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Math Help - Work needed to pump water from inverted cone

  1. #1
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    Work needed to pump water from inverted cone

    Hi, I know

    the gist of how to do these problems but I am getting the wrong answers.

    A tank in the shape of an inverted right circular cone has height 6 meters and radius 3 meters. It is filled with 5 meters of hot chocolate. Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. The density of hot chocolate is

    volume= pi r^2
    r=3, y=6 so r=.5y
    distance is from 0 to 5 meters
    density is given as 1010

    So W= density * volume * distance

    W= 1010 [pi(.5y)^2]dy(6-y)
    w=intregal from 0 to 5 (6-y)1010pi(1/4)y^2dy
    =(1010pi/4) intregal from 0 to 5, 6y^3-y^3dy

    and so on... I got 74367.388 J but it's wrong. any help will be appreciated. Thanks
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  2. #2
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    Re: Work needed to pump water from inverted cone

    W = \delta g \pi  \int_0^5 \left(\frac{y}{2}\right)^2 \cdot (6-y) \, dy

    where g = 9.8 \, m/s^2

    so ... multiply your result by 9.8
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  3. #3
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    Re: Work needed to pump water from inverted cone

    Thank you!

    Also I have a really quick question, if instead of giving you density of a liquid, it give you the weight. Example. the weight of the liquid is 60 lbs/ft^3

    What does that mean? I know F=ma=Density * Volume*acceleration so does that weight of a liquid= density*volume? Thanks
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  4. #4
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    Re: Work needed to pump water from inverted cone

    Quote Originally Posted by makeupgirl1107 View Post
    Thank you!

    Also I have a really quick question, if instead of giving you density of a liquid, it give you the weight. Example. the weight of the liquid is 60 lbs/ft^3

    What does that mean? I know F=ma=Density * Volume*acceleration so does that weight of a liquid= density*volume? Thanks
    that would be the weight density ... \delta g
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