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    it's not 97.18 because the triangle you are using isn't the resolved force and you could draw the forces with that angle and resolve them to find it's incorrect, drawing a line connecting the 2 ends and finding it is equal to 6 is not a resolved force
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    (Original post by physicskid123)
    QQ
    QQ
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    (Original post by ajamesg)
    Attachment 551872this is the proof that it is 97.18 degrees
    You are wrong, put one of the forces due north, and the other with 97.18 and resolve to get the resultant and it won't be 6N.
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    (Original post by sandpitturtle)
    was P 12 on q5?
    Yes
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    (Original post by Furthermaths100)
    You are wrong, put one of the forces due north, and the other with 97.18 and resolve to get the resultant and it won't be 6N.
    As I said previously, I agree now that I am wrong. Do you remember how many marks the whole question was worth?
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    * unrelated question.. but
    IF i got Bs in S1 and M1 but As in C1,C2,C3,C4, can I get an A overall?
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    (Original post by Parallex)
    agree w/ 6.76 for the last one

    tension was 1.81N or something for the pulleys

    can't remember much lol
    yes thats what i got too
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    (Original post by czj1997)
    * unrelated question.. but
    IF i got Bs in S1 and M1 but As in C1,C2,C3,C4, can I get an A overall?
    definitely
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    (Original post by czj1997)
    * unrelated question.. but
    IF i got Bs in S1 and M1 but As in C1,C2,C3,C4, can I get an A overall?
    Yes but you will need higher As to make up for the Bs. You just need to average 80 UMS overall to get an A (480+ total UMS points).
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    Any idea what 68 would be UMS wise?
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    its angle to the vertical not horizontal that's why its 88 degrees. (180-97)
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    (Original post by Buymoria)
    Any idea what 68 would be UMS wise?
    68 marks could get you 100 ums, depends on how everyone did overall. Doubt that will get you less than 96 ums though.
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    I also did the cosine rule but youre forgetting to minus the value of 97 whatever from 180
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    (Original post by Arnold12345)
    its angle to the vertical not horizontal that's why its 88 degrees. (180-97)
    82.7 degrees
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    forgot to work on time on question 1 :unimpressed:, 2 marks off?

    (also what was the question on the contact force asking?)
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    I got the answer right for the 82.8 degrees question, however the method I used was definitely over complicated because when I was checking I realised that I could have done it in a much simpler way i.e. using the cosine rule. Do you think they will mark me down for having used an alternative complex method despite getting the right answer?
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    also does anyone have any ideas as to what the 90 ums boundary will be? around 65-66 maybe....??
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    (Original post by G_8)
    I got the answer right for the 82.8 degrees question, however the method I used was definitely over complicated because when I was checking I realised that I could have done it in a much simpler way i.e. using the cosine rule. Do you think they will mark me down for having used an alternative complex method despite getting the right answer?
    No chance they will mark you down, I was thinking of doing it the complicated way but realised before I started, both are legitemate ways to do it and will both be on the mark scheme
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    (Original post by G_8)
    also does anyone have any ideas as to what the 90 ums boundary will be? around 65-66 maybe....??
    This was quite hard for an M1 paper I would say - papers like this are usually about 63 for 90 ums and this will be no exception.
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    (Original post by Snicherz)
    forgot to work on time on question 1 :unimpressed:, 2 marks off?

    (also what was the question on the contact force asking?)
    Yes 2 marks off. Not too sure on the contact force question.
 
 
 
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