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Yes there is, first consider the original power formula $P = w/t$ Now we know that work is equal to force.displacement (both vectors) $P = (F.D)/t$ $P = F.(D/t)$ Displacement over time is velocity, so $P = F.V$ Hence power can be stated as a dot product of the vectors force and velocity Expanding the dot product $P = FV\cos\theta$ Where ...


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But isn't $I^2R$ equal to $V^2/R$, therefore if $R$ is constant doesn't the power depend on the square of the voltage so surely it doesn't matter whether it is high voltage or high current. Consider the wires connecting the power plant to the appliance; let the effective amplitude of oscillating current flowing through all wires be $I$ and let $V$ be ...


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If you keep the resistance constant, then $V=IR$ means that voltage is directly proportional to current. If you keep the power constant, then $V=\frac{P}{I}$ means that voltage is inversely proportional to current. However, because $V=IR$, we can write that $P=I^2R$. Therefore, if we say resistance is constant, then power must change with current, which ...


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for same voltage supply, the power consumed by two resistances in series connection is less in compare to power consumed by same resistances in parallel connection. Therefore we can say that - P(series) < p(parallel)


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First, when you rewrote the first formula using the "explicit notation", it is very hard to see what you gained. On one hand, you indicated that $y$ is a function of $x,t$ which would be fine (although it's clearly redundant to write this long expression instead of $y$ all the time) but I can't understand why you wrote $x_0,t_0$ instead of $x,t$. Even more ...


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I think this formula would work better F=MxA Since you are hovering you need to consider the acceleration of gravity in your calculations. So you would get F= M x 9.832 meters per second squared. Use kilograms for the mass in this case


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There is a confusion of what that $v$ means. You are thinking about the velocity of the drone, which is stationary this $v_{drone}=0$. But in your equation, to calculate the power needed by the wings, you have to consider the velocity of the motor providing the thrust (propelling air downwards at a certain rate) to keep the drone floating which is making ...


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Power equals work/time. Work is force times distance, so you can simplify to force times speed, when the force is constant and the force is causing the speed. In your situation there is a force due to gravity that would do work on your drone, and what you need is for the drone to do work to counter balance that. In other words, power is always work over ...


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The problem is that finite wings inherently induce drag (called lift induced drag). Even if you could assume the wing infinite you still have parasitic drag. So you will always need to supply energy to keep the plane flying at a constant speed (and as a consequence to keep generating lift). Without drag it could, in principle, keep flying without power. It's ...


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In principle, you are right that it is possible to keep an object in the air without expending any energy. Think about a blimp, you can fill it with helium, and then just let it float. It doesn't have to do anything to float, it just floats because it is lighter than air at ground level (the air gets less dense as you go up). You can even make heavier than ...


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The Peltier and Seebeck effects are each others opposites so to say. The Seebeck effect is described by the electromotive force or voltage $V$ generated at a temperature difference $\Delta T$ across the ends of a material: $$V=S\Delta T$$ $S$ is the Seebech coefficient and is a material constant that depends on charge carriers, material density and much ...


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TL;DR You would eventually reach the higher velocity, but the hill ends before you get close. No Gear Changes First lets dismiss the notion that the power and torque requirements to ascend a hill at constant speed are different. To ascend a constant grade hill at a constant speed requires the same constant force $F$ regardless of velocity $V$. Let's ...


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Both equations are valid. You just made an error in taking suddenly 20 $\Omega$ instead of 40 $\Omega$. $$P = I^2 \cdot R = 25 A^2 \cdot 40 \Omega = 1 kW$$ $$P = U^2 / R = 40000 V^2 / 40\Omega = 1 kW$$ The power is always determind by the current that runs through the part of interest times the voltage drop over it. If there are other parts in a ...


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For turbulent flow, the friction goes up as the square of the velocity. So the force (pressure) goes as $n^2$. And you are moving $n$ times more liquid. The velocity (or volume flow) increases by $n$. The two factors combine to give the cube law since power = force times velocity, or pressure times flow rate.



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