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    (Original post by Jullith)
    I drew a diagram for you. Hope it helps.

    Attachment 228356
    This is great, thankyou!!
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    (Original post by Namod)
    lol
    that aint a question...
    The real question is why would it be equal?
    Eh idk. I'm confused, what is it equal to?
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    (Original post by Namod)
    Mate 1W = 1W, it doesn't equal 1W per second and it doesn't equal 1W per hour.
    But 1W = 1J per second.
    Ooh okay. 1KW = 1000J per second then I think so
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    2b, jan 12, is the max velocity, zero acceleration and minimum PE always like that? will it always be when the graph crosses the acis?
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    (Original post by Stickyelmo)
    Ooh okay. 1KW = 1000J per second then I think so
    yes

    (Original post by eggfriedrice)
    Eh idk. I'm confused, what is it equal to?
    If the net force on a spring is 0
    Then the tension on the spring is equal to it's weight.
    lol
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    Is there anything on the spec. that has NEVER come up?
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    (Original post by Namod)
    CAN ANYONE CHECK PLEASE:
    jan 2013 paper Q2bii

    EDIT:can anyone explain it lol? I didn't get how they got 6.7
    because they only have to go half way around the earth to do one complete earth, if that makes sense? as with one orbit it will cover a strip either side, not just on one side. so it only needs to go half way around (6.7 orbits) before covering the whole earth. As opposed to the 13.4 orbits or whatever the calculation works out, which would be necessary if the satellite only went from north to south on one side of the globe, which it doesnt, it goes back up again to north on the otherside.

    made that a bit wordy, apologies
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    What's everyone doing besides practice papers?
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    (Original post by Supes180)
    What's everyone doing besides practice papers?
    hmmmm practice papers maybe?
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    (Original post by Namod)
    yes



    If the net force on a spring is 0
    Then the tension on the spring is equal to it's weight.
    lol

    But I mean what's the net force equal to when it's undergoing harmonic motion?
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    HELP please
    A car of mass 1000kg is travelling at a velocity of 10ms-1. It accelerates for 15s, reaching a velocity of 24ms-1. Calculate:
    a) The change in momentum of the car in a period of 10s
    b) The average force acting on the car as it accelerates


    I can get part b, i get 930N just putting numbers into equation F=m(v-u/t) but part a i cant get :L any help would be much appreciated
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    (Original post by eggfriedrice)
    But I mean what's the net force equal to when it's undergoing harmonic motion?
    net force constantly changing as the acceleration is changing.
    When the displacement is 0, acceleration is 0, net force is 0. (F=ma, F= m x 0 =0)
    When displacement maximum, acceleration is maximum, net force is maximum.
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    (Original post by 9876543211)
    HELP please
    A car of mass 1000kg is travelling at a velocity of 10ms-1. It accelerates for 15s, reaching a velocity of 24ms-1. Calculate:
    a) The change in momentum of the car in a period of 10s
    b) The average force acting on the car as it accelerates


    I can get part b, i get 930N just putting numbers into equation F=m(v-u/t) but part a i cant get :L any help would be much appreciated
    a=(v-u)/t = (24-10)/15= 14/15
    v-u= at = (14/15)(10) = 28/3
    change in momentum = m(v-u) = 1000 * 28/3 =28000/3 = 9333

    Worked in fraction because dont like to write recurring numbers.
    Tell me if im correct.
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    could someone give me a good method for boyles law. sorry if its been asked.
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    (Original post by Namod)
    a=(v-u)/t = (24-10)/15= 14/15
    v-u= at = (14/15)(10) = 28/3
    change in momentum = m(v-u) = 1000 * 28/3 =28000/3 = 9333

    Worked in fraction because dont like to write recurring numbers.
    Tell me if im correct.
    I got the same answer as you but different methods. Anyhow the answer is apparently 1.4x10^4? its SAQ Q1 in chapter 2 of cambridge ocr advanced series physics 2. i dunno if its a mistake or ive done something wrong as they rarely make mistakes in a text book....
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    (Original post by 9876543211)
    I got the same answer as you but different methods. Anyhow the answer is apparently 1.4x10^4? its SAQ Q1 in chapter 2 of cambridge ocr advanced series physics 2. i dunno if its a mistake or ive done something wrong as they rarely make mistakes in a text book....
    mistake in the book, cant find anything wrong with what i did.
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    Going to head off now. Good luck for tomorrow everyone!
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    Do you actually need to know the charles' and boyle's law experiment? The charles' law experiment isn'e even on the spec.? Is the law itself even on the spec?
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    (Original post by Namod)
    mistake in the book, cant find anything wrong with what i did.
    yeah looking at it i think they must of meant the change in momentum after 15s and not 10s... glad im not wrong then phew . thanks anyway
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    why does a liquid have a generally higher specific heat capacity than solid?
 
 
 
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