Q:A 41.7-kg sign is suspended by two wires, as the drawing shows. Wire 1 makes and angle of 42.2deg with the horizontal and wire 2 makes an angle of 58.8deg. Find the tension in wire 1 and wire 2.

To solve this, you have to find 2 equations, because there are 2 unknowns. Assume, Tension in wire 1 = T1 Vertical component of T1 = T1 sin42.2 Horizontal component of T1 = T1 cos42.2 Tension in wire 2 = T2 Vertical component of T2 = T2 sin58.8 Horizontal component of T2 = T2 cos58.8 Since the sign board is in equilibrium, the forces must balance each other. Horizontally, the horizontal components of both forces are balancing each other. Therefore, T2 cos58.8 = T1 cos42.2 T2 = (T1 cos42.2) / (cos58.8) -------------------(1) For the vertical components of the wire, both are balancing the weight of the sign. T1 sin42.2 + T2 sin58.8 = (41.7)(9.8) T1 sin42.2 + T2 sin58.8 = 408.66N ----------------(2) Combine (1) into (2), T1 sin42.2 + (T1 cos42.2)(sin58.8) / (cos58.8) = 408.66 T1 ( sin 42.2 + cos 42.2 tan 58.8) = 408.66 T1 = 215.66N Subst T1 = 215.66N into (1), T2 = (T1 cos42.2) / (cos58.8) = 308.4N Hope it is correct ^^

Q:Wire runs from breakers to a wall outlet along the garage wall

If you drill through the studs so that the cable does not lay on the face of the structure, studs, you can use NM (non-metallic) sometimes call romex. You should use 12 gage so use 12-2 cable which will have three wires including the ground. Connect to a new or otherwise unused 20 amp breaker and install as many outlets you want to this circuit.

Q:I have a 1989 Nissan 240sx. Somehow the fan caught a hold of the wiring harness and tore all the wires. any idea how much it's going to cost?

Buying a wiring harness will be prohibitive, and even transplanting one from a wrecking yard can be expensive. The harness can be repaired by splicing in pieces of wire where wire is missing and splicing together damaged pieces of wire. It is tedious but not particularly time consuming - I spliced in a section of harness after an engine fire in about 2 hours, working slowly.

Q:I need thick floral wire and they keep saying things like 24 gauge 26 gauge 28 gauge.. like how would i know which one is thiickest if i'm buying it online.

Large number = small wire It is (Usually) American Wire Gauge. There are a number of different wire standards, most standards the larger number is smaller wire, except metric measurements which use actual measurements.

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:Ok. so i wired my cd deck into my 92 buick, but everytime i start the car i have to manually turn on the cd deck. i also have to change the settings everytime! and once i turn off the car the setting will not save and the setting will be lost. Now everytime i turn my car on i have to manually put the settings again! this did not happen in my old car! and i think it might be the wiring! Because the last cd deck i had did save the settings! and would turn thge cd player on everytime i turned on my car!! and it would play the track back, right where i left off. please help! i did attach the yellow cable wire the red cable to the battery current.? is there a problem!

You should have the yellow power wire connected to constant power and the red wired to ACC. You don't and this is why it's bahiving the way it is.

Q:so im doing a custom buggy and i need so different colored wire im looking for a place to get lots of different colored wire for cheap

This Site Might Help You. RE: i needed many different colored electrical wire for cheap? so im doing a custom buggy and i need so different colored wire im looking for a place to get lots of different colored wire for cheap

Q:There is a circuit with a battery connected to two wires in parallel. Both wires are made of the same material and are of the same length, but the diameter of wire A is twice the diameter of wire B. Answer true or false for the following questionsThe curent through the battery is five times larger than the current through wire B.The power dissipated in wire A is 16 times the power dissipated in wire B.The voltage drop across wire B is larger than the voltage drop across wire A.The resistance of wire B is four times as large as the resistance of wire A.The resistance of wire B is twice as large as the resistance of wire A.

the diameter of wire A is twice the diameter of wire B This means the cross-sectional area of A is 4 times that of B, which means the resistance of A is 1/4 that of B. To put some numbers on it, if B is 4 ohms, A is 1 ohm, total R is 0.8 ohms. Which means that if you have 1 amp of current flowing, you have 0.8 volts drop across the two wires, and 0.8 amps flowing through A and 0.2 amps flowing through B. Power in A is 0.8*0.8 = 0.64W, power in B is 0.8*0.2 = 0.16W, total power is 0.8*1 = 0.8 watts 1. true, see above 2. false, 4 times 3. false, they have the same voltage drop, they are in parallel 4. true 5. false, see #4 .

Q:Three parallel wires are each carrying a 4A current. ; wire A) is 6 mm from wire B) which is 3mm from wire C); The current in wires B and C are out of the paper, while A is into the paper. What is the magnitude and direction of the magnetic field halfway between wires A and B?I have tried using Biot-Savart Law but keep getting it wrong can someone please help? I also have an equation for two parallel wires but how do I relate it to two?

You can use the Biot-Savart Law, but it is confusing and kind of a waste of your time and space. For straight wires, someone ALREADY worked out the Biot-Savart Law. See the following link for the result: hyperphysics.phy-astr.gsu.edu/Hba... The formula of interest for us is B = mu0*I/(2*Pi*r) where mu0 is magnetic permeability of free space, I is current, and r is distance from the wire carrying the current. B is the magnetic field due to that PARTICULAR wire. To deal with three wires, use a superposition principle and stack magnetic fields on top of each other. Do be aware of direction: remember the right hand rule. RH rule for magnetic fields in vicinity of wires: point thumb in direction of current, curl fingers to show the magnetic field circulation direction. Use this sign convention: + B is up along page, -B is down along page For Wire A: point of interest is r = d_ab/2 or 0.003 m to the right of wire A Point thumb in to the paper and the finger curl indicates that B_A is downward B_A = -mu0*I_A/(pi*d_ab) For Wire B: point of interest is r = d_ab/2 or 0.003 m to the left of wire B Point thumb out of the paper and the finger curl indicates that B_B is downward B_B = -mu0*I_B/(pi*d_ab) For wire C: point of interest is r = d_ab/2 + d_bc or 0.006 m to the left of wire C Point thumb out of the paper and the finger curl indicates that B_C is downward B_C = -mu0*I_C/(2*pi*(d_ab/2 + d_bc)) Add up: Bnet = B_A + B_B + B_C Bnet = -mu0*I_A/(pi*d_ab) - mu0*I_B/(pi*d_ab) - mu0*I_C/(2*pi*(d_ab/2 + d_bc)) Simplify: Bnet = -mu0/pi*(I_A/d_ab + I_B/d_ab + I_C/(d_ab + 2*d_bc)) data: mu0:=4*Pi*10^(-7) Tesla-m/A; I_A:=4 A; I_B:=4 A; I_C:=4 A; d_ab:=0.006 m;d_bc:=0.003 m; Result: Bnet = -6.667 milliTeslas negative sign indicates downward direction.

Q:Two current carrying wires are perpendicular to each other.Wire 1 is placed on top of wire 2.Wire 1 current direction is pointing North, wire 2 current direction is pointing east. Is there a force on either of the two wires?Does one wire spin due to the produced force?Please explain to me how to do this questionThanks

Take your right hand and point its thumb to the North representing conventional current flow (Plus to Minus) in the top wire. Visualize the curl of your right hand's fingers can now be replaced be visualizing a bike tire rotating from nuckles to fingernail. Now do the same with the wire having current going East. Contact the two bike tires together a some infinitesimal spot. If those tires are exactly perpendicular in the X and Y (Z doesn't seem to matter because it is the point where they touch), it would make sense to predict that on top a vector will point West and on the bottom a vector will point South. The net vector should be SW. So, if you marked the initial point of cross over on the wires and mechanically could keep them free to move but still perpendicular, the cross over point would move SW for as long as equal current flowed in the two wires.