Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    9
    ReputationRep:
    Could you put that transverse waves can be polarised whereas longitudinal cannot? I put that as well as transverse waves oscillatile at 90deg to direction of travel
    Offline

    2
    ReputationRep:
    (Original post by zak_121)
    Gold leaf electroscope and a part Young's modulus; only asked for stress
    If I drew Searle's apparatus then said about a control wire, finding the cross sectional area with the micrometer when the spirit level is levelled with both, then find the weight by multiplying the masses of the hanger , weights added to break the wire and control weight then divide this by the area,
    said about eye-ware protection and putting a cardboard box absorbing the weight of a possibly bouncing weight will I get full marks?
    Offline

    1
    ReputationRep:
    (Original post by dylan1016)
    How can the resultant initial velocity be less the the horizontal component of 30ms^-1?
    Because if you use the exact values it will give you a horizontal component of 28.1 or something like that. Then we got given that vertical was 6.3 if I remember rightly. Then obviously just apply pythag and you get around 28.9. This is the more precise answer.
    • Thread Starter
    Offline

    5
    ReputationRep:
    (Original post by Harrysomers1)
    Because if you use the exact values it will give you a horizontal component of 28.1 or something like that. Then we got given that vertical was 6.3 if I remember rightly. Then obviously just apply pythag and you get around 28.9. This is the more precise answer.
    Yeah sorry I've confused myself.
    Offline

    7
    ReputationRep:
    (Original post by Harrysomers1)
    Because if you use the exact values it will give you a horizontal component of 28.1 or something like that. Then we got given that vertical was 6.3 if I remember rightly. Then obviously just apply pythag and you get around 28.9. This is the more precise answer.
    No offence; I think mine is even more accurate ayyy lmao. But yours is also extremely precise; only reason I say so is because I think you did 18/0.64 which gives you 28.1 to 3sf but i did 18/0.6422 ;P so when i did pythag I got 28.7ms^-1. You think I still got full marks, all the working is correct so surely even if they wanna be pricks about it then I should get the rest of the marks.
    • Thread Starter
    Offline

    5
    ReputationRep:
    You'd have to use 30 and not the horizontal, was the mass 0.160 or 0.180?
    Offline

    1
    ReputationRep:
    (Original post by dylan1016)
    Yeah sorry I've confused myself.
    It's all good pal, we will both get full marks I should think
    Offline

    1
    ReputationRep:
    (Original post by Parhomus)
    No offence; I think mine is even more accurate ayyy lmao. But yours is also extremely precise; only reason I say so is because I think you did 18/0.64 which gives you 28.1 to 3sf but i did 18/0.6422 ;P so when i did pythag I got 28.7ms^-1. You think I still got full marks, all the working is correct so surely even if they wanna be pricks about it then I should get the rest of the marks.
    Hahahah yeah you are more accurate, I do think there should be a range of answers allowed
    Offline

    7
    ReputationRep:
    (Original post by chemari1)
    If I drew Searle's apparatus then said about a control wire, finding the cross sectional area with the micrometer when the spirit level is levelled with both, then find the weight by multiplying the masses of the hanger , weights added to break the wire and control weight then divide this by the area,
    said about eye-ware protection and putting a cardboard box absorbing the weight of a possibly bouncing weight will I get full marks?
    I think so; I also wrote about repeating these measurements and taking an average; I forgot about putting cardboard box/reducing distance to the ground to avoid the weights causing damage.
    Offline

    1
    ReputationRep:
    (Original post by dylan1016)
    You'd have to use 28.7ms^-1 and not 30, was the mass 0.160 or 0.180?
    Pretty sure it was 0.16
    Offline

    2
    ReputationRep:
    (Original post by Parhomus)
    I think so; I also wrote about repeating these measurements and taking an average; I forgot about putting cardboard box/reducing distance to the ground to avoid the weights causing damage.
    thank you
    Offline

    7
    ReputationRep:
    (Original post by Harrysomers1)
    Hahahah yeah you are more accurate, I do think there should be a range of answers allowed
    Yeah, that's the funniest thing, me and my friend; we usually get good grades and the first thing we said when we came out was; I'm pretty sure there should be a range of values for the answers to those question. :P
    Offline

    1
    ReputationRep:
    (Original post by chemari1)
    If I drew Searle's apparatus then said about a control wire, finding the cross sectional area with the micrometer when the spirit level is levelled with both, then find the weight by multiplying the masses of the hanger , weights added to break the wire and control weight then divide this by the area,
    said about eye-ware protection and putting a cardboard box absorbing the weight of a possibly bouncing weight will I get full marks?
    I did pretty much same as you, bar the cardboard box
    Offline

    1
    ReputationRep:
    (Original post by Parhomus)
    Yeah, that's the funniest thing, me and my friend; we usually get good grades and the first thing we said when we came out was; I'm pretty sure there should be a range of values for the answers to those question. :P
    Yeah definitely there was so many show that's, but that's good cause you know if yours right so they're my favourite questions
    Offline

    2
    ReputationRep:
    (Original post by Harrysomers1)
    I did pretty much same as you, bar the cardboard box
    what did you put for the one about the photoelectric effect?
    Offline

    7
    ReputationRep:
    I mean for the kinetic energy at the beginning I got 66J which was a pretty nice answer; for the next part I got 3.14J and the last kinetic energy one I got 62.7J. And When I did 62.7+3.14 it was almost the same as 66J so I think I should get the marks D: By the way what did people write for how to check if the diameter had changed; I wrote about if the reading on the ohm-meter changed; because p and l are constant so if it changed it would mean cross sectional area had changed >>>diameter changed. Only one I wasn't sure about was why they used the metal blocks; I just wrote about it reducing contact resistance :C.
    Offline

    0
    ReputationRep:
    For all the confusion about the velocity question. The question specifically told us to work out the velocity using the answer that they give you for time which is 0.6s not what you get. That should give you 30ms. I am thinking due to the peculiar wording on that question, ocr will not mind if you used your own answer for time.
    Offline

    7
    ReputationRep:
    (Original post by chemari1)
    what did you put for the one about the photoelectric effect?
    For that I think I wrote about how the fact that UV did allow for the leaf to fall but not the white light showed the threshold frequency which the wave like nature couldnt explain because if it was due to the wavelike nature then the continuous flow of energy would allow emission of electrons. Since this didn't happen I said that it showed the wavelike nature; also for that question I wrote about how the increase in intensity which would apparently speed up the process if it was only wavelike in nature didn't make a difference supporting the threshold/particle like nature. Also I wrote about how increasing the intensity when it was above threshold freq showed that there was particulate nature for EM radiation because the process of the leaf falling back into place sped up; this shows the wave particle duality as it meant the number of photons per m^2 increased and so due to the one to one ratio if it was a particle would mean the number of electrons released per second would increase >>>>falls faster as shown in observations.
    Offline

    7
    ReputationRep:
    (Original post by dman2607)
    For all the confusion about the velocity question. The question specifically told us to work out the velocity using the answer that they give you for time which is 0.6s not what you get. That should give you 30ms. I am thinking due to the peculiar wording on that question, ocr will not mind if you used your own answer for time.
    I don't think so; if you used 0.6 s it would give you an answer of exactly 30ms^-1 and the question asked to show it was around 30ms^-1 not exactly 30.
    Offline

    2
    ReputationRep:
    (Original post by Parhomus)
    For that I think I wrote about how the fact that UV did allow for the leaf to fall but not the white light showed the threshold frequency which the wave like nature couldnt explain because if it was due to the wavelike nature then the continuous flow of energy would allow emission of electrons. Since this didn't happen I said that it showed the wavelike nature; also for that question I wrote about how the increase in intensity which would apparently speed up the process if it was only wavelike in nature didn't make a difference supporting the threshold/particle like nature. Also I wrote about how increasing the intensity when it was above threshold freq showed that there was particulate nature for EM radiation because the process of the leaf falling back into place sped up; this shows the wave particle duality as it meant the number of photons per m^2 increased and so due to the one to one ratio if it was a particle would mean the number of electrons released per second would increase >>>>falls faster as shown in observations.
    I put pretty much the same but that about intensity, I did refer to the results comparing them once, that of visible light at 4 cm making it fall slowly and uv light at 10 cm making it fall quick
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?

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

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