describe the basis on which you would decide the spacing of clips supporting pvc/pvc cables striaght runs and on bends
126669
Its Entirely up to you!:)
346344
The stress in the cable is m*g/A.  The Young's modulus of steel is about 200*10^9 N/m^2 (pascal).  Therefore the strain will be\n\nm*g*/(A*200*10^9).\n\nThe elongation is strain*length.\n\nFor the second part, replace g above with (g+a), where a =2.7m/s^2\n\nFor the last, compute the mass from S = m*(g+a)/A, The elastic limit is Sm, so the max mass is m = Sm*A/(g+a)\n\nNote: keep the units straight.  Use kg, m, sec, N in all calcs then convert to the asked-for units last.
560821
http://www.eng-tips.com/index.cfm
336901
Imagine cutting the tube down its length and opening it up into a long, narrow rectangle.  The outside area would be the length times the width.  The width of the rectangle is really the circumference of the circular cross section of the tube.  Thus, since you know the diameter, you have this:\n\nsurface area = length * pi * diameter\n\nwhere pi is approximately 3.14159265358979.\n\nNote, if the tube is not flexible, you have to be careful measuring the length of the bent tube.  it is the length of an imaginary line through the center of the tube, but as the other answerer pointed out, you can get this by taking the average of the "inside the curve" length and the "outside the curve" length.  Of course, if the tube is flexible, just straighten it out and measure it.\n\n\nNote--straightening, cutting and opening up, etc., will change the area slightly as it stretches part of the tube and compresses another part--this is a very small effect (unless you tie the tube into a tight knot or something) and probably can't easily be measured well anyway--should be too small to matter in most applications.
393809
this question was on this yr's midyear exam, ill give u a clue.\nyou need to seperate the forces acting on the wire into two types \n1) acting horizontally (the force to the left equals the force to the right)\n2)acting vertically (the force to the bottom {or the normal force = 20 *9.8} equals the force to the top)\n\nequate them both and you should get the answer
56003
It is the length divided by the radius
324495
number of sides, congurency?, size of angles, distinguishing characteristics
208310
A needle and some fishing line should do the trick...fishing line will support the clip alot better than thread.
154131
Here are the different types of cabling and the maximum distance they can run:  \n\nUse a 10Base5 Thick Coaxial 500 meters\n  \n10BaseT Unshielded Twisted Pair 100 meters\n10Base2 Thin Coaxial 185 meters\n10Base5 Thick Coaxial 500 meters\n10BaseF Fiber Optic 2000 meters\n100BaseT Unshielded Twisted Pair 100 meters\n100BaseTX Unshielded Twisted Pair 220 meters
387042
1. The ratio of h to r is not dependent on the can being made of aluminium. The surface area is just a concept - it cannot be aluminium.\n\n2. Is a cube (assuming that no aluminium is used in forming the edges) \n\n3. You need to do this bit.\n\nFurther Investigation  - you need to do this bit , it looks as though if you limit your investigation to regular prisms that will be O.K., investigating cones would be impressive. \n\nIn so far as the wastage is concerned. If you find out what the area+wastage is for the optimally sized circular cylinders and compare with optimally sized hexagonal prisms then this may be sufficient?
12911
Go to Paragraph in your toolbar and select spacing and then select the box that says single and switch it to double. You should see the sample diagram box go from single to double when you click apply.
555592
I can't say for certain, but I assume the length of the gaps does not increase.  The length of the tubes increase so that the time it takes to traverse the tube stays the same.  I think the only constraint on the gaps is how close the tubes can be to each other.  You'd want it to be as small as possible to get as much accelleration as possible.
461303
since it is a smooth vertical circle i.e.\npie 22/7,the let the highest point of the string to the lowest be h,given: ring=w,angle=2a\ntherefore the ring will go wobbly down and the string will tighten.
474305
length of the rod doesn't matter.\nFind the momentum of the putty.\nThe momentum of the pendulum and putty is the same.\nFind the velocity of the two combined.\nOnce you know their velocity together, find their kinetic energy together.\nTheir potential energy together is the same.\nDetermine how much height they gain.
315572
The best method to solve these type of problems is to sove them through VENN Diagrams. First draw a rectangle which is the sample space, then draw three circles intersecting each other wjich are E, F AND G. The common region between E & F is P[E n F] and so on. It is quite difficult to explain this in writing but I hope you have understood. Now by drawing Venn diagram I have tried to solve ur questions.\n\nPr[EuF'] = P[E] + P[F'] - P[E n F']\n         = 0.35 + (1 - 0.5) - (0.35 - 0.1)\n         = 0.6\n\nPr[F'nG']= 1 - P[F u G]\n         = 1 - ( P[F] + P[G] - P[F n G])\n         = 1 - ( 0.5 + 0.45 - 0.3)\n         = 0.35\n\nPr[E'nG] = P[G] - P[E n G] = 0.45 - 0.2 = 0.25\n\nI think above answers are correct and hope they help you!!
539597
It is related to the length.
300038
To solve, one needs to know the length of the thread.