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Edexcel Physics Unit 2 "Physics at work" June 2013 Watch

  • View Poll Results: The last question - Does resistance increase or decrease?
    It increases ( using V=IR or some other method)
    70.73%
    It decreases using the 'lattice vibrations' theory
    29.27%

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    i can't remember that i divide the time by 2 or not :"( ...... if it is not ..so I will lose the mark the for this question T.T
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    (Original post by LegendX)
    did they ask for new thickness in the question? I thought all they wanted you to figure out was whether or not it had corroded? I put it had corroded but did it in terms of time, as if it had corroded, the time taken for ultrasound pulse to return would be lower...?
    you can show it in either way I first showed it by finding the new thickness which was less. and then I calculated the time it takes to travel 4cm. but I think I did the first method more clearly so might just get marks for that.
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    You definitely needed to divide the time by 2, as for using the decrease in Time to show it had corroded, It could be a valid method, it's certainly not wrong, but as they gave you an initial thickness I thought it'll probably be best to work off that


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    (Original post by tronyeu)
    i can't remember that i divide the time by 2 or not :"( ...... if it is not ..so I will lose the mark the for this question T.T
    even if u do it will be just one mark. u will get mark for using the right formula and saying that it corroded because thickness is less.
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    Okay and also, I still think that 1/2mv2 / VIt was the correct answer >.>, also guys, was the definition for wavelength 1 mark or two marks?
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    (Original post by LegendX)
    Did the question ask you to work out the amount that it had corroded by? I thought all it asked was whether or not it had corroded? So I measured the time taken which was far less than the time given and stated it must have corroded...
    I said that it had corroded and worked the new thickness to be 3.009cm. Meaning it had corroded by 0.991cm.
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    (Original post by LegendX)
    Okay and also, I still think that 1/2mv2 / VIt was the correct answer >.>, also guys, was the definition for wavelength 1 mark or two marks?
    it was one mark. luckily I went through it in the morning. I wrote the distance between the two adjacent points that are in phase.
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    (Original post by jollygood)
    even if u do it will be just one mark. u will get mark for using the right formula and saying that it corroded because thickness is less.
    thanks , sice I am a international student , so this is the first time i do the exam so I worry ..... i think i can get A , but I am not sure , i can full mark or not , i think it is really hard right ??? , in this exam , i can do everythings . i think it is not difficult for me
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    (Original post by CharlieTT)
    I said that it had corroded and worked the new thickness to be 3.009cm. Meaning it had corroded by 0.991cm.
    seems like u didn't divide the time by two that's why u got 3.o1 cm if u divide the time by two the new thickness is 1.5cm.. as thickness is reduced to large extent that's why the time difference was large too.
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    (Original post by jollygood)
    it was one mark. luckily I went through it in the morning. I wrote the distance between the two adjacent points that are in phase.
    what did u say about the 3-d glasses question lol?
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    Who can give me the Unoffical mark scheme please ? , i want to double check the answers T.T
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    Guys what did u answer for the Emf and internal resistance last part of question that asked why using a non negligible resistance ammeter wouldn't affect but using a low resistance voltmeter would affect the readings of EMF and r?
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    (Original post by LegendX)
    what did u say about the 3-d glasses question lol?
    that was so stupid questions I cant even recall what I wrote in them.. for intensity one I said that filter of one eye that has angle of 35 will allow light only parallel to plane and will cut out the rest of light so intensity will be reduced and films appear darker.
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    if we use the voltmeter which has low resistance , so the current can pass through the voltmeter , therefore the voltmeter and the resistor will share the current or the reading in the ammeter would be lower ? ( i'm not sure about this , if i wrong , who can tell me what is right answers ? )
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    (Original post by jollygood)
    that was so stupid questions I cant even recall what I wrote in them.. for intensity one I said that filter of one eye that has angle of 35 will allow light only parallel to plane and will cut out the rest of light so intensity will be reduced and films appear darker.
    I remember talking about how unpolarised contains a variety of different oscillations vibrating in different planes meaning that unpolarised light was brighter but polarised light cuts some of the wavelength out so it effectively redcues that brightness, kind of like reducing the flare in something
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    (Original post by tronyeu)
    if we use the voltmeter which has low resistance , so the current can pass through the voltmeter , therefore the voltmeter and the resistor will share the current or the reading in the ammeter would be lower ? ( i'm not sure about this , if i wrong , who can tell me what is right answers ? )
    I wrote exactly the same thing but I added the line in start that as voltmeter is connected in parallel so if it would have low resistance the current will split and flow through it as well and less current will flow through internal resistance.
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    whereas , the E.m.f has constant potential different , and voltmeter and resistor are parallel so the potential different in the resistor and the reading the voltmeter are the same , and when we use the voltmeter which has high resistance , so the current can not pass through the voltmeter so the reading in the ammeter would be higher ( i think so )
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    (Original post by LegendX)
    I remember talking about how unpolarised contains a variety of different oscillations vibrating in different planes meaning that unpolarised light was brighter but polarised light cuts some of the wavelength out so it effectively redcues that brightness, kind of like reducing the flare in something
    yeah its same as what I wrote that light will be cut out and intensity is reduced.
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    any ideas of what raw marks are needed for full ums?
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    (Original post by tronyeu)
    whereas , the E.m.f has constant potential different , and voltmeter and resistor are parallel so the potential different in the resistor and the reading the voltmeter are the same , and when we use the voltmeter which has high resistance , so the current can not pass through the voltmeter so the reading in the ammeter would be higher ( i think so )
    I thought that it would be because the voltmeter was connected in parallel with the internal resistance the overall resistance would be lower as 1/Rt = 1/R1 + 1/R2 -

    i take it thats wrong?
 
 
 
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