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How was Edexcel physics?? watch

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    (Original post by RazMaTaz)
    the bubble wasn't 6.8 ! thats what i put but then i found out the value for the underwater pressure was about 382 (or something) + 101 (**** !) so it came out at about 8.4 or something, bastered noob i am

    thats right... .the answer is 8.8cm3
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    (Original post by yipee)
    oh yeh...what did everyone get about the delta U = delta W + delta Q , one?
    U=0
    constant temp means no internal energy is lost or gained

    W= -2016 j
    all heat gained by electrical working is lost to the water

    Q = +2016 J
    continuous work done on the heating element by the power supply

    basically all u had to show is U = 0
    then W=-Q
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    (Original post by shift3)
    emf = V + Ir
    0.58 = 0.45 + 0.6r
    r = 0.22 ohms

    I didn't use the gradient.


    YESSSS ... papers werent that gr8... but did get this right... get the emf from the graph.. along with voltage and current.... and use the formula..

    V=E-Ir to get r
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    For the question of the alpha particle graph...this is what i did

    forces at X and Y are away from the nucleus....as the particle moves closer to P the decelerating force is greater... (for those doing A2 physics as well... f=kqQ/r2 from unit 5 can apply here.)... so it slows down.... and as it moves away.. there is an accelerating force... so the speed increases back to the same level at Y.
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    hey guys
    have done unit 2 about 3 hours ago
    for that question with the bubble, ive added the atmospheric pressure to result i got for p, so mmy answer for volume was 8.6cm³
    is it actually correct?

    What did you get for DeltaQ in question 3?

    peace
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    (Original post by mathematician)
    tht phy 3 was generally prett piss
    wot did ppl get the greater surface area for and the greatest intensity for.
    did ppl get gemini to be like 11.1 times greater.
    and disadvantage of on the ground was that radiation would be hindered when it arrives at the telescope
    deneb larger surface area and vega more intensity
    11.1 is rite ..
    i rote the friggin telescope wont move if its underground so it can only focus in one direction ....
    btw .. the advantage of a larger dish was to detect less luminous light sources in the sky ?
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    (Original post by infekt)
    U=0
    constant temp means no internal energy is lost or gained

    W= -2016 j
    all heat gained by electrical working is lost to the water

    Q = +2016 J
    continuous work done on the heating element by the power supply

    basically all u had to show is U = 0
    then W=-Q



    delta W can never be negative.... the question said the law applies to the COIl... work is done on the coil by the power supply...
    so delta W=+2016
    all the heat is loss by the system(coil) so delta Q(which stands for heat GAINED by the system) is -ve 2016.
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    (Original post by infekt)
    deneb larger surface area and vega more intensity
    11.1 is rite ..
    i rote the friggin telescope wont move if its underground so it can only focus in one direction ....
    btw .. the advantage of a larger dish was to detect less luminous light sources in the sky ?
    yer phy 1 and 3 were very nice exams actually,
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    oop my questions have already been answered!
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    (Original post by RussianDude)
    hey guys
    have done unit 2 about 3 hours ago
    for that question with the bubble, ive added the atmospheric pressure to result i got for p, so mmy answer for volume was 8.6cm³
    is it actually correct?

    What did you get for DeltaQ in question 3?

    peace

    thats right... i got 8.8..... u have to add the 101kpa pressure... cuz the question says that the formula given only gives the pressure INCREASE.... so to get the actual pressure u add the 101kpa.....
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    (Original post by [_Z_])
    delta W can never be negative.... the question said the law applies to the COIl... work is done on the coil by the power supply...
    so delta W=+2016
    all the heat is loss by the system(coil) so delta Q(which stands for heat GAINED by the system) is -ve 2016.
    my bad ... it was a typo
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    agh yeh im gona try and forget about it all now :cool:
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    (Original post by [_Z_])
    For the question of the alpha particle graph...this is what i did

    forces at X and Y are away from the nucleus....as the particle moves closer to P the decelerating force is greater... (for those doing A2 physics as well... f=kqQ/r2 from unit 5 can apply here.)... so it slows down.... and as it moves away.. there is an accelerating force... so the speed increases back to the same level at Y.
    yer thats wot i did, as particle approaches p its speed decreases due to the reulsive force from nucleus, then as it moves past it experiences a pushing force so the speed increases
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    (Original post by yipee)
    agh yeh im gona try and forget about it all now :cool:

    im with u bro.
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    WHAT did ppl get for the thickness of the lead? (phy 1)
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    (Original post by [_Z_])
    For the question of the alpha particle graph...this is what i did

    forces at X and Y are away from the nucleus....as the particle moves closer to P the decelerating force is greater... (for those doing A2 physics as well... f=kqQ/r2 from unit 5 can apply here.)... so it slows down.... and as it moves away.. there is an accelerating force... so the speed increases back to the same level at Y.
    there is no decelerating force...there is an accelerating force since both charges are positive and therefore repel each other...alpha has charge 2+ and the nuclues has a positive charge
    if you use F = kq1q2/r² when you reach closer to the nucleus,which means r is reduced and F is greater,therefore a larger force..
    when there is a larger force,there is an increase in acceleration..

    the graph should show an inverse V graph


    and as for the thickness of the graph,i know i had something x 10^-4 m
    i know thats the right answer,cause ive read from it somewhere..repeat from a different exam board
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    (Original post by MalaysianDude)
    there is no decelerating force...there is an accelerating force since both charges are positive and therefore repel each other...alpha has charge 2+ and the nuclues has a positive charge
    if you use F = kq1q2/r² when you reach closer to the nucleus,which means r is reduced and F is greater,therefore a larger force..
    when there is a larger force,there is an increase in acceleration..

    the graph should show an inverse V graph
    NO, thats wrong, your basically saying as the alpha particle approaches nucleus it is ATTRACTED TO IT?, how does that work,
    its repulsed, then as it goes passed the nucleus there is a puching force on the particle, so the line goes down a fraction( as it is very ionising), then the line drops, then almost instantly the line increases and carries along to the same point as the beggining
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    (Original post by MalaysianDude)
    there is no decelerating force...there is an accelerating force since both charges are positive and therefore repel each other...alpha has charge 2+ and the nuclues has a positive charge
    if you use F = kq1q2/r² when you reach closer to the nucleus,which means r is reduced and F is greater,therefore a larger force..
    when there is a larger force,there is an increase in acceleration..

    the graph should show an inverse V graph


    and as for the thickness of the graph,i know i had something x 10^-4 m
    i know thats the right answer,cause ive read from it somewhere..repeat from a different exam board

    the alpha partuicle was travelling from left towards P..... the force from P was towards the right.... so it decelerates..... after passing p.. the force from p is in the direction as its direction of travel..so it accelerates..
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    (Original post by RazMaTaz)
    the bubble wasn't 6.8 ! thats what i put but then i found out the value for the underwater pressure was about 382 (or something) + 101 (**** !) so it came out at about 8.4 or something, bastered noob i am
    yeah i got 8.82 too m8...that is right cos 344331 was the increase in pressure so u got 2.oocm^3*(334331+101000/101000
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    (Original post by MalaysianDude)
    there is no decelerating force...there is an accelerating force since both charges are positive and therefore repel each other...alpha has charge 2+ and the nuclues has a positive charge
    if you use F = kq1q2/r² when you reach closer to the nucleus,which means r is reduced and F is greater,therefore a larger force..
    when there is a larger force,there is an increase in acceleration..

    the graph should show an inverse V graph


    and as for the thickness of the graph,i know i had something x 10^-4 m
    i know thats the right answer,cause ive read from it somewhere..repeat from a different exam board
    i got 7x10^-4m
 
 
 
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