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Math Help - Finding Feet Given Density

  1. #1
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    Finding Feet Given Density

    I am stuck!

    qbkr21

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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by qbkr21 View Post
    I am stuck!

    qbkr21

    I thought the radial value was 8.35 mm = .835 cm

    So V=\pi r^2h\implies 7.14\times10^3~cm^3=\pi(.835~cm)^2h

    Solving for h, we see that h=\frac{7.14\times10^3~cm^3}{\pi(.835~cm)^2}\appro  x\color{red}\boxed{3.26\times10^3 ~cm}

    Does this make sense?

    --Chris

    EDIT: I see now that .835 cm was the diameter...my mistake...

    The equation should be V=\pi r^2h\implies 7.14\times10^3~cm^3=\pi(.4175~cm)^2h

    Solving for h, we see that h=\frac{7.14\times10^3~cm^3}{\pi(.4175~cm)^2}\appr  ox\color{red}\boxed{1.3\times10^4 ~cm}
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    I thought the radial value was 8.35 mm = .835 cm

    So V=\pi r^2h\implies 7.14\times10^3~cm^3=\pi(.835~cm)^2h

    Solving for h, we see that h=\frac{7.14\times10^3~cm^3}{\pi(.835~cm)^2}\appro  x\color{red}\boxed{1.02\times10^4 ~cm}

    Does this make sense?

    --Chris

    EDIT: I see now that .835 cm was the diameter...my mistake...

    The equation should be V=\pi r^2h\implies 7.14\times10^3~cm^3=\pi(.4175~cm)^2h

    Solving for h, we see that h=\frac{7.14\times10^3~cm^3}{\pi(.4175~cm)^2}\appr  ox\color{red}\boxed{3.26\times10^3 ~cm}
    Thanks for the response, but how do you figure the radial value = 8.35 when the problem states that it is the diameter. I divided by 2 (a half) to get the radius. From there I would still be stuck in millimeters. I was thinking about then going over to volume and converting it into millimeters cubed, then dividing it by pi and (4.175 mm)^2. I'd then be left will millimeters of which I could covert back into feet. Does this work or am I making it way to complicated?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Thanks for the response, but how do you figure the radial value = 8.35 when the problem states that it is the diameter. I divided by 2 (a half) to get the radius. From there I would still be stuck in millimeters. I was thinking about then going over to volume and converting it into millimeters cubed, then dividing it by pi and (4.175 mm)^2. I'd then be left will millimeters of which I could covert back into feet. Does this work or am I making it way to complicated?
    I noticed those mistakes once I posted...

    I edited them as you can probably tell now. Sorry about that.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Thanks for the response, but how do you figure the radial value = 8.35 when the problem states that it is the diameter. I divided by 2 (a half) to get the radius. From there I would still be stuck in millimeters. I was thinking about then going over to volume and converting it into millimeters cubed, then dividing it by pi and (4.175 mm)^2. I'd then be left will millimeters of which I could covert back into feet. Does this work or am I making it way to complicated?
    Yes, this would work. I don't see how your way would be complicated. However, do you know the conversion from mm to ft?

    --Chris
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  6. #6
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    RE:

    Quote Originally Posted by Chris L T521 View Post
    Yes, this would work. I don't see how your way would be complicated. However, do you know the conversion from mm to ft?

    --Chris
    Is this correct?

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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Is this correct?

    I get \frac{7,140,000~mm^3}{\pi(4.175~mm)^2}\approx1.3\t  imes10^5~mm

    Converting to feet, I got ~428 ft.

    --Chris
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    Quote Originally Posted by Chris L T521 View Post
    I get \frac{7,140,000~mm^3}{\pi(4.175~mm)^2}\approx1.3\t  imes10^5~mm

    Converting to feet, I got ~428 ft.

    --Chris
    Woops I forgot to square the radius

    How did you only get two sig figs for the 1.3* 10^5?
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  9. #9
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Woops I forgot to square the radius

    How did you only get two sig figs for the 1.3* 10^5?
    Ahaha...I forgot about sig figs [and the rules]...

    but I saw that 130387.326\approx 1.3\times10^5~mm

    I didn't think we could have 3 sig figs...

    --Chris
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    Quote Originally Posted by Chris L T521 View Post
    Ahaha...I forgot about sig figs [and the rules]...

    but I saw that 130387.326\approx 1.3\times10^5~mm

    I didn't think we could have 3 sig figs...

    --Chris
    Hey Chris just one other thing. In using Sig Figs do conversions count as Sig Figs such as multiplication, and also when I divide the diameter by 2 does the 2 count as a Sig Fig?

    Thanks,

    qbkr21
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  11. #11
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by qbkr21 View Post
    Hey Chris just one other thing. In using Sig Figs do conversions count as Sig Figs such as multiplication, and also when I divide the diameter by 2 does the 2 count as a Sig Fig?

    Thanks,

    qbkr21
    erm...I not sure at the moment

    This may help you out.

    --Chris
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