Join TSR now and get all your revision questions answeredSign up now

AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread] Watch

    Offline

    2
    ReputationRep:
    (Original post by Dante991)
    I'm looking for someone to show me how to do them too, absoulte killer paper
    I hope ours is much more straight forward but I doubt it considering how my exam's have gone so far this year, I've just lost all motivation to revise -_-
    Offline

    2
    ReputationRep:
    I'm going to fail ...
    Offline

    3
    ReputationRep:
    Name:  1433964923046-1519872357.jpg
Views: 82
Size:  511.7 KB
    (Original post by C-king)
    BUMP
    Offline

    2
    ReputationRep:
    (Original post by DanielWall96)
    I'll try help link me the questions
    http://filestore.aqa.org.uk/subjects...1-QP-JUN14.PDF

    19, 20 & 21 please, thanks.
    Offline

    0
    ReputationRep:
    (Original post by samlyon)
    http://www.tomred.org/uploads/7/7/8/...b_specimen.pdf
    http://www.tomred.org/uploads/7/7/8/...4_specimen.pdf

    Can anyone explain 2b please, how do you know what the phase difference is from the frequency?
    Anyone? I'm just confusing myself
    Offline

    2
    ReputationRep:
    so each charge is providing half the potential
    25/2 = 12.5

    The potential from P is doubled as the distance is halved so = 25
    The potential from Q is decreased by 1/1.5 so becomes 8.333333333
    25 + 8.3333333= 33.3

    C
    Offline

    1
    ReputationRep:
    (Original post by QueNNch)
    Hey guys does anyone know how to calculate the time period for this:
    January 2011



    The correct answer is B but my calculations can never get that answer.
    Apparently we are supposed to use ω2r = 9.81

    Thanks
    If the object is to appear weightless, then the centripetal force needs to be equal to the weight, so that the resultant force is zero. Therefore mg = mw2r. The masses cancel each other out, so that gives you w2r = 9.81. Then substitute 2pi/T for w and rearrange to get the answer.
    Offline

    3
    ReputationRep:
    Thanks a lot JaySP C-king DanielWall96 Klaxoii AR_95 jh52
    You all were really helpful. Thanks a lot!
    It all makes sense now

    Rep'd
    Offline

    2
    ReputationRep:
    (Original post by JaySP)
    Name:  1433964923046-1519872357.jpg
Views: 82
Size:  511.7 KB
    ahh thanks, I was doing this but thought it was too much working for one mark, these questions are too deceptive
    Online

    19
    ReputationRep:
    (Original post by C-king)
    ahh thanks, I was doing this but thought it was too much working for one mark, these questions are too deceptive
    Don't ever think that with MC. It can vary from plugging into an equation without doing any work, to 3 or 4 step questions
    Offline

    0
    ReputationRep:
    Hi
    Does anyone have any ideas about what the 6 marker could be on?
    Offline

    15
    ReputationRep:
    (Original post by C-king)
    ahh thanks, I was doing this but thought it was too much working for one mark, these questions are too deceptive
    Thanks for this, I started off correctly but didnt know you divided the 2 equations
    Offline

    3
    ReputationRep:
    Name:  14339653568131469626839.jpg
Views: 131
Size:  502.4 KBAttachment 425043425047Attachment 425043425047425000 Sorry some are clearer than others
    Attached Images
      
    Offline

    3
    ReputationRep:
    (Original post by C-king)
    Attachment 424989
    Can someone pls help me on this have no clue
    using E=0.5CV^2 derive two equations (leaving units in microfarads) so you get

    3200 = CVsquared

    800 = C(V-2)squared

    you got two simultaneous divide first eqn by V and the sub into 2 hence you get v = 4 and then get C from eqn 1 again (although I would rather stick with plotting each of those 4 answers in the eqn and check by verification)
    Offline

    1
    ReputationRep:
    (Original post by C-king)
    BUMP
    May have been answered, Is the answer B?
    Offline

    1
    ReputationRep:
    (Original post by samlyon)
    Anyone? I'm just confusing myself
    Think about moving a spring very slowly. You'd barely be affecting the oscillations which is what its getting at with 0.2 Hz.

    1.5hz is resonance as its the same as the natural frequency.

    As for 10Hz, you must remember those resonance curves, right? Think of what happens to the amplitude as the driver frequency exceeds the natural frequency.

    Edit: I didn't pay attention to your question. When in resonance , the driver and the displacement of the oscillation system are 90 degrees out of phase. It's just a fact.
    Offline

    0
    ReputationRep:
    Are we allowed Graphics calculators in the physics exams?
    Offline

    0
    ReputationRep:
    (Original post by AnneOldman)
    Does anyone know any examples of the different types of damping?

    Posted from TSR Mobile
    Light damping:
    person on a swing

    Heavy damping:
    used in some earthquake proof buildings

    Critical damping:
    used in suspension systems, i.e shock absorbers

    Hopefully this helped.

    Offline

    2
    ReputationRep:
    (Original post by JaySP)
    Impulse=∆momentum=kgms-1=Ns
    Thanks mate is there ever a unit kgms-2


    Posted from TSR Mobile
    Offline

    1
    ReputationRep:
    Someone give me a question they can't do, multiple choice is my favourite!
 
 
 
Poll
If you won £30,000, which of these would you spend it on?

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

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