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Please can someone good at physics help me with this past paper? watch

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    http://store.aqa.org.uk/qual/gcse/qp...W-QP-JUN08.PDF

    Can someone please help me on question 7a, i really dont get it thanks?

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    As the toy is hovering the lift force is equal and opposite to the weight of the toy, therefore 0.6N.

    Rearranging the second equation gives
    change in momentum=force x time taken for the change (which is given in the question)

    Once you have calculated the change in momentum use the first equation to calculate the velocity of the air.
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    thank you so much! Sorry but how do you know they are equal?
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    Newton's first law. For an object remain be stationary (or travel at a constant speed) there must be zero resultant force. Therefore there must be an equal and opposite force to the weight of the toy (generated by the movement of the wind) that is preventing the toy from falling and is not larger than the weight of the toy otherwise the toy would accelerate upwards (F=ma).
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    When the toy is hovering the lift force is equal to the weight, (this is becuase, according to Newton's first law, if an abject is stationery the resultant force acting on the body is 0)

    use the second formula to find the change in momentum.... the time is given in the question and the force is equal to the weight....

    Finally divide the momentum with the mass of the toy to get velocity...
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    Excellent, thanks
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    Thanks to DopplerEffect aswell. You guys have really helped me understand this concept.
 
 
 
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