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It is better to stagger them. You want the largest overall effective propeller area for efficiency. This is the same reason that airplanes with wide wingspan are more efficient. Note that sailplanes, where efficiency really matters, have wide and thing wings. The reason is that lift comes from momentum, which is mass times velocity of the air that is ...


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Power is defined as $P = Fv$ where $F$ is the driving force and $v$ is the velocity of the moving object. In this case, determine the values of both $F$ and $v$, and use this to calculate the power. If you need additional help, feel free to ask in the comments.


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Let the battery voltage be $V_S$. Then, the battery current is $$I_S = \frac{V_S}{(100 + 100||100)\Omega} = \frac{V_S}{150 \Omega} $$ The current through the series (left-most) resistor is just $I_S$ and the current through each parallel resistor is just $I_S/2$. Since the resistor power is given by $$p_R = R \cdot I^2_R$$ the power delivered to the ...


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You'd be lucky to even generate 1 Watt. Compare yours to this much larger 10 Watt generator intended to operate off hundreds of degrees of temperature difference: http://www.devilwatt.com/products/17-10-watt-camping-stove-thermoelectric-generator


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The details of how your particular device will perform is very much a function of how that device was constructed and how it will be used, but in general, the physical effect is called the Seebeck effect -- briefly, the conversion of a temperature difference across a device to a voltage. Efficiency for such devices is actually quite low, with $\eta = 0.02$ ...


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If you're looking to produce power from a temperature differential, go with a device optimized for the Seebeck effect. Peltier and Seebeck effects are essentially the same thing, or rather flip sides of the same thing, but thermoelectric generators (Seebeck) are optimized differently from thermoelectric coolers (Peltier).


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For 1, you are spot on. The output voltage and current are multiplied to get the power. Numbers in the press are often optimistic. It may be a peak power, rather than one that can be sustained if they were willing to release enough water over a long period.


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You have to assume a certain shape for the torque curve. For example if the torque is more of less constant then $$ T = \frac{P_1}{\omega_1} = \frac{P_2}{\omega_2} $$ $$ \frac{200}{4000} = \frac{P}{5000} $$ It helps to know the rpm of peak torque, as typically torque varies as a parabola near that point. Depending on the valvetrain, the torque curve will ...


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No, it is not correct. Let $a$ be acceleration; $F$ force; $m$ mass; $v_0$ initial velocity; $v_f$ final velocity; $P$ power required; $x$ distance travelled and $t$ time taken. Hence, $$P=\frac{Fs}{t}=\frac{ma\ x}{t}$$ Then $a=\frac{v_f^2-v_0^2}{2x}=\frac{5^2-0^2}{2*10}=1.25\text{ m s}^{-2}$ and $t=\frac{2x}{v_0+v_f}=\frac{2*10}{0+5}=4\text{ s}$ Hence, ...


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Yes you can transform power from overhead powerlines if you run parallel to them. I directly undernieth them. For example if you were to use an electric fence setup. By using the insulated attachments and you ran a wire back and forwards say a few 100 feet long, this acts as a transformer to the high voltage and current being used in the above HV lines. I ...



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