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Physics Aftermath

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  • Registered Users Posts: 39 Guinea


    Meh, twas hard enough, was only looking for a C. With 45 minutes study yesterday, I happened to study the right things, and I'm now hoping for a solid B. Qs 5 ad 11 were ridiculously simple though.


  • Registered Users Posts: 19 Adam We


    Carlowboy, what school you in???


  • Closed Accounts Posts: 129 ✭✭madnirvana


    DtotheK wrote:
    my omega = 32 ...pretty certain it's wrong tho.

    my K is what you are all gettin 39.somethin..


    madnirvana i see you want that answer pretty badly!! i threw out my paper and dont even know the question so sorry!


    its okay ;)


  • Registered Users Posts: 2,312 ✭✭✭Limerick Dude


    JC 2K3 wrote:
    I took it as being per year and got it wrong, but if you think about it, Watts is joules per second and therefore you multiply it by 60 to get joules per minute.


    do you not divide by 60 to get minutes, considering it will be a smaller number per minute than it would be for seconds.... :confused:


  • Closed Accounts Posts: 129 ✭✭madnirvana


    do you not divide by 60 to get minutes, considering it will be a smaller number per minute than it would be for seconds.... :confused:


    1 watt is 1 joul per second.

    and 60 seconds they asked for..so * by 60.


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  • Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    madnirvana wrote:
    anyone got the answer for

    10 a i???

    i get like 11580473 m/s
    eV = (1/2)mv^2

    v = sqrt(2eV/m)

    v = sqrt(2(1.6022*10^-19)(700*10^3)/(1.6726*10^-27))
    v = 11580473.99 m/s

    Correct you are madnirvana ;)


  • Registered Users Posts: 2,149 ✭✭✭ZorbaTehZ


    Q6 any takers?

    F=ma
    F=-ks

    (0.3)(9.8) = -k(0.085)
    2.94 = -k(0.085)
    -34.59 = k

    w^2 = -k/m
    w^2 = 34.59/0.3
    w^2 = 115.3
    w = 10.74
    T=2n/w
    T=2n/10.74
    T=0.59

    a=-w^2.X
    a(max)=-w^2.A
    a(max)=-(115.3)(0.025)
    a(max)=-2.8825

    Net force is zero at 0.085m (point of equilibrium)

    EDIT: nub


  • Closed Accounts Posts: 93 ✭✭microbiek


    any 1 do Q12 c what happens to the current when its replaced with a coil???


  • Closed Accounts Posts: 129 ✭✭DtotheK


    **** the last part was so easy!!!!!!!!!!!!!!!!!!!!!!!


  • Closed Accounts Posts: 279 ✭✭adam_ccfc


    microbiek wrote:
    any 1 do Q12 c what happens to the current when its replaced with a coil???
    It reduces, cause of self-induction.


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  • Registered Users Posts: 2,312 ✭✭✭Limerick Dude


    madnirvana wrote:
    1 watt is 1 joul per second.

    and 60 seconds they asked for..so * by 60.


    wait what was the question again, because i think i actually did that :p


  • Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    Oh yeah, Q12 (b), how do you get the sound intensity if they don't give you the area of the speaker??


  • Closed Accounts Posts: 129 ✭✭madnirvana


    JC 2K3 wrote:
    eV = (1/2)mv^2

    v = sqrt(2eV/m)

    v = sqrt(2(1.6022*10^-19)(700*10^3)/(1.6726*10^-27))
    v = 11580473.99 m/s

    Correct you are madnirvana ;)


    yayee

    thank u for your hard work.. i know writing maths in this forum is extremely painful:)


  • Registered Users Posts: 2,312 ✭✭✭Limerick Dude


    JC 2K3 wrote:
    Oh yeah, Q12 (b), how do you get the sound intensity if they don't give you the area of the speaker??

    welll they said yer man was 3m away from speaker, so i multiplied the power by 3 metres....


  • Closed Accounts Posts: 93 ✭✭microbiek


    JC 2K3 wrote:
    Oh yeah, Q12 (b), how do you get the sound intensity if they don't give you the area of the speaker??


    W/m^2 so 25/9 = 2.777778


  • Closed Accounts Posts: 51 ✭✭ron-burgandy


    That Q2 isn't in our book, but it was in an edition of Science Plus (the science version of Exam and Career guide) that I read yesterday. €25 well spent!

    The copper was heated by placing it inside a boiling tube covered in boiling water. Anything along those lines should be alright.

    I think they will have to mark some of these questions easy...


  • Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    madnirvana wrote:
    yayee

    thank u for your hard work.. i know writing maths in this forum is extremely painful:)
    No problem, to be honest I have no problem writing maths in text form, I guess I've just gotten used to it :p
    welll they said yer man was 3m away from speaker, so i multiplied the power by 3 metres....
    Yeah... that's wrong.... sure the unit you'd get there would be W m, not W m^-2, so....


  • Registered Users Posts: 2,312 ✭✭✭Limerick Dude


    That Q2 isn't in our book, but it was in an edition of Science Plus (the science version of Exam and Career guide) that I read yesterday. €25 well spent!

    The copper was heated by placing it inside a boiling tube covered in boiling water. Anything along those lines should be alright.

    I think they will have to mark some of these questions easy...

    hmm i think i guessed that :p


  • Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    microbiek wrote:
    W/m^2 so 25/9 = 2.777778
    I thought of that, but it doesn't make sense. Surely you need the area of the loudspeaker....


  • Registered Users Posts: 2,312 ✭✭✭Limerick Dude


    ugh, the more i read this, the more i realise i got loads wrong! aaaah


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  • Registered Users Posts: 2,149 ✭✭✭ZorbaTehZ


    JC 2K3 wrote:
    I got confused there, because the charge is supposed to be on the outside of the dome and the gave the diameter....

    I took d = 0.07....

    It's in my book, but I managed to leave it at school.
    This non-trivial result shows that a uniform spherical distribution of charge acts as a point charge when observed from the outside of the charge distribution; this is in fact a verification of Coulomb's law.

    http://en.wikipedia.org/wiki/Gaussian_surface


  • Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    Ah I see, perhaps I won't be counting that question then...

    (Although, since a Van de Graff generator has a rod through the bottom of it where the belts go, it's not a uniform sphere, is it?)


  • Closed Accounts Posts: 158 ✭✭madgal


    I wouldn't worry too much about it. They'll have to adjust the marking scheme to generate a specific percentage of A's B's and C's so you could do alright. Also, you have other subjects to rely on so one paper can't throw you off just like that.

    No. I need 2 B2's and 3C2s.

    My plans where to get B's in Chemistry and music and C's in french, physics and english.

    I just screwed up physics,so now I need to get an A in Chemistry to make up.


    Why did Xrays not come up? Just a 7 marker question in Q5. I was raging. I had that all learnt off and was hoping that may give me a few decent marks.
    Whoever wrote that paper should be sacked, shot and hung out to dry.

    Btw answer to semiconductor Q - Why is silicon a semiconductor ... I said because it is between a good conductor and a good insulator... please tell me I was right.


  • Registered Users Posts: 2,149 ✭✭✭ZorbaTehZ


    JC 2K3 wrote:
    Ah I see, perhaps I won't be counting that question then...

    (Although, since a Van de Graff generator has a rod through the bottom of it where the belts go, it's not a uniform sphere, is it?)

    Do you realize how difficult the math would be if we had to take that into consideration?


  • Closed Accounts Posts: 1,504 ✭✭✭Nehpets


    Anyone else think physics should be two papers? They could do it like theory and maths or by sections


  • Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    ZorbaTehZ wrote:
    Do you realize how difficult the math would be if we had to take that into consideration?
    :p


  • Closed Accounts Posts: 158 ✭✭madgal


    Nehpets wrote:
    Anyone else think physics should be two papers? They could do it like theory and maths or by sections

    No... perhaps a foundation paper for stupid stupid people like myself who can't even do an honours paper.


  • Posts: 16,720 ✭✭✭✭ [Deleted User]


    madgal wrote:
    No... perhaps a foundation paper for stupid stupid people like myself who can't even do an honours paper.

    Would be a bit silly imo having a foundation paper for a non-compulsary subject.

    /Aside: It's mad how I can't remember most of what was on the Physics course.


  • Closed Accounts Posts: 131 ✭✭Tomlowe


    the principle on which the definition of the ampere is based anyone? is it that a current carrying conductor in a magnetic field experiences a force proportional to the current? i sure hope so


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  • Registered Users Posts: 348 ✭✭nedward


    Current-carrying conductors exert a force on each other is what I said...your answer looks right anyway...


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