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Math Help - Hi my name Jhay from PHL. engr student help please [solid mensuration problem)

  1. #1
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    Hi my name Jhay from PHL. engr student help please [solid mensuration problem)

    Please Give Me the Solution and The Figure of the problem(its very impt. for me to understand)
    Problem
    1. The diameter of a well is 6 feet and the water is 7 feet deep.How many gallons of water are there in the well,reckoning 7.48 gallons to a cubic foot?
    2.A paint manufacturer desires a cylindrical steel drum to hold 50 gal. of roof paint. For convenience in handling, it is found necessary to limit the inside diameter to 2 1/2ft. find the height of the drum desired. (one gal. =231 cu. in.)

    3.Find the waste in making the largest possible cylindrical rod from a bar of iron 3 feet long which has a square cross section whose diagonal is 6 inches.
    4. A cubic foot of water weights about 62.4. What must be the diameter of a cylindrical pail 1 ft.high in order for it to hold the water from 25 lb. of ice? how many square inches of sheet tin are required to make the pan (neglect waste in cutting and lapping )
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    Super Member ebaines's Avatar
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    Re: Hi my name Jhay from PHL. engr student help please [solid mensuration problem)

    For each of these you ned to apply the formula for the volume of a cylinder, which you should know is  \V = \pi R^2h. Here you are given the diameter of each cylinder, and since  D = 2R this makes the volume formula  V = \pi \frac {D^2} 4 h. Finally, given that there are 7.47 gallons/ft^3, you can convert the volume formula to gallons by multiplying by that factor:

     G (gallons) = 7.48V = 7.48 \pi \frac {D^2} 4 h

    Now you should be able to solve each problem. Post back with yuor answers and we'll check them.
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    Re: Hi my name Jhay from PHL. engr student help please [solid mensuration problem)

    1. V=7.48gal/ft^3 *π (d^2)/4 h
    V=π (6ft.^2)/4*7ft
    V= 7.48gal/ft^3 *197.92 ft^3
    V=1480.44gal

    2. V= π (d^2)/4*h
    50 gal * 231inch^3/gal = π(2.5ft * 12inch/ft)^2/4 *h
    11550inch^3 = 706.86 *h
    706.86inch^2 *h/706.86inch^2 = 11550inch^3/706.86inch^2
    h=16.34inch
    3. i dont know how to solve please help
    4. i dont know how to solve please help...

    hey can u give me the figure of problem 1 to 4 please tnk u in advance
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    Re: Hi my name Jhay from PHL. engr student help please [solid mensuration problem)

    thank u very much !
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    Super Member ebaines's Avatar
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    Re: Hi my name Jhay from PHL. engr student help please [solid mensuration problem)

    Parts (1) and (2) are correct.

    For (3) you need to compare a square bar that has diagonal length of 6 inches across its face with a cylindrical bar cut out of that square bar. If the diagonal is 6 inches, then the length of each side of the square bar face is  \w=6/\sqrt {2} inches, which is also the value of the diameter or the cylindrical bar that can be cut from it. Now it's simply a matter of comparing the volume of the square bar versus the cylindrical one.

    For (4) you're again using  V = \pi (\frac {D^2} 4) h, where V is 25 pounds water divided by 62.4 pounds per cubic foot. You have the value for h, so you can solve for D. Then the area of the pan will be the area of the bottom of the pan  A_1= \pi D^2/4 plus the area of the sides of the pan  A_2 = \pi D h.
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