron. So a hadron is responsible for weak since they decay through weak interaction. Not interact.
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    (Original post by Zakee)
    What about a neutron-neutrino interaction where the W- boson causes the neutrino to change into an electron? Or a proton - antineutrino interaction where the W+ boson causes the antineutrino to form a postiron?
    Exactly, hadrons do interact through both the strong and weak interaction.. also a decay is just a type of interaction.
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    (Original post by Zakee)
    It all depends on the country. What people must realize is because the people who usually comment here roughly A/B boundary, they're going to say 'oh the exam was easy/moderately easy/fine'. So the grade boundaries here may be augmented slightly. I remember last year during my GCSEs, people here were saying, exam was easy, so 90 UMS will be 54/60. Surprisingly, 50/60 was 90 UMS. It all does truly depend.

    My personal belief is that this paper will be 67-68/70 for 120 UMS.
    Yes that's seems about right. Here everyone says it was easy but more people will probably find it difficult! We'll only find out when the grade boundaries come out I suppose
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    http://www.thestudentroom.co.uk/show....php?t=2357624

    Unofficial mark scheme
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    (Original post by SamHedges)
    The question asked what affect would decreasing the frequency so does that agree with what I wrote, I think that's what I was trying to say. If you imagine the threshold frequency of a metal was 6\times10^{14} for example, at this frequency the photons would only have energy to remove the electrons nearest to the surface as the photons absorbed by electrons further in would only cause them to collide with other particles within the metal and lose their KE.

    Increasing the frequency then to 7\times10^{14} say, will mean that electrons further from the metal's surface will have energy sufficient to leave the metal so more electrons emitted when freq increased?

    I looked at a quick interactive animation and increasing frequency meant higher photoelectric current.
    Oh I thought you were talking about the max KE question. Decreasing the frequency would mean the emitted electrons have a lower kinetic energy. If you decreased it enough, then yes, less electrons would be emitted and then eventually zero when it goes below f0.
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    (Original post by Blayney)
    Exactly, hadrons do interact through both the strong and weak interaction.. also a decay is just a type of interaction.
    Not sure about that, I think an interaction requires 2 parties, whilst a single particle can decay solo.
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    Thought it was a great exam
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    this exam was so nice, wasn't too difficult
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    (Original post by .raiden.)
    Why do you specifically need 114 may I ask?
    I hope to apply to Cambridge for medicine and need 95% ums according to my careers advisor
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    (Original post by Charlottecl)
    this exam was so nice, wasn't too difficult
    Anyone have the actual paper, if so I will do it then upload it?
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    (Original post by GeneralOJB)
    Did you calculate the power transformed by the battery?
    yes
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    (Original post by pilotluke1)
    yes
    As long as you showed it was equal to the sum of the power of the components in the circuit that you put in the table, you should get full marks.
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    Walking out of the exam and hearing people's answers: :sigh:
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    It's AS physics, you won't get marked down for writing interaction or force. In this syllabus, hadrons are affected by both the strong and weak nuclear forces, whereas leptons are only affected by the weak nuclear force.
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    I put a similarity as that both leptons and hadrons are made up of fundamental particles, which is true... however I feel that's a bit fluffy...
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    (Original post by StalkeR47)
    Interaction is always either w+ or w - in decay. Proton ---- turns into a neutron. So a hadron is responsible for weak since they decay through weak interaction. Not interact.
    I don't know why you are so keen on trying to flaw other people's answers, but I'm 99.9% sure that if you wrote," both Hadrons and leptons interact using the weak force", that you will be right and you will get a mark (if there is 1 for it anyway).

    Electron capture is an obvious example where a hadron and a lepton interact through the weak interaction force.

    In both the AQA official textbook and the CGP revision guide, it says that all particles experience the weak interaction.

    Yes it is true that hadrons decay through the weak interaction but it's still correct to write that they interact/experience the weak interaction.
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    I really can't tell how I've done. I flew through it but that means its gone either brilliantly or terribly
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    (Original post by Mr.Suhk)
    1) 20 p, 28 n, 18 e
    2) + 3.2 x 10 ^-19 J
    3) 3.99 x 10 ^ 6 C/kg
    4) feynman stuff ( d quark, W+ Boson, positron, electron neutrino)
    5) weak interaction
    6) B
    7) mass/charge/range etc
    8) I did baryon number and electron lepton number conservation but there are mpre
    9) big 6 marker, pretty easy
    10) particle/ antiparticle: i did electron and positron and said charge was opposite

    11) max and range of KE: photons energy dependant upon freq. as E= hf. Ekmax = hf - phi thereofore max KE exists. Range of ke's because more energy needed to remove deeper electrons etc.
    12) prove 1.8eV (pretty easy)
    13) what is threshold freq. : i got like 4.4 x 10 ^14 Hz
    14)increasing frequency increases max KE of emitted electrons as by equation or by photon energy route, your choice
    15) Double intensity doubles number of photons per second which doubles number of emitted electrons per second

    16) peak to peak = 128V
    17) peak = 64V
    18) Vrms = 45 V
    19) freq. = 100 Hz
    20) horizontal line at 45V
    21) time base: 2ms/div y sensitivity: 16V/div dunno about 3rd mark

    22) 4.2 x 2.5 = 6.3V
    23) 12- 6.3 = 5.7V
    24) 5.7 / 2 = 2.85A
    25) 4.2 - 2.85 = 1.35 A
    26) R = 5.7 / 1.35 = 4.2
    27) I said total R is 1.4 but if it incuded internal resistance then it was 2.9 ohms
    28)think something after but forgot sorry

    29) current = 6/ (50 + 35 + 5 ) x 1000 = 6.7 x 10 ^-5 A
    30) V = 0.33V
    31) increas intensity decreases toatl resistance which increases total current thereofre voltemeter value is greater (alternatives possible)
    32) 30000 ohms
    Are you sure about the answers to 30/31? I agree with the 31 and I got 30,000 ohms for 32. But for 29 I got like 1x10^-4 A, and for 30 I got 0.51V. Any one else agree?
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    When it said shows how the reaction is conserved with that feynam diagram what did people put I did:

    Charge and lepton number and the showed it by writing the equation of:

    P ----> n + e+ + neutrino


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    for question 7a i did 12 divided by 90000, which led me to get part B wrong will i still get marks for part B?
 
 
 
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