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3

A few thoughts to help you on your way. When an elevator is moving, you have to do work against gravity. You are changing the potential energy of the system. The faster the elevator moves, the more work per unit time is needed (because power = work times velocity). If you are changing the velocity of an object, you are changing its kinetic energy: if it's ...


2

As for the helicopter problem, theoretically, arbitrarily low power can be sufficient to float a load, if you push down a lot of air with very low speed. However, you need longer and lighter (and maybe wider) blades for that, so the problem you'll have to solve is that of structural strength. Let me note that recently a muscle-driven helicopter ...


1

As radio amateurs we've all learned the various relationships of power, voltage, current and resistance as expressed in Ohm's Law Ohm's law is: $$ E = IR \tag{1} $$ This doesn't directly say anything about power. There is the related Joule's first law, which relates to electrical power converted to heat in resistive materials: $$ P = I^2 R \tag{2} $$ ...


3

As Kevin Reid aptly explains, the circuit you have drawn is not realizable. But, let's take the closest physical thing you could build, assuming: your voltage source can supply enough energy that we don't hit its limits like all physical things, this apparatus has non-zero size Then, the circuit you actually built is this: simulate this circuit ...


0

I doubt this is your answer: what is the frequency $f$ doing, in the problem of a magnet sliding down an incline? What you should ask yourself is: how is changing the thickness of the aluminum foil going to modify the time down the incline? Increasing the thickness of the aluminum foil increases/decreases/leaves unaltered the foil's resistance (you pick ...


4

… an ideal power source capable of providing infinite current with no drop in the voltage it supplies. … Let's ignore the effects of current density on superconductors for now. … In these phrases is the explanation for the contradictory possibilities you have computed: you have supposed an impossible circuit. As a mathematical model, the behavior of ...


0

The measure $ R $ of resistance is an invented one. It was deduced long ago by experiment that many materials had a constant ratio $ \frac {V}{I}$ between the voltage applied and the current flowing. Thus the quantity 'resistance' was defined to be precisely this ratio. Later, when inductive and capacitive effects were observed, 'reactance' $ Z $ was defined ...


0

Math already done by others, but my first impression would be yes and no. Yes, because any load requires energy from somewhere; No, because electrical loads less than a major sound system are so insignificant you will never notice outside of a lab. Driving up one decent hill will consume more gasoline than the added load of your laptop over the entire tank, ...


3

This is really just a footnote to alemi's answer. The electricity for your car is supplied by the alternator, and the torque required to turn the alternator depends on the current it's supplying. As you draw more current more torque is required to rotate the alternator and the car has to use more power to do it. So yes, plugging in your laptop will increase ...


5

$$ 110 \text{ hp} = 82 \text{ kW} $$ This is 1000x what the laptop draws. You won't notice. To put that change in perspective, you would see a similar increase in the power (85 W) used by the car if your speed changed from 65.00 mph to 65.02 mph, since $P \propto v^3$ (at high speed, the power goes as the velocity cubed) as per this answer, so $ ...


1

The part that I don't get is mathematically how to get from one to the other. $$\frac{d}{dt}U_L(t) = p_L(t) = v_L(t) \cdot i_L(t) = L \cdot i(t) \cdot \frac{di}{dt}$$ but $$\frac{d}{dt}\left\{\frac{1}{2}Li^2(t)\right\} = L \cdot i(t) \cdot \frac{di}{dt} $$ thus $$U_L(t) = \frac{1}{2}Li^2(t)$$


0

No real transformer is lossless, the output power is always smaller than the input power. One can, however, build transformers that have extremely precise voltage and current ratios (but none of those are transferring any power). Such transformers are being used in calibration and measurement applications.


2

$P = IV$ applies to all circuit branches. $P = I^2R$ or $P = V^2/R$ are restatements of the general rule that apply when we are considering power delivered to an ideal resistor that behaves according to Ohm's law $V = IR.$ I have seen in some circuit $V^2/R$ is not equal to $I^2R$ (like when there is capacitor or inductor). Why is that? Those ...


4

The trouble is that your table, or whatever object it is, will act as a waveguide. That's because the sound waves will (partially) reflect of the wood/air surface then travel back into the table and interfere with other waves. The result is going to be hideously complicated to calculate. As Luboš says in a comment, if the thickness of the table is much less ...



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