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    Will the boundaries for this exam be low? I had a bad teacher and I don't know as much as I would like to...
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    (Original post by Puffin111)
    Will the boundaries for this exam be low? I had a bad teacher and I don't know as much as I would like to...
    Probably lower


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    I've got a terrible feeling about this exam, the other two weren't as bad compared to last year's so this one's most likely going to be a toughie D: but hey, low grade boundaries!
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    Does anybody have a diagram to show constructive and destructive interference? I know that constructive happens when the waves match crest to crest or trough to trough, and that destructive happens when crest matches trough. But how is this shown on a diagram?
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    (Original post by BrokenS0ulz)
    Does anybody have a diagram to show constructive and destructive interference? I know that constructive happens when the waves match crest to crest or trough to trough, and that destructive happens when crest matches trough. But how is this shown on a diagram?
    Excuse my terrible paint skills:

    http://gyazo.com/3afda0c8316ceef17b7df259357b23fd
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    (Original post by andersson)
    Excuse my terrible paint skills:

    http://gyazo.com/3afda0c8316ceef17b7df259357b23fd
    haha! thankyou but that's not quite what I meant!

    http://www.ocr.org.uk/Images/79263-q...igher-tier.pdf

    It's on question 7(b)
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    I've got 3 six markers in my workbook- does anyone want me to post them here?


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    (Original post by BP_Tranquility)
    I've got 3 six markers in my workbook- does anyone want me to post them here?


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    yes, please
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    I've attached some six markers.

    The P5 one looks to be the most difficult out of the three!

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    Is finding the vector sum different to finding the relative speed?

    E.g. It says in my revision guide that a car moving at 10m/s to the right and a car moving 10 m/s the left has a relative speed of 20m/s.
    10+10=20.

    But then, if there was someone pushing the door at 200N, and there was someone pushing the door on the other side at 200N as well, the resultant force would be 0N.

    200-200=0

    But I would have thought the resultant force would be 400N (200+200=400) if we used the same idea as the above example...So, how do you explain this?




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    Can someone explain to me how to calculate resultant velocities with parabolic trajectories? I have no idea what to do! Would be very much appreciated
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    (Original post by BP_Tranquility)
    Is finding the vector sum different to finding the relative speed?

    E.g. It says in my revision guide that a car moving at 10m/s to the right and a car moving 10 m/s the left has a relative speed of 20m/s.
    10+10=20.

    But then, if there was someone pushing the door at 200N, and there was someone pushing the door on the other side at 200N as well, the resultant force would be 0N.

    200-200=0

    But I would have thought the resultant force would be 400N (200+200=400) if we used the same idea as the above example...So, how do you explain this?




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    I think this is because:

    To find the relative speed, you take the first velocity and minus the second velocity. To find the vector sum, you get the first force and add the second force.

    i.e.

    Velocity of car 1 = 10m/s
    Velocity of car 2 = -10m/s
    Relative velocity = 10m/s - -10m/s = 20m/s

    Force 1= 200N
    Force 2 = -200N (other direction)
    Resultant force = 200N + - 200N = 0N

    EDIT: If the door doesn't move, it means all the forces acting upon it are balanced, so it couldn't be 400N.

    Something like that, if I've explained it ok
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    (Original post by Ani97)
    Can someone explain to me how to calculate resultant velocities with parabolic trajectories? I have no idea what to do! Would be very much appreciated
    Resultant velocities are quite easy. So assuming you were pushing a cart, then hooked it on to another one. The speed you were pushing the one cart at was 10m/s lets say, and the additional cart was travelling at 15m/s before you hooked onto it. Your resultant velocity would be 25m/s forward in the direction you were travelling in. If they are in opposite direction, take away the greater velocity from the lower velocity, and the resultant velocity will be in the direction of the greater speed. Think of it like when your playing tug of war, or trying to pull a friend, you have to push harder to get them to go in your direction. A different analogy would be when your standing still and someone comes running towards you, you don't stay still, you travel at the same speed as them ( before you fall flat on your face). I hope my way of explaining it is okay. And as for parabolic trajectories, you don't need to work them out, just know their shape and why it has that shape. If you have any other questions I'll start a final questions physics thread.
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    If I got 35 out of 48 on controlled assessment and 118 on the first exam (an A) how many marks will I need for an A overall?

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    Hi can someone please explain question 7a on the specimen paper. Thanks
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    Is anyone else screwed for this exam tomorrow?
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    (Original post by Randomz96)
    Hi can someone please explain question 7a on the specimen paper. Thanks
    Vertical velocity = 12m/s
    Horizontal velocity = 9m/s (wind)

    So what's the total velocity of it? Think of it like a triangle - so you use pythagoras:

    root(12 squared + 9 squared) = 15m/s.
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    (Original post by dazzer19)
    Is anyone else screwed for this exam tomorrow?
    Pretty much
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    (Original post by Knowing)
    Vertical velocity = 12m/s
    Horizontal velocity = 9m/s (wind)

    So what's the total velocity of it? Think of it like a triangle - so you use pythagoras:

    root(12 squared + 9 squared) = 15m/s.
    Thanks I get it now. Just one more question can you explain part b as well. Thanks
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    (Original post by Randomz96)
    Thanks I get it now. Just one more question can you explain part b as well. Thanks
    No he is not right, because:

    It takes 5 seconds to fall. Because the horizontal velocity is 9m/s, if dropped, it will travel 5 x 9 = 45 metres, not 20 metres.
 
 
 
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