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19

The device is apparently working as a heat pump, for which I give a brief theoretical analysis here. In the example given, the $P_h=69{\rm pW}$ light output comprises the $W=30{\rm pW}$ input by the researchers together with $P_c=39{\rm pW}$ of heat that was formerly in the chip. We can model the process by ideal heat pumping as follows. Heat drawn from ...


6

Essentially, yes. The derivation of the escape velocity is based only on the energy balance and energy does not depend on the direction. This follows from the property of the gravitational field to be conservative, so work required to move between any 2 points is independent from the path you take. Vague intuition: the vertical speed you estimate is about $...


5

Newton's third law tells us that the momentum imparted on one body is equal and opposite to the momentum imparted on another if they interact. We then have $$ \Delta \vec p_1~=~-\Delta\vec p_2. $$ The change in momentum is $\Delta \vec p_i~=~m\vec a_i\Delta t$, $i~=~1,~2$. The change in momentum is with Newton's second law due to a force so that $$ \vec F_1~...


4

Before mixing the average kinetic energy of the molecules which make up the tea is greater than the average kinetic energy of the molecules which make up the milk. This is restatement of the fact that the tea is hotter than the milk as the temperature of a substance depends on the average kinetic energy of the molecules which make up the substance. When you ...


4

The underlying reason for OP's flawed argument is that a premature use of EOMs in the stationary action principle $$ S~=~\int\!dt ~L(r,\dot{r};\theta,\dot{\theta}), \qquad L(r,\dot{r};\theta,\dot{\theta})~=~\frac{1}{2}m(\dot{r}^2 +r^2\dot{\theta}^2) -V(r),\tag{A}$$ destroys the variational principle. Concretely OP is implicitly assuming that (3) is a ...


4

Given that the device is extracting heat (vibrational energy) from the lattice that the LED is embedded in, the conservation of energy issue can be understood. The question you really should be worried about is, "does this violate the second law of thermodynamis?" Is the overall free energy increasing? You can use thermoelectric coooling, for example, ...


4

In newtonian mechanics the angle does not matter, but in relativity it does. For example: An object close to the speed of light launched horizontally will orbit circular at a distance of 3GM/c² from the center of mass (the so called photon sphere), but it will escape if launched vertically. When you launch it at a distance just above 2GM/c² (the so called ...


3

The classical interference pattern is explained by the equations governing the behavior of light, and energy there is treated as a collective phenomenon, using the Pointing vector Energy transfer in a light beam can be best understood as an emergent phenomenon from the underlying quantum mechanical level. Innumerable photons create the visible interference ...


3

The conceptual problem seems to be however there is no way to define the friction for the rope since it has two parts, with apparently distinct potential energies. Don't worry about the "distinct potential energies". You can just compute the gain in potential energy for each part separately. For the bit already hanging down, as the rope slides by a ...


2

Effective potential is defined by the formula $E=T_{radial}+V_{eff}(r)$. Your calculation shows that once you make this identification it is not true that $\mathcal L = T_{radial}-V_{eff}$, but that is fine. This happens because centrifugal term (i.e. the one with angular momentum) is really kinetic term and not a true potential. Hence it must enter the ...


2

@dvij gave the equation $$g\sin \theta =R\frac{d^2\theta}{dt^2}=R\frac{d\omega }{dt}$$ If we multiply this by omega, we obtain: $$g\sin \theta \frac{d\theta}{dt}=R\omega\frac{d\omega }{dt}$$ If we integrate this equation between 0 and t, we obtain: $$g(1-\cos \theta)=\frac{R}{2}\omega^2$$ So we have $$mg\cos\theta-2mg(1-\cos \theta)=N=mg(3cos\theta-2)$$ I ...


2

If you accept that no external work was done, then if there is a change in the state of a system through which the kinetic energy changed, there must be a corresponding change in potential energy. The key to understanding the (rather poorly narrated) video is that the lecturer implies (at T=2:30) that $\Delta E=0$ from which it follows that $\Delta KE= - \...


2

Newton's second law, force f is $$f=m\frac{d^2 x}{d t^2}$$ x is position vector of the particle. $$f=-\frac{d v}{dx}$$v is the potential energy. $$m\frac{d^2 x}{d t^2}=-\frac{d v}{dx}$$ Multiply both sides with $\dot x$ $$\frac{m}{2} \frac{d\dot x^2}{dt}=-\frac{dv}{dt}$$ $$ \frac{d}{dt}(\frac{1}{2}m\dot x^2+v)=0$$ i.e., $$\frac{dE}{dt}=0$$Energy is ...


1

You are talking about two different cases. At first you say: We notice that our liquid is no longer as hot as before adding milk. And your meaning about "our liquid" is the tea. Then you say: The total thermal energy of the system (the cup) is conserved. And your meaning about "system" is tea + milk. The total thermal energy of tea and milk is ...


1

When you say that the system (ie the contents of the cup) is not as 'hot' as before, you are talking about temperature. However, the system was initially at two different temperatures (hot tea, cold milk). There is no obvious way of comparing the two initial temperatures with one final temperature. To decide if there has been a change in temperature, you ...


1

A simple pendulum performs simple harmonic motion when it is displaced very slightly. You can say that a simple pendulum performs a periodic motion which can be treated as simple harmonic motion in small oscillation Now lets move forward supposing it to be pure SHM. The time period of a simple pendulum does not depend on how much it is displaced( but it ...


1

The idea of a "Zero-Energy Universe" is a theory held by a limited number of scientists. There are several stackexchange question that expand on the theory and may help you. Zero energy universe Total energy of the Universe


1

We put the circular orbit of the particle on a straight line and convert the motion to a 1-dimensional rectilinear motion as follows : The arc length, the natural parameter $\:s(t)\:$ is the distance travelled on the straight line till time $\:t\:$. The speed $\:v(t)\:$ on the straight line is the magnitude of the tangent to the circle velocity. Now, on ...


1

The efficiency is being miscalculated because they left out the energy being used to heat up the device to 135 degrees! This is similar to claiming that an amplifier that takes a 1 mW input signal and "converts" it into 1 W output signal, as having an efficiency of 1000%! They are leaving out the energy being supplied to the amplifier "external" to the ...


1

When you reading the plot in the paper, when efficiency is above 100%, the temperature is 135 degC (275 F) and the light output power is low. This makes me think, when you heat an steel bar to hot red, it emits light without electricity. You can pretend to inject electricity into the bar, and your efficiency is infinite. Energy is from hot environment. ...


1

Feynmann's machines are idealised devices and we don't know or care about how they work internally. All we know is that we can connect the weights to the machines and they will move them according to the rules. The 1kg and 3kg weights start at height zero. We connect them to machine B and it lowers the 1kg weight to $h=-1$ and raises the 3kg weight to $h=Y$....



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