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anonymous

  • 5 years ago

Eight turns of a wire are wrapped around a pipe with a length of 20 cm and a circumference of 6cm. What is the length of the wire?

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  1. dumbcow
    • 5 years ago
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    my guess would be to think of the pipe as being rolled out so it looks like a rectangle of length 20 and width the circumference 6. Now each time the wire is wrapped around it travels one revolution of the pipe or 6cm but it also goes down the length of the pipe 20/8 or 2.5 cm, going back to the rectangle that would look like pulling the wire diagonally across the rectangle to a point 2.5 cm down forming a right triangle. Find the length of the hypotenuse of that triangle which is length of the wire for first revolution and multiply by 8. _>c^2 = 6^2+2.5^2 = 42.25 ->c = 6.5 _>length of pipe = 6.5 * 8 = 52

  2. dumbcow
    • 5 years ago
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    hope that made sense

  3. anonymous
    • 5 years ago
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    SO IF THE RECTANGLE IS ROLLED OUT WOULD THE 6 BE THE BASE

  4. dumbcow
    • 5 years ago
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    yes

  5. dumbcow
    • 5 years ago
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    wait it depends if the pipe is standing up then its the base :)

  6. anonymous
    • 5 years ago
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    What would be the equation without the the numbers only using (b) (H) or whatever, so I can try it

  7. dumbcow
    • 5 years ago
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    Given length of pipe H, circumference C with wire wrapped N times. length of wire = sqrt(C^2+(H/N)^2)*N

  8. anonymous
    • 5 years ago
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    How did you get 6.5 c

  9. dumbcow
    • 5 years ago
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    sqrt(C^2+(H/N)^2), C=6,H=20,N=8

  10. anonymous
    • 5 years ago
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    I forgot to sqrt it, then 52 will be the length of the wire?

  11. dumbcow
    • 5 years ago
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    yes

  12. anonymous
    • 5 years ago
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    thanks so much

  13. dumbcow
    • 5 years ago
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    welcome, i dont get your railroad problem, is there more info?

  14. anonymous
    • 5 years ago
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    thats was the whole question the teacher gave me, I emailed her today hopefully she get give me a hint.

  15. anonymous
    • 5 years ago
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    what number did you get in the end for the length?

  16. dumbcow
    • 5 years ago
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    ok sounds like you need the initial height of the rail, im prob just missing something

  17. anonymous
    • 5 years ago
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    I get about 507.5cm

  18. dumbcow
    • 5 years ago
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    how do you get that? there are only 8 turns of the wire

  19. anonymous
    • 5 years ago
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    8 complete turns stretched out over 20cm of length.

  20. anonymous
    • 5 years ago
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    A turn is 2pi radians.

  21. anonymous
    • 5 years ago
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    It's a spiral.

  22. anonymous
    • 5 years ago
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    You have to find a parametric equation for the spiral in terms of the information you have and find the arc length of the spiral.

  23. anonymous
    • 5 years ago
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    ok so what equation would use to slove the question?

  24. anonymous
    • 5 years ago
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    The equation for a spiral is \[u(\theta) = (r \cos \theta , r \sin \theta , a \theta + b)\]

  25. anonymous
    • 5 years ago
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    I've jumped a couple of steps because I don't have much time.

  26. anonymous
    • 5 years ago
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    You know what the radius of the spiral should be, so you need to find a and b. This last coordinate is the height, z. When theta =0, the height of the spiral is 0, so b is 0. When the height of the spiral is 20, theta = 8 x 2pi = 16 pi (since you're told there are 8 complete turns in 20cm), so 20 = 16pi.a --> a = 5/(4pi).

  27. anonymous
    • 5 years ago
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    \[u(\theta) = (r \cos \theta, r \sin \theta , \frac{5 \pi}{4}\theta)\]

  28. anonymous
    • 5 years ago
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    The arc length is given by\[L=\int\limits_{\theta = 0}^{\theta = 16 \pi}\sqrt{r^2 \cos^ 2 \theta + r^2 \sin^2 \theta + \frac{25 \theta^2}{16\pi^2}}d\theta \]\[=\int\limits_{\theta = 0}^{\theta = 16 \pi}\sqrt{r^2 + \frac{25 \theta^2}{16\pi^2}}d \theta =\frac{1}{4\pi}\int\limits_{0}^{16 \pi}\sqrt{16\pi^2 r^2+25 \theta^2}d \theta\]

  29. anonymous
    • 5 years ago
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    \[r=\frac{3}{\pi}\]so substitute that in and integrate.

  30. anonymous
    • 5 years ago
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    You'll have to make a substitution in your integral in order to solve it. I don't have anymore time, sorry. If this isn't something due very soon, I can come back to it.

  31. dumbcow
    • 5 years ago
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    hmm what class is this for?

  32. anonymous
    • 5 years ago
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    Sorry if I've confused matters :s

  33. dumbcow
    • 5 years ago
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    no you're correct i just doubt parametric equations is required for the class

  34. dumbcow
    • 5 years ago
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    rosette what class was this problem for?

  35. dumbcow
    • 5 years ago
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    anyway 507 is obviously way too much you would have to wind the wire around the pipe like 80 times

  36. anonymous
    • 5 years ago
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    this is for quantitavtive literacy

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