Galvanized Iron Wire In Coil With High Quality

Galvanized Iron Wire In Coil With High Quality System 1

Galvanized Iron Wire In Coil With High Quality

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Loading Port:: Tianjin

Payment Terms:: TT OR LC

Min Order Qty:: 5 m.t.

Supply Capability:: 50000 m.t./month

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Specification

Material:

Galvanized Iron Wire

Cross Sectional Shape:

Round

Application:

Construction Wire Mesh

Type:

Galvanized

Wild in Width:

1.1M

Surface Treatment:

galvanized

Metal Wire Drawing:

Cold Drawing

Status:

In Hard State

Thickness:

Metal Thick Wire

Galvanized Technique:

Hot Dipped Galvanized,Electro Galvanized

Product description
High Quality Galvanized Iron Wire
Material: High quality low carbon steel

Processing And Character: Have been gone through the process of wire drawing, acid washing, rust removing, annealing and coiling, it offers excellent flexibility and softness.

Usage: Used in weaving wire mesh, construction, handicrafts, express way fencing mesh, packaging of products and other daily uses.

Specification: Hot dipped galvanized iron wire, BWG24-BWG8; Electric galvanized iron wire: BWG36-BWG8

Product show
High Quality Galvanized Iron Wire

Galvanized wire
Wire gauge size	SWG			BWG
	Inch		mm	Inch	mm
8	0.160		4.06	0.165	4.19
9	0.144		3.66	0.148	3.76
10	0.128		3.25	0.134	3.40
11	0.116		2.95	0.120	3.05
12	0.104		2.64	0.109	2.77
13	0.092		2.34	0.095	2.41
14	0.080		2.03	0.083	2.11
15	0.072		1.83	0.072	1.83
16	0.064		1.63	0.065	1.65
17	0.056		1.42	0.058	1.47
18	0.048		1.22	0.049	1.25
19	0.040		1.02	0.042	1.07
20	0.036		0.91	0.035	0.89
21	0.032		0.81	0.032	0.813
22	0.028		0.71	0.028	0.711
23	0.024		0.61	0.025	0.64
24	0.022		0.56	0.022	0.56
25	0.02		0.51	0.020	0.51
26	0.018		0.46	0.018	0.46
27	0.016		0.42	0.016	0.41
28	0.015		0.38	0.014	0.36
29	0.014		0.35	0.013	0.33
30	0.0124		0.32	0.012	0.31
31	0.012.		0.30	0.010	0.25
32	0.011		0.27	0.009	0.23
33	0.010		0.25	0.008	0.20
34	0.009		0.23	0.007	0.18
35	0.008		0.21	0.005	0.13
36	0.0076		0.19	0.004	0.10
37	0.07	0.17
38	0.006	0.15

Q: A 0.500 g wire is stretched between two points 95.0 cm apart. If the tension in the wire is 600 N, find the wire's first, second, and third harmonics.______Hz (1st)______Hz (2nd)______Hz (3rd): Kia ora In order to answer this question, we first need to ascertain the speed of a wave in this wire. The speed of a wave on a wire depends upon the tension 'T' and the linear density (mass per unit length) 'μ' of the wire. v=√(T/μ) T=600 N The linear density of your wire is 5.00E-4 kg/0.95=5.263E-4 kg/m. So for this wire, v=√(600/5.263E-4) =1068 m/s Now we have the speed, we need to find the wavelengths that correspond to the harmonics we are interested in. The wire is fixed at both ends. Nodes occur at fixed ends. Therefore the first harmonic will occur when there is a node at each end and the longest possible wavelength that satisfies this condition is λ=2L (you get half a wavelength on the wire). So λ=2*0.95=λ=1.90m. If v=1068m/s and λ=1.90 then by the wave equation v=fλ f=v/λ=1068/1.90=562.1 Hz. So that is the first harmonic. The second harmonic has twice the frequency of the first; the third harmonic has three times the frequency of the first. The second harmonic will therefore occur at 562.1*2=1124 Hz and the third at 562.1*3=1686 Hz. Because your data was given to 3sf you need to round your answer to 3sf: 1st harmonic: 562 Hz 2nd harmonic: 1120 Hz 3rd harmonic: 1680 Hz

Q: How do I wire an outlet that has two black and to white wires. It this case the white being the HOT wires?? Don't know why all the white wires in my house are HOT.. And the black are nuetral???: Seen it before in residential, commercial and industrial apps. Why? Haven't a clue other than folks are just backwards that way. *L* I laugh now but it's usually not funny in the field. I've heard arguments between electric power and electronic folk over black v. red color codes but don't remember similar arguments over black v. white. Perhaps someone older than me can shed some light...

Q: Some states allow the use of aluminum wire in houses in place of copper. If you wanted the resistance of your aluminum wire to be the same as that of copper, would the aluminum wire have to have a greater diameter than, smaller diameter than, of the same diameter as the copper wire? Calculate the thickness ratio of aluminum to that of copper.: Are you using 12 volt batteries to do this..?

Q: which guage power wire will handle 100 amps: 4 or 2 gauge wire running 10 to 20 feet for 100 amps.

Q: If you put clay over wire and bake it would the wire melt and ruin the sculpture?Details on wire:I'm not sure what kind of wire it is but it says Bright Floral Wire..Wire is silverDetails on Clay:Again, im not sure what clay...its from polyform products and it says Premo Sculpey so im guessing Polymer? :#92;Clay is Black.Bakes for 275F (130C)30 minutes per/par/por1/4 in (6mm): Copper Chicken Wire

Q: What do all the wires do: Constant 12V+Orange Switched 12V+Yellow GroundBlack IlluminationGray DimmerBrown Antenna TriggerPink AntennaRight Rear Front Speakers5 1/4 Dash Left Front (+)Tan Left Front (-)Gray Right Front (+)Light Green Right Front (-)Dark Green Rear Speakers6 x 9 Rear Deck Left Rear (+)Brown Left Rear (-)Yellow Right Rear (+)Dark Blue Right Rear (-)Light Blue

Q: Two wires, 2.44 m apart, both carry current 2.2 A toward the bottom of the screen. The right wire is extremely long, and the left wire is 0.36 m long. What is the magnetic force on the left wire?: Magnetic force, F = BIL --------(1) where B is the magnetic flux density of the magnetic field caused by the right wire, acting perpendicularly to the left wire in Tesla I is the current in the left wire in Amperes L is the length of the left wire in metres To find B (due to the right wire), you have to use the formula B = (?0)(I)/(2πr) -----(2) where the I in this case is the current in the right wire, r is the distance between the two wires, and ?0 is the permeability of free space, numerically defined as ?0 = 4π×10?7 using eqn (2) so B (due to right wire) = (?0x2.2)/(2πx2.44) = 1.8x10^-7 T using eqn (1) and the value of B calculated above F = (1.8x10^-7 x 2.2 x 0.36) = 1.43x10-7 N

Q: So I know that the neutral wire is the 'return wire'. But why does my textbook say that it has the potential to have the same amount of electricity as the live wire (if the wiring is faulty)? But essentially, if the wiring is not faulty it is technically safe to touch it, it has 0 volts.Could someone explain this to me?: a million. whilst that is working regularly, Voltage could be present for the time of stay and independent twine. cutting-edge could flow from stay to independent twine. 2. whilst there's a fault, The voltage would be present interior the stay twine and since the circuit isn't closed, No cutting-edge could be flowing. 3. No that is no longer ultimate. yet at times, the earth twine could be transmitted to the sub station the place in its joined with the independent. 4. green 5. The Casing could be linked to earth. for this reason the independent is shorted with earth.

Q: One hundred meters of a certain type of wire has a resistance of 7.2 Ω. What is the resistance of 2.5 m of this wire?I figured it would just be a proportion so (7.2 * 2.5) / 100 = 0.18 Ω. Is this the correct approach?: yes, it's a proportion. (2.5/100)7.2 = 0.18 ohm A little more analytical approach is dimensional analysis. I sometimes get proportions backwards, but dimensional analysis prevents that. 7.2Ω/100 m = 0.072Ω/m 0.072Ω/m x 2.5m = 0.18Ω .

Q: Whenever I'm making jewelry, I typically wind up using crimp beads (with toggle closures, in most cases). Whenever I cut off the beading wire at the end, after enclosing the crimp bead around it, there's always this itsy bitsy part of the wire that irritates me to no end when I wear the jewelry. It seems that I can only cut the wire so much before I wind up snipping at the crimp bead. How can I remedy the irritation fact of the left-over wire?Maybe put glue on the wire piece and let it dry? Idk. Grr.: Depending on the size of the hole in the beads, I usually run the wire down throw about 2 inches worth of beads; then crimp and clip. This also ensures that if the necklace if caught and pulled you have about 2 inches of slack before you are chasing beads on the floor. If you hole in the beads are small you can do two different things. 1) find a coordinating bead with larger holes that you can end the necklace with on both sides of the catch 2) Be sure your wire is running straight back down the original wire after going through the crimp, then crimp and clip.The end of the wire will be straighter and not tend to go out to the side and punch your neck. And unfortunately there is not a solution every time.

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