# Tag Info

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If you want to create a generator then this might help. And if you really want to use the energy which is produced while the magnets repel you can use a piezoelectric generator.

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Hint:Another possible way to solve it would be to observe that $$a=\frac{dv}{dt}=\frac{dv}{dx}*\frac{dx}{dt}$$ Hence $$a=v\frac{dv}{dx}$$ Now according to question power is constant Hence P=k(say) $$Fv=k$$ $$\Rightarrow mav=k$$ $$\Rightarrow mv^2\frac{dv}{dx}=k$$ Solve the differential equations with the given limits to get the an equation of v in terms of ...

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Hint : You assumed the force to be constant by using $W=Fx$ which is wrong. It is $W=\int Fdx$ Use $P=\frac{dW}{dt}$ Use work energy theorem. Use calculus. Find distance covered when speed is $6ms^{-1}$. That should eliminate your variables if the question has sufficient information.

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I think your question is either: what FREQUENCY of alternating current, as in the case of a local open core transformer, or else what frequency of RF energy (as in microwave, radio wave, etc.) could be used to power or provide charge to laptops batteries 'wirelessly'? Not just any frequency you want, that's for sure. In the United States, the FCC takes a ...

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In the simplest case, and some unmentioned assumptions, the minimum charging time would be given by the formula already mentioned: $$T = \frac{V \times Ah}{W}$$ this gives 12x150/25 = 72 hours. However, this case makes at least two major assumptions, the battery is fully discharged and there are no losses of any kind! In the real world, the battery ...

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Power is usually just quoted as energy/time, but this is actually a bit vague: Which energy, and which time? When talking about energy, we either reference a system/physical object which the energy is a property of, or we are talking about the energy exchanged between two systems. The time when used to talk about power implies a process occurring over some ...

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All the equations are really the same thing. Which one you use is dependent on the question given. Power is the rate at which electrical energy is supplied to a circuit or consumed by a load. $$Power=\frac{energy}{time}$$ $$P=\frac{QV}{t}$$ where $QV$ is the electrical energy consumed by the load with potential difference $V$ when $Q$ charge passes ...

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All are correct. Note that that $I$ is current passing through the resistor and $V$ is potential difference across resistor. Use those equations in which you know the values of variables and not have to calculate them. Why not use all equations in a question and satisfy yourself that all are correct? These are just using ohm's law in $P=VI$. You must ...

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You may think you are not moving when you plank, but your body maintains plank by pushing you back up imperceptibly after you droop imperceptibly. Your muscles do work against gravity. I can't imagine how to estimate how much.

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The first equation is the definition of power which is, in words, the rate of energy conversion: $$P \equiv \frac{dE}{dt}$$ Where it is understood that $E$ is the amount of energy converted. For example, in a mechanical system where gravitational potential energy is converted to mechanical kinetic energy and vica versa. In electrical circuits, the power ...

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Picking up on jinawee's answer, current is charge per unit time: $$I = \frac{Q}{t}$$ So substituting for $I$ in your second equation gives: $$P = V \frac{Q}{t}$$ But $VQ$ is just the work done, i.e. the energy, in moving a charge $Q$ through a voltage difference of $V$. So substituting $E$ for $VQ$ gives us: $$P = \frac{E}{t}$$ which shows that ...

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Power is defined as: $$P(t)=\frac{dE(t)}{dt}$$ This is valid for any system. If energy is constant, then: $$P(t)=\frac{E(t)}{t}$$ If you're dealing with a resistance in a circuit, the dissipated power is given by Joule's law: $$P=VI$$ So the last one is a particular case of the first one.

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The mistake comes from averaging V and then squaring the average. If you squared the voltages and then found the weighted average of the squares, the result would match your initial calculation. There is another, more basic way to calculate the average power. Find the energy released during each of the two periods; add the two amounts of energy and ...

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Maybe this can help. Consider a body accelerating under constant power from the initial conditions $t=0$, $x=x_1$ and $v=v_1$. $$a(v) = \frac{P}{m v}$$ $$t(v) = \int \frac{1}{a(v)}\,{\rm d} v = \int \frac{m v}{P} \,{\rm d}v = \frac{m}{P} \left( \frac{v^2}{2} - \frac{v_1^2}{2} \right)$$ $$v(t) = \sqrt{ \frac{2 P t}{m} + v_1^2}$$  x(v) = \int ...

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There would be real problems if as you said " run a really long wire from the array to the ground" if you did that you would have alot of voltage drop. There is a significant voltage drop even for a run of wire of 100 ft. But if you put a pv system on say a mountain top with few sun obstructions and gathered the power there. But you would have to live ...

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In general a larger antenna gives a larger signal at the input of the receiver. Since the receiver is not noise-free the signal to noise ratio is increased, resulting in better reception. The antenna itself is not noise-free (unless strongly cooled down) but this is usually of minor importance. Actually amplitude modulation is not so much used, rather a ...

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