High Quality Galvanized Iron Wire Roll
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Quick Details
| Place of Origin: | Standard of Zinc Coating: | Model Number: | |||||||
| Surface Treatment: | Galvanized Technique: | Type: | |||||||
| Function: | Wire Gauge: | Standard: | |||||||
| Material: | Elongation: | Package: | |||||||
| Zinc Coating: | Application: | Used in wire mesh, artware , metal hose , binding wire | |||||||
Tensile Strength Range
Size (mm) | Tensile Strength (mpa) |
0.15-1.60 | 290-550 |
0.65-1.60 | 400-550 |
1.61-6.00 | 400-1200 |
- Q: I dont understand it I am trying to get a 1ohm load on a 2ohm dvc speaker but cant figure out how to tell if the wire you connect in the postive terminal on the sub in a neg wire or positive how do you tell what the gray line is on this diagram positive or negetive and how to tell. thanks
- You are wiring the sub in parallel. from the positive terminal of the amp you connect both positive terminals of the sub, and Viceversa for the negative. The wire is either positive or negative depending on which terminal it touched. Make sure your amp can handle an impedance of 1 ohm.
- Q: A street lamp weighs 200 N. It is supported by two wires that form an angle of 130° with each other. The tensions in the wires are equal.What is the tension in each wire supporting the street lamp?If the angle between the wires supporting the street lamp is reduced to 100°, what is the tension in each wire?
- Problem 1: If the tension in the two wires are equal, the angle they make with the horizontal must also be equal. Always remember that a horizontal line is a straight angle; hence, it has a measurement of 180°. Let's look for the angle the wires make with the horizontal. 130 + x + x = 180 130 + 2x = 180 2x = 180 - 130 2x = 50 (Divide both sides by 2) x = 25° Each wire makes a 25° angle with the horizontal. We can now find the tension. The mass must be divided by 2. Using trigonometry, let's find the tension force. The divided weight of the lamp shall be the opposite side, and the tension force shall be the hypotenuse. sin 25° = (200 / 2) / Ftens Ftens = 100 / sin 25° Ftens = 236.62 N (Answer) The tension is 236.62 N in each wire. Problem 2: We use the same approach. 100 + x + x = 180 100 + 2x = 180 2x = 180 - 100 2x = 80 (Divide both sides by 2) x = 40° Each wire makes a 40° angle with the horizontal. Calculate the tension force. sin 40° = (200 / 2) / Ftens Ftens = 100 / sin 40° Ftens = 155.57 N (Answer) The tension is 155.57 N in each wire. Hope this helps!
- Q: What good would it do to change my spark plugs and wires would it be a good idea for a 86 year of a car they've never been changed before I was wondering would it help the car out thanks for the help
- yes, i would change spark plugs and wires. the spark plugs will fire better an make you get better fuel milage, and same with the wires. the wires after so much age depending on how close the wires are can have small arcs go between the wires and cause miss firing/pre ignition making it harder on the engine.
- 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.
- Wiring that previous could desire to get replaced. Insulators wreck down over the years. In Edison's day insulation became into, rope and tar, or fabric, with glass insulators to guard lumber. you like a rectifier to create pulse ac (see image voltaic Panel grant) i hit upon it not straight forward to have faith anybody might grant DC power. The generator could desire to be on the threshold of factor to apply, do to severe voltage drop in DC cutting-edge over long distance. through fact of this AC is used now. till that's an Edison Museum image voltaic power is DC too. I advise studying image voltaic power for ideal DC wiring practices. I actual have been a electrician my total existence and know little or no approximately DC wiring and contraptions. in case you presently have electrical energy, get multimeter attempt for ac voltage first. then DC to steer away from unfavorable the meter, to confirm what power you have
- Q: I'm trying to install a ceiling fan but I'm having trouble with the wiring. Coming out of the fan I've got a black, white, black and white striped, and a green wire. coming out of the cieling i've got a black, a white and a red wire.The fan's green wire is connected to the bracket as a ground, so that's out of the way. It's the other wires that confuse me. I've tried, white to white, black to black, and BW striped to red, but that didn't do anything.Any ideas? Thanks.
- Be careful when with just hooking Black to Black or White to White as has been suggested. In some configurations the white wire is part of the switch leg and is not nuetral. Being that this is a 50 year old house, the Hot wire from the breaker box could very possibly be coming in at the light box instead of the switch box which would mean that there is a wire that is using both the Black and the White wires for power. One of them is taking power to the switch and the other is bringing power back to the light from the switch when the switch is turned on. In this scenario, the white wire would not be nuetral!!! The brown wire was typicallly used for hot up into the 70's and then they changed it to black. Your best bet is to hook it up exactly the way you took it off and hook the Blue and Black wire from your fan to the same wire. I assume that this fan has pull chain controls for both the light and the fan?.....
- Q: Hi,I'm trying to install a ceiling light with a 3 wire configuration (white, black, ground) into a wire configuration of 4 wires (white, red, black, ground). This is configured for two on/off switches.So far, I have white to white, black to black, and ground to ground with the red capped off. This works with one switch perfectly, however when I hit the other switch, the circuit breaker blows.Please help, and thank you!
- I think according to your description of the wire configuration what you have is a power supply at the switches and 2 lead wires (red and black) coming up to power the light and fan separately. The white wire, or neutral wire is for a return path while the ground wire is a backup in the event the neutral fails.At the switch in the wall, you should note a black wire (power source) coming to the bottom terminal of any switches. From there, you will find another black or red wire going to the light itself. The neutral wires are all connected together and tucked away in the switch box. If the 4 wire configuration is from the fixture itself, then you would have a black wire for power to the light, a red wire for power to the fan unit while the white is a neutral return. If there is only one wall switch, you could in theory, connect the black and red from the fixture to the power supply coming up from the switch and therefore use the pull chains located on the fixture to turn on the light or fan or both. Failing this, it would be best to call a qualified electrician and protect yourself from what could be a nasty fire. Good Luck.
- Q: I currently have a ceiling fan wired to a dimmer wall switch that controls the fan and the light, not good I know. I would like to set up the dimmer switch where it controls the light only and then i can use the fan pull switch for the fan. but i am not sure how to wire this with my current wires. In the wall box where the dimmer switch is I have two sets of wires coming from two different locations. one set has a black, red, white, and green. the other location has black, white and copper. currently the green and copper are connected and capped, the two whites are connected together and capped, two blacks are and connected to the black wire coming from the dimmer and the red is connected to the other black wire from the dimmer. how would I wire the dimmer switch to only control the light and not the fan? I want to make sure i do this safely.any assistance is greatly appreciated.
- enable me initiate out by making use of asserting that the cord's popping out of the ceiling bypass like this, Black cord is alway the warm cord the White cord is often the common floor cord the bare cord is the earth floor cord. if there's a purple cord popping out of the ceiling it is so which you would be able to turn the mild off at extra then one change . i'm uncertain what a hunter fan is ,yet what ever you do do no longer hook the black (warm) cord with the WHITE (floor) cord, in case you probably did try this ,you will could reset your breaker , it form of seems such as you may have 2 purple cord popping out of the ceiling , in case you do you may hook them to the Black / White cord, it is so which you would be able to paintings the mild from the two change.
- Q: The new light has 1 live, 1 neutral 1 earth wire, do i remove the earth connect the reds if so what about the 2 blacks?
- There should be black and white wires in the ceiling. Black is hot, white is neutral. If there is no ground wire, that's OK. You just wouldn't connect the ground wire anywhere.
- Q: i have a 3mm diameter wire that has only a single thick conductor and it's really really hard?how do i know if it is a magnetic wire or not?are thick magnetic wires hard or soft to be wrapped?are there other types of wire that have a single conductor thick conductor?note: single thick conductorgt;gt; i mean they are not thin parallel collected wires like usual
- All okorder /
- Q: Say you have a 5 cm current wire carrying 10 A going from left to right. Directly 1 c.m below the left end of this wire is a long wire that is perpendicular to the first wire and goes out of the page. What is the net force on the 5 cm wire?I've tried using F=ILB with the I of the first wire and the B of the second wire.
- I didn't read the question so I was carefully working out the force. The wire is perpendicular to the first wire, so using the right hand rule you discover that the field it creates is PARALLEL to the first wire at this left end. The magnetic force is caused by the component which is PERPENDICULAR to the wire which is in fact zero. So there is no magnetic force at this point. As you move along the wire you get a diminishing amount of magnetism caused by the wire which is going out of the page but that field has a component which is DOWN the page. Therefore that part of the wire experiences a force which is into the page. ( take your right hand, put the thumb along the wire pointing to the right, the fingers point down the page, the palm points into the page which is then the direction of the force) I would be surprised if you were required to work out the magnitude of the force in this context. You can't use F= ILB because both the magnitude and the direction of the field varies at different points along the wire. If the perpendicular wire had been directly below the middle of the other wire there would have been no net force. If you were of a level where working out the force was appropriate you would need to set up the formula for B at various points along the wire, taking the vertical component only and integrate this over the range from 0 to 5 cm. Not a trivial mathematical task.
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High Quality Galvanized Iron Wire Roll
- Ref Price:
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- Loading Port:
- China Main Port
- Payment Terms:
- TT OR LC
- Min Order Qty:
- -
- Supply Capability:
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