Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie

**Physics...Before/After**

Options
1…1012141516

Comments

  • Options
    #332
    clio94
    Registered Users Posts: 18 ✭clio94


    Wesc. wrote: »
    This is how I did it.
    First I got the initial speed by using v=u+at where u=0, a= 9.7552 (I think?) and t=13
    Then I subtracted the total distance from 5000 and my answer for (iii)
    Then I used s=ut + 1/2at^2 to get a (wasn't sure on this)
    Then I used v=u +at again to get final speed and then added that to initial and divided by 2

    Yeah I did that exactly the same .. But wasn't the acceleration changed because of the parachute?
    Think we over complicated it .. Must have just been displacement/time :confused:


  • Options
    #333
    Tankosaur
    Registered Users Posts: 102 ✭✭Tankosaur


    mathstalk wrote: »
    God, why couldn't you have been my Physics teacher!


    I can't possibly be, I'm just a figment of your imagination.


  • Options
    #334
    SD021
    Registered Users Posts: 31 ✭SD021


    Tankosaur wrote: »
    The question said after the first 13 seconds " he fell 5km in the next 4 minutes and 36 seconds"

    To find average speed you just divide distance by time

    5000/2760 = 18.1 metres

    Where are you getting that? O.o

    AFAIR it brought him down to 5km. He didn't fall 5km.. I hope so anyway. Lemme find the paper.


  • Options
    #335
    mathstalk
    Registered Users Posts: 162 ✭✭mathstalk


    subz3r0 wrote: »
    Yes it does.



    You draw the line of best fit on the thing itself, shove it in your booklet and voila.

    It doesn't say line of best fit, it says "appropriate graph". Not only does the question not specify, it makes no sense.


  • Options
    #336
    finality
    Registered Users Posts: 1,763 ✭✭✭finality


    Wesc. wrote: »
    This is how I did it.
    First I got the initial speed by using v=u+at where u=0, a= 9.7552 (I think?) and t=13
    Then I subtracted the total distance from 5000 and my answer for (iii)
    Then I used s=ut + 1/2at^2 to get a (wasn't sure on this)
    Then I used v=u +at again to get final speed and then added that to initial and divided by 2

    I think I did something along those lines too, I ended up doing (v + u)/2

    I think that's wrong though?


  • Advertisement
  • Options
    #337
    Tankosaur
    Registered Users Posts: 102 ✭✭Tankosaur


    SD021 wrote: »
    Where are you getting that? O.o

    AFAIR it brought him down to 5km. He didn't fall 5km.. I hope so anyway. Lemme find the paper.

    Ara feck I misread it.

    Edit: so the answer is just 26000/ ( 4x60 + 13 + 36)

    Same method just different values


  • Options
    #338
    donvito99
    Registered Users Posts: 5,818 ✭✭✭donvito99


    Thought the paper was grand apart from Q6, and looking at these answers now I'm thinking I did alot worse in that question than I thought.


  • Options
    #339
    DesmondGAF
    Registered Users Posts: 51 ✭✭DesmondGAF


    Tankosaur wrote: »
    The question said after the first 13 seconds " he fell 5km in the next 4 minutes and 36 seconds"

    To find average speed you just divide distance by time

    5000/2760 = 18.1 metres

    I hope I didn't make a stupid error, but Isn't 4 mins 36seconds just 276 seconds?


  • Options
    #340
    Wesc.
    Registered Users Posts: 715 ✭✭✭Wesc.


    finality wrote: »
    I think I did something along those lines too, I ended up doing (v + u)/2

    I think that's wrong though?

    I think it's right. At one stage I tried to do it like applied maths Q10 but it didn't really work out :P But no I think what I did was right anyway.


  • Options
    #341
    SD021
    Registered Users Posts: 31 ✭SD021


    Tankosaur wrote: »
    Ara feck I misread it.

    Edit: so the answer is just 26000/ ( 4x60 + 13 + 36)

    Same method just different values

    You had to away the value you got in the previous part from the distance aswell


  • Advertisement
  • Options
    #342
    ECM2012
    Registered Users Posts: 10 ✭ECM2012


    SD021 wrote: »
    Tankosaur wrote: »
    The question said after the first 13 seconds " he fell 5km in the next 4 minutes and 36 seconds"

    To find average speed you just divide distance by time

    5000/2760 = 18.1 metres

    Where are you getting that? O.o

    AFAIR it brought him down to 5km. He didn't fall 5km.. I hope so anyway. Lemme find the paper.

    I got 18.1 too, assumin he fell from rest


  • Options
    #343
    michaelm82803
    Registered Users Posts: 56 ✭✭michaelm82803


    'Does the resistance of the diode remain constant during the investigation? Justify.'
    Anyone know the answer for this? I'm not 100% sure on it...


  • Options
    #344
    iLaura
    Registered Users Posts: 619 ✭✭✭iLaura


    michaelm82803 wrote: »
    'Does the resistance of the diode remain constant during the investigation? Justify.'
    Anyone know the answer for this? I'm not 100% sure on it...

    That part caught me out too when I was reading the paper, so I just didn't do it, so unfortunately I can't help. For some reason I think not though, as it'll heat up and resistance will fall? idek


  • Options
    #345
    Wesc.
    Registered Users Posts: 715 ✭✭✭Wesc.


    michaelm82803 wrote: »
    'Does the resistance of the diode remain constant during the investigation? Justify.'
    Anyone know the answer for this? I'm not 100% sure on it...

    No, I just subbed it the I and V values into R=V/I and got various resistance values and that's how I proved it.


  • Options
    #346
    ECM2012
    Registered Users Posts: 10 ✭ECM2012


    michaelm82803 wrote: »
    'Does the resistance of the diode remain constant during the investigation? Justify.'
    Anyone know the answer for this? I'm not 100% sure on it...

    I said no because the graph wasn't a straight like through the origin which would imply constant resistance


  • Options
    #347
    finality
    Registered Users Posts: 1,763 ✭✭✭finality


    michaelm82803 wrote: »
    'Does the resistance of the diode remain constant during the investigation? Justify.'
    Anyone know the answer for this? I'm not 100% sure on it...

    No it doesn't. if it did the graph would be a straight line through the origin as V=RI so I/V = 1/R.


  • Options
    #348
    BL1993
    Closed Accounts Posts: 341 ✭✭BL1993


    michaelm82803 wrote: »
    'Does the resistance of the diode remain constant during the investigation? Justify.'
    Anyone know the answer for this? I'm not 100% sure on it...
    Said no because line is not straight line through origin. Current only properly flows through it when junction voltage and depletion layer is overcome meaning a great hange in resistance...

    Or something like that. :P


  • Options
    #349
    finality
    Registered Users Posts: 1,763 ✭✭✭finality


    BL1993 wrote: »
    Said no because line is not straight line through origin. Current only properly flows through it when junction voltage and depletion layer is overcome meaning a great hange in resistance...

    Or something like that. :P

    Yeah I said something about junction voltage too, I'd say the marks are for the observation based on the graph though


  • Options
    #350
    clio94
    Registered Users Posts: 18 ✭clio94


    But weren't they looking for the average velocity for the time taken to drop the 276 seconds after the first parachute was opened ..

    Like wouldn't you need the velocity when it was opened & then again when he had dropped for 276 seconds .. And then just v-u / 2 ??


  • Options
    #351
    Tankosaur
    Registered Users Posts: 102 ✭✭Tankosaur


    michaelm82803 wrote: »
    'Does the resistance of the diode remain constant during the investigation? Justify.'
    Anyone know the answer for this? I'm not 100% sure on it...



    resistance = voltage/current which is the slope of the graph.

    The slope is not a straight line therefore the resitance is not constant.


  • Advertisement
  • Options
    #352
    Wesc.
    Registered Users Posts: 715 ✭✭✭Wesc.


    clio94 wrote: »
    But weren't they looking for the average velocity for the time taken to drop the 276 seconds after the first parachute was opened ..

    Like wouldn't you need the velocity when it was opened & then again when he had dropped for 276 seconds .. And then just v-u / 2 ??

    Yes, which was what I did? :L And did you mean to say v+u/2 ?


  • Options
    #353
    An0n
    Registered Users Posts: 187 ✭✭An0n


    I'm delighted with that paper.
    Not quite the A but I reckon I got a high B.

    That was my last exam too!
    Just back from town got a few bottles chilling in the fridge; off to a friend's house for predrinks then into town for a piss up.

    Too bad it's f*cking raining haha.

    Hope everyone did well; don't dwell on questions that you may have messed up on though. There's no point! It's over now! :] Good luck for August.


  • Options
    #354
    Physics FTW
    Registered Users Posts: 14 ✭Physics FTW


    Is it just me or was that entire paper phrased badly?

    Also what did you guys say for Q4 the last part asking about if the diode was reverse biased?


  • Options
    #355
    BL1993
    Closed Accounts Posts: 341 ✭✭BL1993


    Physics FTW wrote: »
    Is it just me or was that entire paper phrased badly?

    Also what did you guys say for Q4 the last part asking about if the diode was reverse biased?
    reverse direction of diode, replace milliammeter with microammeter and connect voltmeter between microammeter and diode. Drew a diagram too. :)


  • Options
    #356
    Wesc.
    Registered Users Posts: 715 ✭✭✭Wesc.


    Physics FTW wrote: »
    Is it just me or was that entire paper phrased badly?

    Also what did you guys say for Q4 the last part asking about if the diode was reverse biased?

    You had to place the ammeter beside the diode. I don't think it was phrased too bad tbh.

    Wait **** I think BL1993 is right ****


  • Options
    #357
    Physics FTW
    Registered Users Posts: 14 ✭Physics FTW


    damn I said you should change to A.C. because I though current wouldn't flow in reverse biased


  • Options
    #358
    Wesc.
    Registered Users Posts: 715 ✭✭✭Wesc.


    Ok so it should be a microammeter but shouldn't it be beside the diode? And the voltmeter stays in the same place?


  • Options
    #359
    An0n
    Registered Users Posts: 187 ✭✭An0n


    Physics FTW wrote: »
    damn I said you should change to A.C. because I though current wouldn't flow in reverse biased

    That would defeat the purpose of it being in reverse biased silly!


  • Options
    #360
    clio94
    Registered Users Posts: 18 ✭clio94


    Wesc. wrote: »
    Yes, which was what I did? :L And did you mean to say v+u/2 ?

    Yeah sorry v+u/2 .. The answer seemed pretty big though so wasn't positive ?


  • Advertisement
  • Options
    #361
    BL1993
    Closed Accounts Posts: 341 ✭✭BL1993


    Wesc. wrote: »
    Ok so it should be a microammeter but shouldn't it be beside the diode? And the voltmeter stays in the same place?
    Know because you only want to measure the current going through the diode.


Advertisement