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    (Original post by sameehaiqbal)
    Here I did the questions for you

    Attachment 536603

    Attachment 536603536605

    When you say the angle of water needs to be smaller than the angle or air. Is the angle you are talking about measured from the normal line to the ray ?

    Wow Wow Wow, you never fail to impress me, haha

    Thank you very much !
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    (Original post by Hey234)
    I need help with Sqa 2014 paper 2 qu2 please.
    Okay well you want to find out the resistance of the thermistor but to do that you need the current which we don't have.

    You can attain this by doing:

    V2 = Vs - V1 = 3.0

    To find the current: I= V2/R
    = 3.0/1050
    = 0.00285714

    Then
    R= V1/I
    = 2.0/0.00285714
    = 700 OHMS
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    (Original post by Ethan100)
    When you say the angle of water needs to be smaller than the angle or air. Is the angle you are talking about measured from the normal line to the ray ?

    Wow Wow Wow, you never fail to impress me, haha

    Thank you very much !
    Yes. Thats correct
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    I got 65% on the specimen paper. What is that as a grade?
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    (Original post by sameehaiqbal)
    I got 65% on the specimen paper. What is that as a grade?
    The notional grade boundaries are:
    A 70% +
    B 60% - 69%
    C 50% - 59%
    D 45% - 49%
    No Award 0% - 44%

    Note that each exam is different and so grade boundaries can be adjusted accordingly, just like last years higher maths exam, and probably this year nat 5 exam, the grade boundary was adjusted to 34% for a C.
    So to answer your question, 65% is a B.
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    (Original post by Tayls102)
    The notional grade boundaries are:
    A 70% +
    B 60% - 69%
    C 50% - 59%
    D 45% - 49%
    No Award 0% - 44%

    Note that each exam is different and so grade boundaries can be adjusted accordingly, just like last years higher maths exam, and probably this year nat 5 exam, the grade boundary was adjusted to 35% for a C.
    So to answer your question, 65% is a B.
    Urghh. Really wanted to get an A but oh well I'm happy with B. Guess i'll do more questions and go over the topics again Thank you xx
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    Im so exhausted. Been spending all day doing endless questions and note taking. I dunno how I'm going to revise for uni exams if I find this revision so tiring!

    My brain is literally fried. I think after the exam I may pass out :rofl:
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    (Original post by sameehaiqbal)
    Urghh. Really wanted to get an A but oh well I'm happy with B. Guess i'll do more questions and go over the topics again Thank you xx
    Maybe it has something to do with doing papers at 2am in the morning??? working too hard won't help a lot.
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    (Original post by Tayls102)
    Maybe it has something to do with doing papers at 2am in the morning??? working too hard won't help a lot.
    Omg you're right lmao :rofl:
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    Does anyone know how to do Question 3 on the 2015 past paper. It's so obscure...
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    (Original post by scottishguyy)
    Does anyone know how to do Question 3 on the 2015 past paper. It's so obscure...
    Multiple choice or section two?? x
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    (Original post by scottishguyy)
    Does anyone know how to do Question 3 on the 2015 past paper. It's so obscure...
    For Q3 on multiple choice you have to use the resistance in parallel rules.

    So 1/RT = 1/R1 +1/R2 ....

    The two resistors with 4 ohms are in parallel so use the formula above to find the resistance IN PARALLEL.
    Then the 4 ohm resistor at X should be added as it's in SERIES.
    Then that's you final answer

    For Q3 on section 2, what part are you stuck on?
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    (Original post by scottishguyy)
    Does anyone know how to do Question 3 on the 2015 past paper. It's so obscure...
    I'm guessing it's section two Question 3

    3a. (i) - As you can see from the diagram, position Y is more closer (in length/distance) to the steel sample than Position X therefore Position Y will be transmitted in less time than Position X. Therefore you can easily identify that the time taken between the pulse being transmitted and received at Position X is 15microseconds

    (ii) - You want to calculate the length of the steel sample i.e it's width at position X (i don't know if I'm making sense but just let me know if i'm confusing you). You are given the speed of ultrasound in steel which is 5200ms-1. And you have also concluded from the previous question that the time for Position X to be transmitted and received is 15microseconds. Therefore you can use d=vt to calculate the distance of the steel sample.

    Note: the time is in microseconds so do d=5200 x 15x10-6. This equals to 0.078m.

    Then to calculate the thickness you divide 0.078 by 2 to get 0.039m.

    b. So from the graphs and diagrams you have concluded that Position X has a longer distance than Position Y. However position Z is in between them. So you would just draw a line that is between Position X and Position Y to get the mark.

    c. (i) - Simply do f=1/T. i.e f= 1/4.0x10-6 (because 4.0 is in microseconds) and you should get the answer 2.5x10^5

    (ii) - You would use v=fWavelength
    wavelength=v/f
    = 5200/2.5x10^5
    = 0.021m

    d - The speed of ultrasound in brass will be less than steel because in brass it takes longer for it to travel the same thickness.

    Hope this helps
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    (Original post by sameehaiqbal)
    I'm guessing it's section two Question 3

    3a. (i) - As you can see from the diagram, position Y is more closer (in length/distance) to the steel sample than Position X therefore Position Y will be transmitted in less time than Position X. Therefore you can easily identify that the time taken between the pulse being transmitted and received at Position X is 15microseconds

    (ii) - You want to calculate the length of the steel sample i.e it's width at position X (i don't know if I'm making sense but just let me know if i'm confusing you). You are given the speed of ultrasound in steel which is 5200ms-1. And you have also concluded from the previous question that the time for Position X to be transmitted and received is 15microseconds. Therefore you can use d=vt to calculate the distance of the steel sample.

    Note: the time is in microseconds so do d=5200 x 15x10-6. This equals to 0.078m.

    Then to calculate the thickness you divide 0.078 by 2 to get 0.039m.

    b. So from the graphs and diagrams you have concluded that Position X has a longer distance than Position Y. However position Z is in between them. So you would just draw a line that is between Position X and Position Y to get the mark.

    c. (i) - Simply do f=1/T. i.e f= 1/4.0x10-6 (because 4.0 is in microseconds) and you should get the answer 2.5x10^5

    (ii) - You would use v=fWavelength
    wavelength=v/f
    = 5200/2.5x10^5
    = 0.021m

    d - The speed of ultrasound in brass will be less than steel because in brass it takes longer for it to travel the same thickness.

    Hope this helps
    JHEEEEEEEEEEEEEEEEEEEZ!

    Why are you only getting 65% in the past paper?

    You should be getting A's, you seem very well prepared

    Smart.
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    (Original post by Ismee)
    JHEEEEEEEEEEEEEEEEEEEZ!

    Why are you only getting 65% in the past paper?

    You should be getting A's, you seem very well prepared

    Smart.
    Haha i'm really not, I'm quite nervous for the exam tbh.
    I got 65% because I missed out an entire question on half life - forgot to revise it, i'm doing it now.
    And also I didn't like one of the open ended questions in that specimen paper, skipped that too :/
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    Is this correct ....??

    Volume increases = Pressure decreases
    Temperature Increases = Pressure increases
    Tenperature increases = Volume increases
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    (Original post by Ethan100)
    Is this correct ....??

    Volume increases = Pressure decreases
    Temperature Increases = Pressure increases
    Tenperature increases = Volume increases
    Yeah thats right
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    How do you change a volume from litres to m^3 ?
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    (Original post by Ethan100)
    How do you change a volume from litres to m^3 ?
    What question is that?
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    1 litre of Helium gas stored at a pressure of 800k Pa in a cylinder is released to fill a 5 litre party balloon. What is the pressure of the gas inside the balloon?
 
 
 
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