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How to improve performance! The electric motor?
The ore is extracted from the mine by conveyor belt. The conveyor is driven by a DC electric motor. The engine has an internal ohmic resistance, but is also a perfect engine. http://www.crusher-in-china.com/img/belt-conveyor-1.jpg state Operating current motor constant voltage U = 100V load ore Lo = 3 kg / m depth belt tree H = 100 m (including the additional height above ground) g = 10 m / s ² mineral extraction of Mo = 1 kg per second electric current Io = 40 amps cost of extracting ore (price electricity) Co = $ 1 per tonne engineer proposes to increase the load L4 = 4 kg per meter of tape, the engineer of other proposals to reduce the load L4 = 2 kg per meter of belt. What is the annual extraction of new M4 and M2 and the new costs for C4 and C2 extraction? question Bonus: What is this curve represents: http://en.wikipedia.org/wiki/File:Laffer-Curve.svg
Perfect resitstance engine only ohmic, that is, reactive (Not having to worry about losing after the finals. Current Requirements: = 1 kg / 100 sec * 9.8 m / s = 980 watts of power used = V * I = 40 amps * 100 volts = 4000 watts ...... Did anyone take a screwdriver inside? Search internal resistance RP = I ^ 2 = RR (4000W - 980 W) / (40 amps) ^ 2) R = 1.888 Ω ********* OK lets generalize P 0 = P = VI - I ^ 2R - Load Load = 100VI - I ^ 2 .... 1.888Ω (revised edition of this from 40V to 100V) in the relationship between the load or loss 100VI Electrical resistance = - I ^ 2 1.888Ω / I ^ 2 = 100 1.888Ω VI / (I ^ 2 1.888Ω) - 1 = 100 V / (I 1.888Ω) - 1 = 53.0 / I - 1 ^ %^%&* - It is late and I have not reached the correct results. But for part 2, the Laffer curve represents the maximum efficiency. Increase or decrease the bearing capacity of production is received less net energy. *********** Try to sleep last night, I understood why it stalled. A real engine would be a sort (RPM for example) sweet spot and I tried to extract some form of the ideal ... Scythe also notes that you can not do because there is not enough information. In addition, in real life, I would also like to see if I could increase the tension a little, or in this case, get a new engine ... Well, it's just a hypothesis, so assume that the engine runs a constant speed. Case 1: L = 4 kg / m, the extraction rate = 4 / 3 kg / s = 980 W load * 4 / 3 = 1307 watts of load = 100 V * I - I ^ 2 = 0 1.888Ω 1.888Ω I ^ 2 * - * 100V + I 1307 Watt has 2 roots: I = 29.4 and 23.5 GB I = Cost amplifiers with 23.5 = 23.5/40 * 3 / 4 = [$ .44/ton Note: use the broader term, the extraction cost is $ .55/ton load = 2 = 2 / 3 * 980W = 653 0 = I ^ 2 * 1.888Ω - 100V Watts 653 * I + = I = 45.3 amps and 7.6 amps Obviously you'll want to run your engine of 7.6 amps. Cost * = $ * extraction 7.6amps/40amps 1/ton 3 / 2 = $ 0.285/ton But wait, the engine, instead of running at 40 amps with a load of 980 watts, capable of handling the same load of 13 amps! C3 = 0.325 U.S. $ / ton of interest: If you have things right, ie, hit the bass amp reading, go with 2 kg / m, but if you manage the bad things, go with 4 kg / m, because it is more efficient than today. Regarding the Laffer curve: You are more effective recovery rate of the marginal tax low tax, but you can collect more than a "sweet spot". If you pass, you pay less taxes. As the Laffer curve is a complex problem which has been simplified. lol (Laffer aloud). *********** Interesting thought alone was related to Laffer curve. There are some advantages of a government needs to increase revenue growth and provide longer term. These include education, courts police, fire, military, and the rule of law, transport, R & D, and environmental regulations. If you put all your faith in the Laffer curve, you have no income and no government services and ending with an economy like Somolia. Thus, even if they are more effective in the short term to collect taxes can not be minimal long-term sense.
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how to increase height wikipedia