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Math Help - Average weight of water

  1. #1
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    Average weight of water

    Water is run at a constant rate of 1 ft/min to fill a cylindrical tank of radius 3 ft and height 5 ft. Assuming that the tank is initially empty, find the average weight of the water in the tank over the time period required to fill it. (Take the weight density of water to be 62.4 lb/ft.
    If p is the density of a liquid, then the weight w of a volume V of the liquid is w=pV.

    Here is my solution, but I'm not sure if it is correct???

    V=\Pi\times3^2\times5=141.37

    Average weight:
    1/141.37 \int 62,4 \times (t\times1) dt <0 to 141.37>= (1/141.37)\times(62.4 \times \frac{\t^2}{2} = (1/141.37)\times(62.4\times(141.37^2/2)= 4,410.744 lb


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  2. #2
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    Quote Originally Posted by sun1 View Post
    Water is run at a constant rate of 1 ft/min to fill a cylindrical tank of radius 3 ft and height 5 ft. Assuming that the tank is initially empty, find the average weight of the water in the tank over the time period required to fill it. (Take the weight density of water to be 62.4 lb/ft.
    If p is the density of a liquid, then the weight w of a volume V of the liquid is w=pV.

    Here is my solution, but I'm not sure if it is correct???

    V=\Pi\times3^2\times5=141.37

    Average weight:
    1/141.37 \int 62,4 \times (t\times1) dt <0 to 141.37>= (1/141.37)\times(62.4 \times \frac{\t^2}{2} = (1/141.37)\times(62.4\times(141.37^2/2)= 4,410.744 lb


    looks fine to me.
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  3. #3
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    Quote Originally Posted by sun1 View Post
    Water is run at a constant rate of 1 ft/min to fill a cylindrical tank of radius 3 ft and height 5 ft. Assuming that the tank is initially empty, find the average weight of the water in the tank over the time period required to fill it. (Take the weight density of water to be 62.4 lb/ft.
    If p is the density of a liquid, then the weight w of a volume V of the liquid is w=pV.

    Here is my solution, but I'm not sure if it is correct???

    V=\Pi\times3^2\times5=141.37

    Average weight:
    1/141.37 \int 62,4 \times (t\times1) dt <0 to 141.37>= (1/141.37)\times(62.4 \times \frac{\t^2}{2} = (1/141.37)\times(62.4\times(141.37^2/2)= 4,410.744 lb


    Since the water is coming in at a constant rate, the "average" weight is just the initial weight (0) plus the weight at the end, divided by 2. That is, the volume of the entire tank time the density of water, divided by 2: \pi(3)^2(5)(62.4)/2. That is the answer you got.
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