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    Can someone explain 7b

    https://e3ab10733179873e9fd16d54c56b...%20A-level.pdf
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    (Original post by Acrux)
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    Yes it's zero because there is no acceleration. So you are right.

    For 3bii I think it's 9*10^3*sin7
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    How do I do part ii not sure why its not 9x10^3cos7
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    (Original post by Acrux)
    How do I do part ii not sure why its not 9x10^3cos7
    9x10^3cos7 would give you the support force from the ground.
    Since it wants the component acting down the slope you have to find the force acting parallel to the slope in a sense. Using sohcahtoa that 9x10^3sin7
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    (Original post by darkrep97)
    9x10^3cos7 would give you the support force from the ground.
    Since it wants the component acting down the slope you have to find the force acting parallel to the slope in a sense. Using sohcahtoa that 9x10^3sin7
    Thanks
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    (Original post by darkrep97)
    9x10^3cos7 would give you the support force from the ground.
    Since it wants the component acting down the slope you have to find the force acting parallel to the slope in a sense. Using sohcahtoa that 9x10^3sin7
    What about part iii what formula are they using to work out work done
    and part iv power
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    (Original post by Acrux)
    How do I do part ii not sure why its not 9x10^3cos7
    The question needs to be read very carefully.

    The forces need to be resolved so that they are described as:

    a) perpendicular to the plane of the slope (reaction of the car on the slope)

    b) parallel to the plane of the slope. force of gravity component acting down (parallel) to the slope

    Which means the force vectors are at right angles to each other.

    i.e. the gravitational force (weight) acting vertically downwards becomes the hypotenuse of the resolved forces acting on the car.

    So far so good.

    The question asks for the component of the weight of the car acting down the slope. i.e. The component of the weight of the car which acts parallel to the slope, not normal to it.

    Hence mgsin\theta

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    (Original post by Acrux)
    What about part iii what formula are they using to work out work done
    and part iv power
    For part iii you need to find the the work done per second which is basically power.
    You can do that using P=Fv so in this case P=300x18

    For part iv you need the resultant force on the car to maintain constant speed so it would be 300+your answear in ii. That would give you F then use P=Fv to find the power of the car
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    (Original post by darkrep97)
    For part iii you need to find the the work done per second which is basically power.
    You can do that using P=Fv so in this case P=300x18

    For part iv you need the resultant force on the car to maintain constant speed so it would be 300+your answear in ii. That would give you F then use P=Fv to find the power of the car
    Questions 7b
    Why is increased by a factor of 30^2?
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    (Original post by Acrux)
    Questions 7b
    Why is increased by a factor of 30^2?
    Because the ratio between the diameters is 30, but the breaking force is proportional to the area. Area=radius squared. The ratio between the radii is the same. So the break force increases by a factor of 30^2 because of the areas of the cables.
 
 
 
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