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    how did u lot find it....i messed up the first q and the q about replacing the resistor....were we meant to replace the voltmeter (v1) or the resistor???
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    The resistor, i figured they weren't giving us 2 marks for just redrawing the circuit, so i created a potential divider as otherwise how would v2 be 3v
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    (Original post by sequence123)
    The resistor, i figured they weren't giving us 2 marks for just redrawing the circuit, so i created a potential divider as otherwise how would v2 be 3v
    Ye thats what i figured as well
    I drew a normal resistor + R connected to the power supply.
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    Surely the normal resistor is just v1,as it has a resistance of 10MOhms?
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    my answers for phy2

    1. base unit - ampere
    derived quantity - resistance
    derived unit - ohm
    base quantity - current

    2.
    a) 14.5 sec
    b) 2.7 x 10^26
    c) away from jupiter, towards IO

    3.
    a) 6.0 V
    b) 12 ohms
    c) 1 ohm

    4.
    a) 2 A
    b)current on y axis, voltage on x axis, s type shape line curving towards straight at both ends
    c) temp increases, so resistance increases etc

    5.
    a) temp rate = 0.1 K per sec
    b) water is hot lossing heat to surrounding, energy used to prevent drop in temp etc
    c) i) same shape but steeper start going to higher temperate
    ii) acts as an insulator, wont loose heat to surroundings as rapidly

    6.
    a)-how many moles of gas there is
    -molar gas constant
    b) zero on the kelvin temperature/ -273 degree C
    c) n = 2665.2 mol

    7.
    a) V1 should be the ammeter
    b) high resitance V1 results in tiny current flowing to the resisotor
    c) circuit is same but with R resistor in place of the old one
    R= 10Mohms

    8.
    a) T1 - initial temperature
    T2 - final temperature
    b) initial temp = 213 degree C
    c) make T2 = 0 degree C

    9.
    a) brownian motion, smoke diffuses, mixes completely with air etc, evidence for random motion

    b) diagram of partical going in several directions on after the other.
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    Me silly! for the question tht asks from which direction to where does the charge moves, i wrote from negative to positive! :eek:
    reckon i gota repeat phyiscs all the units!!
    man i dn knw why but i was tooo sleeepy during the paper
    ask me for tips on how to revise next time! i can give 100's of DONT'S!
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    (Original post by StyleLover)
    Me silly! for the question tht asks from which direction to where does the charge moves, i wrote from negative to positive!
    thats what i put as well.
    Current does flow from negative to positive dosent it?
    I mean i wrote from IO to jupiter
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    yes current is defined as the flow of POSSITIVE charde, as they originally thought electrons went the other way, stupid i know but it has stuck, look it up in your physics text book
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    (Original post by tktaylor6)
    my answers for phy2
    5.
    c) i) same shape but steeper start going to higher temperate

    7.
    c) circuit is same but with R resistor in place of the old one
    R= 10Mohms

    8.
    b) initial temp = 213 degree C
    i'm not sure about the ones still remaining above:

    5(c)(i) should the temperature not rise to the same level??

    7(c) nearly 100% sure R = 50 Mohms

    8(b) 521°C
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    for 7c) I got 5*10^6 ohms...or 5 million ohms
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    (Original post by fisfos815)
    for 7c) I got 5*10^6 ohms...or 5 million ohms
    I got that too
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    (Original post by tktaylor6)
    8c) make T2 = 0 degree C
    I said you had to raise the temperature of the hot source, becuase water boils at 100°C so the temperature of the cold sink couldn't be lowered any more.
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    how about the intial temp of the heat engine: 521 degC??
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    7c) I=V1/R
    I=(9-3)/10M
    I=6*10^7 A

    R=V2/I
    R=3/6*10^7
    R= 5 M ohms
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    (Original post by Scottus_Mus)
    I said you had to raise the temperature of the hot source, becuase water boils at 100°C so the temperature of the cold sink couldn't be lowered any more.
    how about condense the steam, so that the temp of the cold sink becomes lower???
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    I got -1.9 but thats wrong....I did everything right except I forgot to divide the 53% by 100...
    How many marks would I loose for that?
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    i think it was out of three: you should get both method marks assuming you put the temp in the correct place and converted to K... you will only lose the answer mark.
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    some people approached it as a potenttial divider question, i approached to find the current in the resistor V2 component, then used this and the p.d to find total R, then a simple reistors in parallel approach, and the intial temp was just sopme simple algebra , also i put a higher temp , because it appeared to still be rising in the previousgraph,
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    i did it as a potential dividor question, but you end up with the same answer either way: 5 M ohms
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    well heres how i did it

    1. realises if a p.d of 3 volts is across the reistor, V2 combo, then 6V is across the voltmeter of resistance 10Mohm, use I = V/R to obtain the current of 6 x 10^-7 A
    2. the current into the resitor V2 combo is 6 x 10^-7 and the p.d is 3V, so the total resistance of the combo is 5 x 10^6 ohms

    1/ (5 x 10^6) = 1/R + 1/ (10 x 10^6) = 2 x 10^-7

    1/ R = 1 x 10^-7

    R = 10 Mohms
 
 
 
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