a copper telephone wire is strung,  with liitle sag betwenn two poles. how much longer is the wire on a summer
439386
Standard copper coaxial CABLE TV wires use a drip-loop to compensate.\n\nThe drip loop is approximately 12 inches long, so the wire probably doesn't move more than six inches.
954336
I looked it up for you - \n\nget the distance from your jump to the ground, then add 3 feet - that's your cord length...
847006
how  long  is a piece  of  string  ...depends on  were you are  what country   your in   etc etc etc    ...best  Idea  is  to speak  to  your local  fire   prevention  office  ...
420078
Here's a link to the standard American Wire Guage table.  It has it all listed:\nhttp://www.eng.auburn.edu/~tdenney/ENGR1110/wireawg.html
30686
Let the length of the equal ropes be x ft\n    the length of the longer rope is (x+2)ft\n\ntherefore,\n\nx+x+x+(x+2) = 44\n4x = 44-2\n4x = 42\nx = 10.5
631579
2 electrodes and a battery
680693
Summer depending on the wind LOL\nYour swing in cold weather will be the main factor for the shorter distance.
910217
There are two issues to consider, cable heating and voltage drop.  Of the two, voltage drop is by far the easiest.  You need to determine what the allowable voltage drop in you cable is.  The voltage drop is just I x R where I is the current and R is teh round trip cable resistance.  The specification of 100 kW isn't enough.  If its a high tension 100 kV line, the current is only 1 amp and a run of 16 AWG wire would deliver about 99,999 volts out of the original 100,000 volts to your load.  If its a 100 volt system then a 4 AWG wire would only deliver only 20 volts to the load.  \n\nThe formula for wire resistance (in AWG) is that 10 AWG has 0.001 ohm/ft.  Each increase of 1 in the gauge increases resistance by the cube root of 2 (about 25%).  You calculate by picking what voltage drop you can live with, for the known current figure the resistance, knowing the length figure the resistance per foot, and then pick the AWG using the above.\n\nThe second concern is wire heating.  Although you could do the heat transfer calculations, instead look up the allowable ampacity for any wire in an electrical engineers or NEC handbook.
1108611
east/west.\n\nto remember, longitutde has lines that are all as "long" as each other (one pole to the other.)
672243
your question doesn't make sense and is unclear - can you clarify it and then maybe i can help you?\n\ni got that a 100ft tower is erected, but is 100ft the actual length of it or the vertical distance it covers?  or is the tower erected straight vertically?  or normal to the hill? and the two wires (i assume you mistyped "guy")?  i don't really get it...
914639
T = mg.\n\nThe length of the cable is irrelevant in this example. The beam is hanging off the cable.\n\nThe tension in the cable therefore is m*g, which is 2450 N.
365352
Wire is typically sold by length off of a spool. Decide what gauge you need (depends on the voltage and how many amps you expect to carry), how many wires (single, double, triple), and what length. Go to a hardware store and they will cut it for you.\n\nRegarding whether it is live or not, the best way to tell is to use a voltmeter. Touch the wire with the probes and see whether there is any current going through it.\n\nHOWEVER, a mistake when dealing with electricity may very well be your last mistake. Don't underestimate the dangers. Not to be rude, but if you have to ask how to know if a wire is live (hot) or not, you may want to forgo doing this yourself and call an electrician. (I realize that this is in homework help, but just in case you were considering a project, I added this warning)
935014
If the square has sides of 12, the perimeter of the square is 4x12 = 48 inches. This is the length of the wire.\n\nTo find out the sides of the equilateral triangle, divide 48 by 3, since the triangle has 3 equal sides.
960889
Circumference of spool = 2PiR\n\nSince the length of the cable is l, the number of times the cable can be wrapped around the spool = l/(2PiR)\n\nEach "wrap" will cover the spool by 2r.\n\nTherefore the total length that can be covered = 2r*l/(2PiR)\n=rl/(PiR)
1169146
At least 12 feet.\nI could be wrong though.
430412
The wires are placed on the poles according to their voltage - highest voltage on top (farthest away from people) and lowest on the bottom. This means the bottom one(s) is usually telephone (12 volts), then cable tv (30 volts) then the "secondaries" which are what your house runs on (220 volts) and then all the way up top, the "primaries" which carry electricity over long distances, and are usually 13,600 volts.
762824
The answer to this is actually pretty simple - and it applies to the Christmas lights, too.\n\nThe cable (or wire) that the signal passes through is formed of multiple strands of much finer conductor, twisted to form a (relatively) uniform outer surface to which the insulating jacket is bound.\n\nOver time, you'll tend to pull the cord into longer and longer lengths ... with one of two results. Either the twist gets tighter (and the cord appears slightly thinner), or the twist gets looser.\n\nIn either instance, the "memory" of the wire will want to try and return it to its initial configuration (including tightness of twist). Once the stress is off the conductor, back it goes! And probably snarls itself in the occurrence, because you've tried to pack it away in the EXTENDED configuration.\n\nKnotting is another phenomenon entirely. In this case, relative motion of the gathered conductor bundle tends to "slip" one loop against the next. In the extreme, what used to be the outside loop becomes the inside loop ... and you have a potential knot. Unknowingly, you reach in and grab the "new" inside loop and pull ... thereby completing the slippage and you now definitely have a knot!