New answers tagged

1

Typically in such questions, the 1st interpretation is intended, that is, they want you to calculate the total energy delivered by the battery up to time $t$, divided by $t$. The instantaneous power at time $t$ (usually) isn't as important, and if they do want you to calculate it you should be able to tell that from the context. IMHO, good textbooks (and ...


3

The parabolic equation for a beam propagating in the $z$ direction comes from inserting $$ \varphi(x,y,z,t) = \psi(x,y,z)e^{i(kz-\omega t)} $$ into the wave equation $$ \frac{\partial^2 \varphi}{\partial x^2}+\frac{\partial^2 \varphi}{\partial y^2}+\frac{\partial^2 \varphi}{\partial z^2}- \frac 1{c^2}\frac{\partial^2 \varphi}{\partial t^2}=0 $$ to get $$ ...


1

Any combustion engine is capable of operation at a variety of RPM's and torque output combinations that depend on the nature of the load that the engine is driving. Since delivered power is the product of torque and RPM, this means that there are many different combinations of the two across the engine's operating regime that will yield the same net power ...


0

Why does the rate at which capacitor in an RC circuit charges only depend on R and C But the rate at which a capacitor charges does depend on $V_\mathbf{emf}$ if, by this, you mean the rate of change of $Q$. For a series RC circuit with zero initial voltage, we have $$\frac{dQ}{dt} = i_C(t) = \frac{V_\mathbf{emf}}{R}e^{-t/RC}$$ Clearly, the rate at ...


0

Of course, in an ideal RC circuit the capacitor never fully charges. This is because the factor of $1-e^{-t/RC}$ can never be exactly equal to $1$ in the expression for the charge on the capacitor $$q(t)=CV_\text{emf}\left(1-e^{-t/RC}\right)$$ Therefore, we need to define some fraction $\rho$ of $CV_\text{emf}$ as the threshold of "the capacitor is charged"....


0

The type of diagram that measures this behaviour is called Current-voltage characteristic. There is actually no single formula because with increasing voltage, the filament inside the light bulb heats up which increases the resistance. This is different for every type of lamp, so you'd need to do an experiment and measure the current for different voltages ...


2

Note that value you are given in the question for $R_{earth}$ is the radius of the Earth's orbit, not the radius of the Earth itself. The expression $\frac P A \times 4 \pi r^2$ is calculating the power radiated by the sun across the area of a sphere as large as the Earth's whole orbit, not just the area of the Earth.


-1

Kilowatt-hours without time information are useless "1 kW⋅h" does not give any information about how much time was needed in order to transform this amount of energy. It could have been either 1000W during an hour or ~114mW during a year. Similarly, my household could also consume 12 gigawatt-hours worth of electricity. It would just need 4000 years to do ...


3

The way you phrase the question is slightly ambiguous, it doesn't double count units per time. One is per time, the other is $\times$ time. And therefore cancels to get joules. However just using joules, you get no idea on the time period that amount of energy was used, just the amount. Using $kW\cdot h$ gives you a reference for how long you used an amount ...


11

There is historical reason behind using kWh as well as matter of general convenience. Many people around the world pay their electricity bills using the "per unit of electricity" concept that energy provider companies have defined as 1kWh. Generally, the supplier has a base charge/connection charge for a certain number of electricity units (say "$b$"). Any ...


4

Practically speaking, maybe you want to know how much energy it takes to keep your refrigerator running for one hour. On the back of a fridge you can find the power in watts. If you measure energy in units of $kW\cdot h$, it's easier to figure out how long you can keep it running. You can call it the "natural unit for a thrifty man".


0

I know this is very old, but I'd like to reply to the analogy about a water bottle. The reason the water doesn't flow any more quickly or slowly when you move your bottle up and down is because of the size of the Earth and how close we are to it. The gravitational force equation is $F_g= \frac{G.m_1.m_2}{r^2}$. You can see from this equation that the force ...


0

You can, using kinematics and there are some third party libraries to do this using factors about the vehicle within 10% of the engine controller it also uses the gradient of terrain. I saw someone also refine this approach to also use average wind speed. I have some python code to do this.


2

https://www.cdc.gov/nceh/hearing_loss/how_does_loud_noise_cause_hearing_loss.html The hairs in your inner ear that respond to sound can be damaged by loud sounds. But each hair responds to one frequency. That frequency vibrates the hair. Loud sounds vibrate them too hard. But no hairs respond to ultrasound. This is different from light. UV light is highly ...


2

The answer is no, because you can't disregard frequency. The echolocation sounds used by bats, for instance, can be extremely loud: they can have SPLs of something like $140\,\mathrm{dB}$ (see here). Yet bats do not deafen us. The reason they don't deafen us is because the sound they produce is very far outside the frequency range to which our ears are ...


1

Work energy theorem states "work is said to be done when there is change in Kinetic energy", i.e. Work = ∆Kinetic energy This is not a very accurate statement of the work energy theorem. A more correct statement is that the “net work” is equal to the change in kinetic energy: $W_{net}=\Delta KE$ This is an important distinction because the term “net work” ...


2

Work energy theorem states "work is said to be done when there is change in Kinetic energy", i.e. Work = ∆Kinetic energy Actually, the theorem is more accurately stated as: The net work done on an object equals its change in kinetic energy. Notice the emphasis on the word "net". That's because work can be positive or negative and if the amount of ...


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