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Math Help - problems involving volumes and surface areas

  1. #1
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    problems involving volumes and surface areas

    Hey, need some help with the steps needed to solve these problems.

    1) calculate the total surface area (inside and outside and both ends) of a gas pipe. length 3.5m, outside diameter 55mm, inside diameter 50mm.

    2) determine the volume of metal used in the making of the gas pipe.

    3) a) determine the number of complete revolutions a motorcycle wheel will make in traveling 5km if the wheel's diameter is 80.2cm.

    b) water flows in a 1000mm diameter pipe to a depth of 800mm calculate:

    i) the wet perimeter of the pipe.
    ii) the area of cross section of the water.

    thanks.
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  2. #2
    Super Member craig's Avatar
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    Quote Originally Posted by clayson View Post
    1) calculate the total surface area (inside and outside and both ends) of a gas pipe. length 3.5m, outside diameter 55mm, inside diameter 50mm.
    First we'll deal with the ends. I assume that you know the area of a circle is \pi r^2.

    The outside of the pipe has radius 27.5. So the area of this cross section, if it were a whole circle, would be \pi(27.5)^2. However there's a whole in this circle with radius 25, ie with area \pi(25)^2. Subtracting the smaller area from the larger one will give you the surface area of the end of the pipe, remember to multiply by two as there are two ends!

    As for the outside and inside of the pipe, we'll do this in a very similar way.

    The formula for the circumference of a circle is 2\pi r. Therefore the circumference of the larger circle would be 55\pi. Now imagine if you cut the pipe down along it's length and unrolled it, you would have a rectangle with the circumference of the circle as the width, and the length obviously as it's length. Multiply the circumference of the larger circle by the length of the pipe to find the surface area of the outside of the pipe.

    Repeat this for the inside using the smaller radius.
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  3. #3
    Super Member craig's Avatar
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    One more thing, remember to make sure you are using the same units. The pipe would be 3,500 mm.

    Now it's your turn. Show us your work for the other parts and we'll help you when you get stuck.
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  4. #4
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    Thumbs up

    Thanks a lot mate, that explanation really helped me understand the question.
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  5. #5
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    Can anyone help on questions 2 and 3. Hit a blank with with them

    My answer fro question 1 is 1154773 mm, does it seem about right.

    Thanks.
    Last edited by clayson; December 16th 2009 at 07:33 AM.
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  6. #6
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    1.
    i get roughly 1155360 by doing the following;

    2(\Pi(27.5^2)-\Pi(25^2))+2\Pi(27.5)3500+2\Pi(25)3500=1155360

    you figure out which of us made the mistake.
    2.

    first calculate the volume using the smaller radius then subtract it from the volume using the larger radius. This gives you the volume of the actual metal and not the emptiness.

    3.
    (b)
    looking through the end you see a circle of radius 500mm and it is coloured in 800mm from the bottom.
    we draw a line(chord) across the circle to represent the height the water reaches which is 800mm so it passes the centre of the circle by 300mm.
    now draw lines from the centre of the circle to the ends of the chord. we need to find the angles these lines make with each other.
    this makes 2 triangles with height 300mm and hypoteneuse 500mm. so the angle at the centre is arccos(300/500).
    2 triangles so 2arccos(300/500)=1.854590436 radians.

    2\Pi-1.854590436=4.428594871 for the perimeter under the water (wet)
    where 2Pi represents 360 degrees
    so perimeter=(4.428594871)(500) cm

    as for the last part you need to subtract the area of the top bit not in the water from the area of the circle. have a crack at it.you can start by finding the length of the chord going across.
    Last edited by Krahl; December 16th 2009 at 09:36 AM.
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