Turn on thread page Beta
    Offline

    0
    ReputationRep:
    6 i think.....anywayz wat did every1 write down for the energy transfer between the battery and lamp?
    Offline

    0
    ReputationRep:
    CRTs scan from left to right :eek: , whaa ! this wasnt in any of the physics modules syllabus, that is a bit unfair how do they expect us to know it!

    Unless of course we watched a tv program caled the history of television 8 years ago
    Offline

    0
    ReputationRep:
    (Original post by rastaman)
    how many marks was the inelastic scattering question?
    5 iirc

    that paper was quite easy overall IMO... im confident that ill get my A overall now
    scattering one was tough... and so was why does shadow on tv shift right... but rest was very easy imo...
    Offline

    0
    ReputationRep:
    (Original post by ---Master A---)
    6 i think.....anywayz wat did every1 write down for the energy transfer between the battery and lamp?
    Chemical potential--->Electric----->heat and light
    Offline

    2
    ReputationRep:
    I thought it was generally ok, there were just a couple of dodgy bits. I haven't studied deep inelastic scattering for ages so couldn't remember the details although I knew roughly what the differences were. What did everyone put for the difference between a beam of He ions from a Van de Graaff generator and a beam of alpha particles for a radioactive source?

    With the co-axial cable I could have done with 2 right hands!!:rolleyes: I was using the right and grip rule but with both hands so I ended up thinking that inside the insulator the magnetic fields from the sheath and the core would cancel out, rather than reinforce each other.

    For the radioactive power I got 10.9 years (I think) to drop to 15W - anyone else get this?
    Offline

    1
    ReputationRep:
    (Original post by Kabeer)
    Also the question about how that power is 20W :eek: , i just played arond with the numbers and the closest thing i got to 20 was 31, so i just wrote that.
    The next part i gueused as 9 years (3/4 of total power therefore half of half life).
    I got 7.47 years. I hope that's acceptable using powers of 2 rather than e^whatever it was.

    (Original post by Kabeer)
    The last quesiton was terrible, especially with the cable :confused: anyone know the answer to that one?
    I think the electric field is a radio field. The field outside the cable is in one direction; the field inside the cable in another direction. Don't know whether the copper wire is +ve or -ve, they probably would take a mark off me for that one. Then again, how was I to know which is positive.

    For the magnetic field inside the cable I said something about the fields close to the wire and the sheath being in opposite directions. and there will be a region where there's no apparent fields. The field outside the cable is main influnced by the current in the sheath.

    God I hate that paper.
    Offline

    0
    ReputationRep:
    (Original post by Camford)
    I got 7.47 years.
    yeah i got that too so did one of my mates
    Offline

    0
    ReputationRep:
    (Original post by Kabeer)
    Unless of course we watched a tv program caled the history of television 8 years ago
    lol
    Offline

    0
    ReputationRep:
    how did yu get 7... years?
    Offline

    1
    ReputationRep:
    (Original post by Kabeer)
    CRTs scan from left to right :eek: , whaa ! this wasnt in any of the physics modules syllabus, that is a bit unfair how do they expect us to know it!

    Unless of course we watched a tv program caled the history of television 8 years ago
    Apparently it has got something to do with the lawn. :eek:

    I think it's something to with the cathode ray osciliscope being scanning from left to right. But I agree it is an unfair question.
    Offline

    2
    ReputationRep:
    (Original post by S1M)
    yeah i got that too so did one of my mates
    Hmm. How did you all do that question?
    Offline

    0
    ReputationRep:
    i got the time to be 9 years, i assumed the power droped to 10W in one halflife, so 15W would be half the half life?
    Offline

    0
    ReputationRep:
    well i cant remember it all so i cant replicate my answer here...

    but the steps were
    work out power = energy per second
    work out lambda
    and use energy released per decay with lambda somehow
    then work out half life (convert to seconds...)
    and i did a few more things to piece it all together but i cant remember...
    Offline

    1
    ReputationRep:
    (Original post by Bezza)
    Hmm. How did you all do that question?
    you could use the energy per decay to find out how many atoms you'd need to be decaying at that instant, and then find out how many atoms would need to be present at that time. then used the decay constant to find out how long it would take to decay to that amount of atoms, got in the region 7.3 to 7.7 cant remember exactly. i think i did a long winded method
    Offline

    1
    ReputationRep:
    (Original post by Bezza)
    Hmm. How did you all do that question?
    Which should come out to be 7.47 or I would lose some marks.
    Offline

    0
    ReputationRep:
    (Original post by Hovick)
    i got the time to be 9 years, i assumed the power droped to 10W in one halflife, so 15W would be half the half life?
    no that doesnt work as the decay is not constant or in a linear relationship...
    Offline

    1
    ReputationRep:
    btw what did anyone put for the analogy of A=(lamda)N ? i had something stupid, mind went blank
    Offline

    1
    ReputationRep:
    (Original post by zoidberg)
    you could use the energy per decay to find out how many atoms you'd need to be decaying at that instant, and then find out how many atoms would need to be present at that time. then used the decay constant to find out how long it would take to decay to that amount of atoms, got in the region 7.3 to 7.7 cant remember exactly. i think i did a long winded method
    You could use the power of 1/2 method or the P=(P0)e^(lambda *t1/2)
    Offline

    1
    ReputationRep:
    (Original post by zoidberg)
    btw what did anyone put for the analogy of A=(lamda)N ? i had something stupid, mind went blank
    I=Q/(RC), I is dQ/dt, i.e the rate of discharge.

    A is dN/dt, i.e. the rate of decay.
    Offline

    0
    ReputationRep:
    (Original post by zoidberg)
    btw what did anyone put for the analogy of A=(lamda)N ? i had something stupid, mind went blank
    yup that one with V=(1/C) x Q i think
    edit... actually i think i put what the above pst says.... for some reason i forget it all... well anyway i should be focusing on next exam now... p4...
 
 
 

University open days

  1. University of Cambridge
    Christ's College Undergraduate
    Wed, 26 Sep '18
  2. Norwich University of the Arts
    Undergraduate Open Days Undergraduate
    Fri, 28 Sep '18
  3. Edge Hill University
    Faculty of Health and Social Care Undergraduate
    Sat, 29 Sep '18
Poll
Which accompaniment is best?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.