0.1 VDC\n\n0.4 VDC\n\n0.6 VDC\n\n0.9 VDC
201876
The same,\n 1. An equivalent circuit of a NPN transistor is two diodes tied anode to anode; one cathode being the emitter, the other the collector, and the junction of the anodes is the base.  \n\n  2. When a NPN transistor is doing-its-thing, there is always a constant 0.6 volt drop between the base and emitter, i.e., the base is always ~ 0.6 volts more positive than the emitter--always! \n\n  3. There is no output at the collector, until the base has reached ~ 0.6 volts and the base is drawing current, i.e., any signal that appears at the base that is not up to ~ 0.6 volts (and not drawing base current), is never seen at the collector.  \n\n  4. The base requires a current, not a voltage to control the collector current.  \n\n  5. The collector is a current source: it does not source a voltage.  \n\n  6. The collector appears to output a voltage when a resistor is connected between it and power.  \n\n  7. The collector is a high impedance when compared to the emitter.  \n\n  8. The transistor can output an amplified signal either from the collector or the emitter (or both).  \n\n  9. When operating with a collector resistor (RL): the output voltage from the collector is an amplified voltage.  \n\n10. When operating with only an emitter resistor (Re): the output voltage from the emitter is not an amplified voltage, because it is always ~ 0.6 volts, below the input (base) voltage--hence the name voltage follower. But because the emitter can source large amounts of current to the "LOAD," it can be said, there was CURRENT amplification. \n\n11. The collector--being high impedance--cannot drive a low impedance load.  \n\n12. The emitter--being a low impedance--can drive a low impedance load.  \n\n13. The voltage gain from the collector is greater than one (Gv > 1).  \n\n14. The voltage gain from the emitter is less than one (Gv < 1).  \n\n15. Both the collector and the emitter: output ~ the same power: E x I = P.
230398
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/posfe.html#c3\n\nThe link above may be of some help. Input the values you want or need.
456290
4V +O2 ---> 2V2O   Vanadium oxide, or divanadium oxide\n\nBoth are correct.  The first is the preferred.
224349
175.76 Amperes...\nusing m=z*i*t\nz=63.546/96500\nt=24*60*60\nm=10*1000
85807
V = Er = 3E6 x 0,30 = 9E5 volt = 900 kV
403039
dC/dt = 4 m/s\nC = 2*PI*R\ndC/dt = 2*PI*dR/dt = 4\nSo dR/dt = 2/PI\n\ndA/dt = d(PI*R^2)/dt = 2*PI*R*dR/dt = C*dR/dt = 32 * 2/PI = 64/PI m/s
67841
P=IV\n =12.5*120\n =1500W\n =1.5KW
176028
here is one of the datasheets available on the internet:\nhttp://www.ee.washington.edu/stores/DataSheets/74ls/74ls73.pdf
359720
3.99, 4.1, 4.12, 4.214
172835
1 MA\n\n1 Megaampere.
449292
It depends on the applied voltage.  Divide the voltage by the resistance which gives you the amperage.  Now multiply the amperage  with the voltage, and that is your wattage.\nW = V²/R\nThis is straight forward for DC; with AC things become a little more difficult, as because of the fluctuation of the voltage, the average is only .707 of the nominal voltage.
90043
.01uF / 1210/ 50v/ avx /digikey.com/\nMIKE
489255
Nominal= 36ohms\nTolerance=5% \nMultiply nominal by .05 , then add that figure to nominal for Max,.subtract it from nominal for Min.,,and study your ohm's LAW!! ;-)=
404659
16. C 150 Ohms not Watts\n\n17. A\n\n18. D Current Divider Rule in a parallel Circuit\n\n19. C\n\n20. A  6V time 0.04A = 0.24 Watts
551965
The voltage induced is proportional to the velocity of the magnet relative to the coil so the answer to a is 15 volts.  It's proportional to the number of turns of wire so the answer to b is 10 volts.
446787
The answer is e) 121\n\ndelta s= delta h/ t in K\n           = 51000/(150+273)
129714
Since m*C*dT=Q;\n\nHeat capacity=135000 J/(5100 g*(31.5K - 18K))= 1.96 J/gK\n\n(K for Kelvin)