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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?
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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dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0my 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0SO IF THE RECTANGLE IS ROLLED OUT WOULD THE 6 BE THE BASE

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0wait it depends if the pipe is standing up then its the base :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What would be the equation without the the numbers only using (b) (H) or whatever, so I can try it

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0Given length of pipe H, circumference C with wire wrapped N times. length of wire = sqrt(C^2+(H/N)^2)*N

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How did you get 6.5 c

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0sqrt(C^2+(H/N)^2), C=6,H=20,N=8

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I forgot to sqrt it, then 52 will be the length of the wire?

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0welcome, i dont get your railroad problem, is there more info?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats was the whole question the teacher gave me, I emailed her today hopefully she get give me a hint.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what number did you get in the end for the length?

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0ok sounds like you need the initial height of the rail, im prob just missing something

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0how do you get that? there are only 8 turns of the wire

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.08 complete turns stretched out over 20cm of length.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A turn is 2pi radians.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have to find a parametric equation for the spiral in terms of the information you have and find the arc length of the spiral.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so what equation would use to slove the question?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The equation for a spiral is \[u(\theta) = (r \cos \theta , r \sin \theta , a \theta + b)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I've jumped a couple of steps because I don't have much time.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[u(\theta) = (r \cos \theta, r \sin \theta , \frac{5 \pi}{4}\theta)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The arc length is given by\[L=\int\limits_{\theta = 0}^{\theta = 16 \pi}\sqrt{r^2 \cos^ 2 \theta + r^2 \sin^2 \theta + \frac{25 \theta^2}{16\pi^2}}d\theta \]\[=\int\limits_{\theta = 0}^{\theta = 16 \pi}\sqrt{r^2 + \frac{25 \theta^2}{16\pi^2}}d \theta =\frac{1}{4\pi}\int\limits_{0}^{16 \pi}\sqrt{16\pi^2 r^2+25 \theta^2}d \theta\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[r=\frac{3}{\pi}\]so substitute that in and integrate.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You'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.

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0hmm what class is this for?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry if I've confused matters :s

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0no you're correct i just doubt parametric equations is required for the class

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0rosette what class was this problem for?

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0anyway 507 is obviously way too much you would have to wind the wire around the pipe like 80 times

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is for quantitavtive literacy
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