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9

I worked at the Wind Energy Project Group at the TU Delft in the Netherlands for a summer. One of the most interesting ideas was a wind turbine which only had one blade, with a massive ugly blob on the other side as a counterweight. The idea was, "if we spend three times as much money manufacturing just one really good blade, could we make the system better ...


9

It's the same reason that most airplane propellers have only two or three blades. As the blade moves through the air (or the air moves over the blade), it leaves a wake. If the next blade encounters the wake, it will be moving through disturbed air and will be less efficient. The more blades you have, the more likely each blade will pass through air ...


5

The issue is that there are multiple crater scaling laws, each with different assumptions, as Horedt & Neukem (1984) show (title of paper is Comparison of six crater-scaling laws). Your first equation is a naive approach that the volume of the hole is linearly related to the kinetic energy, hence the $d= k\cdot E^{1/3}$ relationship. This particular ...


4

Technically, both solids and gasses have quantized energy levels. The difference is that molecules of a gas interact with other molecules very weakly, so the energy levels observed in emission or absorption of a collection of gas molecules are almost exactly the same as the energy levels that would be observed if you had a single gas molecule in isolation. ...


3

"Rest mass" is probably a bit more abstract in modern science than what you seem to be thinking. It is simply this: if you can put yourself into an inertial frame such that a body is at rest relative to you, then that body's rest mass is defined as that body's total energy measured in this particular inertial frame - multiplied by $c^2$, if you want to ...


3

The answer depends on whether the wheels skid. When you brake with just the rear wheel, it's quite possible to skid; if you apply the front brake, the increase in normal force on that wheel tends to prevent skidding (although in extreme cases it could make you fly over the handlebars). Applying the rear brakes hard enough to block the wheel would generate ...


3

So the earth won't gain any significant energy There's your error. The Earth will gain a significant amount of energy, enough to compensate for the loss of energy of the ball. But because the Earth is so massive, its speed won't significantly change.


3

Higher derivatives of position don't have their own corresponding types of energy, because they're not independent quantities. (See this or this or any of several similar questions.) $F = ma$ relates the second derivative ($a$) to the position itself ($F(x)$), so if there were a type of energy that depended on acceleration, you could just re-express it as a ...


2

Imagine wind blowing along a plane with the air by the ground all a nice and steady temperature. Now this wind encounters a mountain range, so is forced upwards. The pressure is lower at higher altitude since there is less remaining atmosphere above it. The temperature of gas decreases when the pressure is lowered, which is why this same air gets ...


2

An atmosphere in absolute equilibrium in fact is isothermal (see below for more detailed analysis of your cannonball). However, if the atmosphere is mixed by wind, gas expands and contracts adiabatically. If the mixing is fast enough, it obeys relatively well the adiabatic invariant, which multiplied by suitable form of ideal gas law ($(T/(pV))^\gamma = ...


2

Why it is colder in mountains, at high altitudes? One answer is that mountains on Earth aren't all that tall. An impossibly tall mountain would see temperatures vary with altitude as depicted below. Tall as it is, even Mount Everest doesn't extend into the stratosphere. This is a question about the lowest layer of the atmosphere, the troposphere. ...


2

Work is transfer of energy from one system to another OR transformation of energy from one form to another. Either way, work does not create energy. When I lift an object, I am transferring energy from my body/muscles to the object-earth system. The energy goes into potential energy of the object-earth system because the separation between the object and ...


2

The energy of a relativistic body is given by: $$ E^2 = p^2c^2 + m^2c^4 $$ where $m$ is the rest mass and $p$ is the relativistic momentum, which is given by: $$ p = \frac{mv}{1 - v^2/c^2} $$ Using this you can easily calculate the energy as a function of velocity. As you have already worked out, it doesn't make any difference whether you are ...


2

What is mass of elementary particles? It is the "length" of the four vector (p_x,p_y,-_z,E), for complex systems it is called rest mass. As the length of three dimensional vectors is not additive ( think adding two opposite momenta), rest masses are not additive, vector algebra has to be used. The invariant mass of your two gammas must be larger than the ...


2

Induction cooktops contain electromagnets below each pot or pan station. When a station is switched on, electric current flows through wire wrapped around an iron core. In order for magnetic flux to be induced in the iron core, the electric current must constantly change, so the current must alternate. The iron core concentrates the magnetic flux ...


2

Quantization is an experimental fact that forced physicists to consider theories that could explain the data. This happened in the beginning of the twentieth century. 1) black body radiation could only be explained by assuming that the radiation came in quanta, i.e. not in a continuous spectrum.) 2) The photoelectric effect showed that light behaved as a ...


2

The quantization of energy levels appears both in quantum and classical mechanics, and it is not a consequence of the Schrödinger equation. It is a consequence of confinement. In fact, anytime that a wave equation (any quantum equation for the wavefunction, or a classical equation for a classical field, e.g., EM field) has periodic boundary conditions in ...


2

Generally, in linear systems modes are independent. Energy does not flow from one mode to another. What causes the coupling is a nonlinearity. The nonlinearity reveals itself at higher amplitudes (nonlinear terms are small at small amplitudes). Thus, when you drive the rod just a little bit the energy DOES go to the higher harmonics, but the coupling is weak ...


2

It is not that easy. Now one engine is delivering about twice as much torque and was probably not designed for that. An engine specifically designed to deliver more torque would like be more efficient than the two combined. Lets look torque alone. Ignore the tow rope and assume the second is far enough back to not get any draft. In this case wind ...


2

No-one knows what dark energy is, but we can be fairly sure it is not related in any way to the electromagnetic force for the simple reason that dark energy is uniformly distributed and charged matter is not. If the two were related we would expect some correlation between them. Incidentally, The Magnetic Universe is an excellent book about cosmological ...


2

I am sorry to disappoint you, but there is no such formula that you can just apply. This is because it strongly depends on how and under what exact conditions and with wwhich tools you did the experiment. Think of this: If every methanol molecule reacts (burns) at once all at the same time, then the exact same amount of energy is spent, but it went really ...


2

Individual photons will not lose any energy as long as they do not interact with any other particle. If you are referring to the intensity of the EM emission, that depends of the angle incidence from their source. So basically, if you imagine a laser that could emit just a single photon in the vacuum of space, that photon would maintain its frequency, and ...


2

In quantum mechanics the equation of motion is the Schrödinger equation $$ i\hbar\,\frac{\partial}{\partial t}\,|\psi\rangle = H|\psi\rangle $$ where the (self-adjoint) operator $H$, the Hamiltonian, determines its evolution. The energy levels are, by definition, the eigenvalues of such operator in its domain of definition $\mathcal{D}_H$. Spectral theory ...


1

In fact, the water would act as a neutron moderator, speeding up the reaction. However, reactor pressure vessels are quite sturdy, and it would be very unlikely for the salt water to enter the pressure vessel.


1

Observe that $v\dot v=\frac12\frac{\text d}{\text dt}v^2 = \frac12\frac{\text d}{\text dt}\mathbf v\cdot\mathbf v = \mathbf v\cdot\dot{\mathbf v}$ and therefore the integral of the OP becomes (assuming a constant mass $m$, e.g. a point particle) $$\Delta\left(\frac12mv^2\right)=\int_a^b mv\ \text dv = \int_a^b \frac{\text d}{\text dt}(m\mathbf v)\cdot\mathbf ...


1

If Rob Jeffries numbers are right (Here) and it seems as good a baseline as any, you get about 3*10^15th Joules per KG of Neutron Star matter. The Hiroshima bomb was about 6*10^13th Joules, so just 1 KG of degenerate matter, you'd be looking at about 50 Hiroshima bombs released in kinetic energy, in the form of (I think) mostly Neutrons. 1 KG of hydrogen ...


1

A possible answer for that might be that if you have a rope with the length $L$, you have a frequency $f$ as the first harmonic frequency with $T=\frac{1}{f}$ as the time between two amplitude maxima. This time is determined by the frequency how fast the wave can propagate in the rope, and is therefore bound to the speed of sound in the rope $$\nu = ...


1

I disagree with the other two answers, or at least believe they are not very clear. The important point is that, as you suggested, a induction cooker only consumes significant amounts of energy when a pot is actually on top of it. There are, of course other losses, but without any metal object in the vicinity, the cooktop is like a transformer without a ...


1

Where does mass come from in pair production? From the energy-momentum of the gamma photons. Think of photon momentum as resistance to change-in-motion for a wave propagating linearly at c. You've maybe heard of electron spin and the Einstein-de Haas effect which "demonstrates that spin angular momentum is indeed of the same nature as the angular ...


1

vsz, we are talking about relativity here, aren't we? Therefore, acceleration and deceleration are relative as well. Not the very fact, obviously, because you can always measure your acceleration with an onboard accelerometer without referring to the outside world. But in order to measure your initial and final velocity you do need a frame of reference ...



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