There is no Roman numeral W. There is I, V, X, L, C, D and M but no W.
I'm assuming you mean the symbol (u) and not 'you'So:u = vw + zWhere vw means v * w (v times w)
yes. not sure of the proof though.
"Vectorial angle" is not a standard term in mathematics. The angle between two vectors v and w is defined as the value theta such that w dot v = v times w times cosine theta. Definitions: w and v are vectors (ordered lists of numbers) of the same dimension n (i.e. both contain n terms). The jth term in w is denoted wj (similarly for v) w dot v = w1v1 + w2v2 +... + wnvn w = square root of w dot w (similarly for w) We always choose theta to be between 0 and pi (i.e. between 0 and 180 degrees), including 0 and pi as possible values of course. If w is a positive multiple of v, theta will be zero. If w is a negative multiple of v, theta will be pi.
V=1/3(l+wl) v=1/3l(1+w) 3v=l(1+w) 3v/(1+w)=l
If it's a 100 W bulb that does not say everything about it. It could be a 12 v, 24 v, 120 v or a 230 v bulb, or other voltage. Look on the case or the packet.
The apparent answer to the question would be (100 W)/(120 V) = 0.8333 A, assuming that, as a pure resistance load, the light bulb has a power factor close to 1.0.
Power= V * A so 120 V * 5 A = 600 W (J/s)
A 240 v 14 w cfl bulb uses about 0.14 amps.
A 60 W light bulb consumes more power (produces more light) than a 40 W light bulb in a given time interval when both are connected to a 120 V circuit. This is because power is directly proportional to the wattage rating of the light bulb.
I = p/v = 100w/9v = 11.11ai = p*v = 100w*9v = 900ai = (p/2)*v = 50w*9v = 450ai = p*2*v = 100w*(9v + 9v) = 8100a
To determine the voltage required to pass 0.5 A of current through a 4.5-W bulb, you can use the formula P = V * I, where P is power, V is voltage, and I is current. Given that power is 4.5 W and current is 0.5 A, you can rearrange the formula to solve for voltage: V = P / I. Plugging in the values, V = 4.5 W / 0.5 A, which equals 9 volts. Therefore, 9 volts is required to pass 0.5 A of current through a 4.5-W bulb.
To do any replacements of different bulbs you have to consider what the new current will be and if the wire size and fuse will be able to take the new current (amps). Use this formula, W = A x V or A = W/V.
The formula you are looking for is I = W/E. Amps = Watts/Volts.
40W Bulb will spiol due to over current passing through its coilAnswerSince the 40-W lamp has a higher resistance than the 100-W lamp, the greater voltage drop will appear the 40 W lamp. As a result the 40 W lamp will be subjected to a voltage beyond its 100-V rating, and the 100-W lamp will be subjected to a voltage below its 100-V rating. Therefore, the 40-W lamp will burn much more brightly than the 100-W lamp.Incidentally, the symbols for the 'watt' and 'volt' are upper, not lower, case: W and V.
it means that a bulb is burned out
A 1000 W heater would have more resistance compared to a 100 W bulb. The higher the power rating of an electrical device, the lower its resistance, as resistance is inversely proportional to power. So, the 1000 W heater would have lower resistance than the 100 W bulb.