Edexcel Unit 1 (IAL): Physics on the Go, WPH01 (24th May 2016)

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    (Original post by djmans)
    unit 3 forum was way more active
    Can you send me a link to the unit 3 forum


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    (Original post by khassan)
    Can you send me a link to the unit 3 forum


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    its already over
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    So i saw a question in qp asking to find the horizontal distance travelled by a projectile at an angle. I used the range equation but the mark scheme used another format. Like v=s/t and finding t from vertical speed.

    But using the range formula, will i get full mark? Answer did match.
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    Need help for question no a(ii) thanx

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    (Original post by djmans)
    its already over
    Yea, I did that paper..wanted to check my answers
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    (Original post by Saad69)
    How to calculate part ii?

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    i just saw the thread i think u have to use 1/2fx or Kx^2(K x squared) they are both supposed to give the same answer...it gives the energy stored in the spring
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    (Original post by Komol)
    Need help for question no a(ii) thanx

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    Name:  Photo on 23-05-2016 at 21.30.jpg
Views: 126
Size:  125.5 KB I'm sorry the picture is inverted but what my teacher said is that the
    maximum force the truck will have is 576N which is lower than the friction therefore the lorries acceleration is limited
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    (Original post by Sulzaae)
    Name:  Photo on 23-05-2016 at 21.30.jpg
Views: 126
Size:  125.5 KB I'm sorry the picture is inverted but what my teacher said is that the
    maximum force the truck will have is 576N which is lower than the friction therefore the lorries acceleration is limited
    Thanx.....

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    Jan 2014 IAL question 17 whole question help
    EUSHA Sulzaae djmans
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    (Original post by Komol)
    Need help for question no a(ii) thanx

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    Simply when the truck accelerates, the box to stay on it must accelerate at the same rate. In order to do so there must be a force accelerating the box at the same rate as the truck. This the friction. The friction is 630N max.So the max acceleration of the box must be 630 dividied by 200kg the mass of the box. Which is therefore the max acceleration of the trucck
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    (Original post by Dr.strange12)
    Simply when the truck accelerates, the box to stay on it must accelerate at the same rate. In order to do so there must be a force accelerating the box at the same rate as the truck. This the friction. The friction is 630N max.So the max acceleration of the box must be 630 dividied by 200kg the mass of the box. Which is therefore the max acceleration of the trucck
    Thanks.....

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    (Original post by Saad69)
    Jan 2014 IAL question 17 whole question help
    EUSHA Sulzaae djmans
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    the graph is correct, in the calculation you must find the gradient of the 2nd graph use the middle part to find the gradient
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    (Original post by Saad69)
    How to calculate part ii?

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    In case anyone wondering this is how to solve it it was only extended to 3.4 cm

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    Is it just my school where physics is just a train wreck compared to all the other subjects?
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    The paper was tough
    Hoping for low boundaries
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    (Original post by Saad69)
    The paper was tough
    Hoping for low boundaries
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    Yea it was i didnt ans all of them for the time limit damn....

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    I thought so too. Stuggled with the Stoke's law question where you find the radius of the raindrop at terminal velocity, and the Vector Diagram question


    (Original post by Komol)
    Yea it was i didnt ans all of them for the time limit damn....

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    (Original post by its_samuel)
    I thought so too. Stuggled with the Stoke's law question where you find the radius of the raindrop at terminal velocity, and the Vector Diagram question
    radius of the raindrop at terminal velocity?which one was that?
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    no unofficial mark scheme?
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    Did any1 else get 7.9x10^-14 ohm metres for the resistivity on the last q?

    And did any1 get 0.77 for the efficiency of the electric motor q? The current and voltage were given along with the mass and height the weight was lifted by...
 
 
 
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