Galvanized Steel Wires For Chainlink Fencing

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Galvanized Wire/Galvanized Steel Wire/ Gavanized Iron Wire

Diatmeters:0.15mm-6mm

Application	Weaving, braiding, fencing, cable armoring, knitting, tie wire, for redrawn, for binding or forming etc.
Material	low carbon steel wire
Diameter	0.15mm~6.00mm
Tensile strength	350-500MPA, or higher
Zinc coating	Min 12g/m2, min 25g/m2, min 200g/m2,min 300g/m2,610g/m2
Surface treatment	Hot dipped or electro galvanized
Packing	On spools In small coils of 25-50kg/coil, 10kg/coil In rosette coil of 100-800kg/coil
Standard	ASTM 641, EN10257-1& EN10244-2

All can be produced according to customers’ requirements.

Q: alright, 5 speakers with 4ohm svc, what would my wiring options be?: With 5 speakers there won't be a symmetral wiring pattern for even power distribution. With 4 it would be 2 pair in parallel with both sets in series for a 4 ohm final load or all 4 in parallel for a 1 ohm final load.

Q: 1994 crown vic but it seemed like only 3 or 4 wires matched up exactly and this one wire was way to short so i had to route it the best way i could the guy said it was the right set of wires but the car runs way better then it did just wondering should i change couple of wires around to make it fit or just leave it what does crossing the wires mean: Are the short wires touching any hot parts(exhaust manifolds)?If not, you'll be fine. There is a chance you used the wrong length wire for a few cylinders but as long as shes running fine and the wires don't touch hot parts you should be safe.Crossing the wires to me means running cyl 1 wire to cyl 5, this would cause your engine to misfire and idle really rough.

Q: A long horizontal wire carries a current of = 52 . A second wire, made of 1.00--diameter copper wire and parallel to the first, is kept in suspension magnetically 5.0 below.What is the magnitude of the current in the lower wire?: The lower wire is kept in suspension magnetically means that the downward force of gravity is equal to the upward magnetic force. Its NOT an easy problem so we need to solve it in small steps. Let's find the two forces acting on the lower wire: (*) Downward force of gravity = mg = (volume) x (density) x (gravitational acceleration) = (pi r^2 L) x(8920)x(9.8) = 0.06896L We assume the radius of the wire is 0.5 mm or 0.0005m, copper density is 8960 N/m^3, gravitational acceleration = 9.8 m/s^2 (**) Upward magnetic force = mu(zero) x i(one) x i(two) x L / (2 pi d) = 2.08x10^(-4) L i(two) mu(zero) is a constant = 4 pi 10^(-7), i(one) is the current in the first wire = 52A, i(two) is the unknown current in the second wire, d is the distance between the wires = 5 mm or 0.005m The forces are equal, therefore 0.06896L = 2.08x10^(-4) L i(two) We divide both sides by L (the unknown length of the lower wire) and now we can solve for the current i(two) ANSWER: i(two) = 331.5 A

Q: which guage power wire will handle 100 amps: A okorder /

Q: Phone wire outside was old is anybody done that, do it your self put a new wire on?, Just curious to know..: The wire comes down or up to a connection box. This is the place separating yours from theirs. Anything on your side of that is yours and you are responsible for maintaining it. Changing it is not difficult. You need some wire, fittings and a tool to attach the fittings. The peiople who sell that should be glad to explain how to do what. Most of the connection boxes now use a plug-in connector so it becomes a piece of cake. If the wire toward the street from there is the problem, just report it and a staticky phone. That is the phone company's problem.

Q: The drawing shows two long, straight wires that are suspended from the ceiling. The mass per unit of length of each wire is 0.050kg/m. Each of the four strings suspending the wires has a length of 1.2m. When the wires carry identical currents in opposite directions, the angel between the strings holding the two wires is 15 degrees. What is the current in each wire?? Im not sure what equations to use for this problem.: You need two basic equation: 1° Force acting on conductor in magnetic field: F = B I λ ..........(1) where B is magnetic field (also called magnetic induction) I is current λ is length of wire 2° Magnetic field near straight wire carrying current I B = μo I / (2πd) ..........(2) where d is distance from the wire μo is the magnetic permeability constant of vacuum (approx. same for air) μo= 4π*10^-7 Tm/A When there are two parallel wires on distance d carrying currents I1 and I2, each of them produce magnetic field causing forces acting on wires. Magnetic field from second wire creates force on first wire: F1 = B2 * I1 * λ Magnetic field from first wire creates force on second wire: F2 = B1 * I2 * λ These forces are equal in magnitude: F = μo I1 I2 λ / (2πd) ..........(3) When currents flow in same direction forces are attractive, and for opposite direction forces are repulsive. When I1=I2, F = μo I? λ / (2πd) ..........(4) Each wire, suspended on strings, decline from vertical position for angle θ=15°/2 = 7.5°, so that the angle between the strings holding the two wires is 2θ=15°. We find F from condition of static equilibrium of torques (relative to point where strings are attached to ceiling): F L cos θ = G L sin θ ..........(5) where G is weight of wire; G = m'gλ where m' is the mass per unit of length of each wire. L is length of strings from (5) we get relation F = G tan θ ..........(6) or F = m' g λ tan θ ..........(7) and from (4) and (7) : μo I? λ / (2πd) = m' g λ tan θ ..........(8) λ on both sides cancels, and we find current I as I = √(2 π d m' g tan θ / μo) ..........(9) You didn't say if there was some initial distance between wires, so we'll assume they are suspended in same points. In that case d = 2L sinθ ..........(10) and finally I = 2 √(π L m' g sinθ tan θ / μo) ..........(11) I = √[1.2 * 0.05 * 9.81 * sin 7.5° * tan 7.5° / 10^-7 ] I = 318.034 A

Q: How does this work? Does the missile drag a wire?: In a wire guided missile system, the missile drags out a wire, or one is spooled out from within the missile, this wire is connected to a targeting system, which is being controlled by an operator from the launch point of the missile. For most systems, as long as the operator maintains a line of sight on the target throughout the flight of the missile, then it will hit wherever the crosshairs are from the targeting system.

Q: There are two switches – the left is for the fan, the right is for the light. They operate independently. I want to replace the light switch with one that has a motion sensor.There are four sets of bundled wires coming out of the back of the box. Each bundle has a white, a black, and a copper wire.Two black wires come out of each switch. The top wires go into two of the bundles in the back of the box, the bottom wires are connected with each other and with two more black wires coming out of the other two bundles coming out of the back of the box.All the white wires are twisted together with a red cap. All of the copper wires are twisted together with another red cap.My motion sensor has a red, black, green and yellow wire. The directions say the yellow wire is not used, because this is not a 3-way light. Is there a straightforward way to do this?: Sure is!. You wire it exactly the way the instructions say to. Which is to use the 2 wires on the existing switch. The bare wire on the switch (if present) is the ground and goes to the green. Just like the instructions say. You are apparently letting all those other wires scare you. The motion sensor is nothing more than an automatic switch. It does the same thing the existing switch does, using the exact same wires. Do Not touch any of the existing splices. Look again at the instructions. Take a deep breath and have at. I've been in control cabinets with Hundreds of wires in them. I ignored all but the ones I need to work on. Other wise I would still be there trying to figure it out! (chuckle)

Q: Why would I want to upgrade the wires on my airsoft gun? What does it do?: cheap wires tend to break over time, short out, etc. It's nice to have quality stuff in you gun.

Q: i already tried wiring this light, but it didn't work. I always thought black wire to black, white to white and ground to ground. i connected the light fixture black wire to the three black wires (as shown in image) and the light fixture white wire to the three white wires, and connected ground wires together.when i flipped on the breaker, the light came on with the switch on off and the breaker tripped. i'm assuming this is wrong and i provided an image as i see in my ceiling mount. Can somebody kindly direct me how to properly connect the light fixture to these wires. thanks!: This should have worked. Any chance you have one of the black wires touching the side of the box or one of the white or ground wires? If the fixture that was removed worked before the new one should work now. Also check that one of the screws holding the fixture didn't touch one of the black wires when you tightened it. .

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