# Tag Info

1

For (1): the energy required to vaporise a liquid is known as the enthalpy of vaporisation, or alternatively as the latent heat of vaporisation. For water at room temperature this is about 44kJ per mole, and a mole of water is 18g. For (2): the density of air at room temperature and pressure is about 1.2 kg per cubic metre, so 1kg of air is 1/1.2 or about ...

2

PhotonicBoom is correct in saying that the airflow created by blowing across the top of the coffee will replace the coffee-heated air with cooler air that will absorb more heat from the coffee. It also allows more of the coffee to evaporate (which might seem like a bad thing, but evaporation is simply the hottest molecules becoming gaseous and leaving, so it ...

1

I am assuming that zero (rest) mass particles don't interact gravitationally with each other and other particles. That's not a valid assumption in general relativity. Particles with zero invariant mass have energy and momentum and, thus, gravitate. Essentially, in general relativity, the density and flux of energy and momentum are the sources for ...

1

The trajectory of an object, whether massless or not, in a curved spacetime is given by the geodesic equation. $${d^2 x^\mu \over d\tau^2} + \Gamma^\mu_{\alpha\beta} {dx^\alpha \over d\tau} {dx^\beta \over d\tau} = 0$$ Solving this is a formidable problem unless there is some helpful symmetry that simplifies the working, but in the particular case of ...

1

No, a massive body is able to bend light around it, which is called gravitational lensing. This has been observed multiple times. EDIT Photons are massless. Otherwise, they would not travel at the maximum speed, which is called speed of light. Keep in mind, that gravitational lensing is not a part of Newtonian mechanics. You need general relativity for ...

1

John Rennie's answer is good already, but I want to add a single point: These fluctuations are very very short. In quantum mechanics you've got Heisenbergs uncertainty principle, which is often stated as $$\Delta x \cdot \Delta p \le \frac \hbar 2$$ and which means, that for any quantum object (think of an electron or a positron created in such a vacuum ...

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I think the key conceptual hurdle is that the vacuum state is not nothing. Quantum field theory describes matter as excitations in quantum fields. These quantum fields are very strange things, and I don't know of any easy way to explain to a non-physicist what a quantum field is. The key thing is that the quantum fields fill all of spacetime. So a vacuum is ...

1

In principle, the gravitational potential energy should be included into total internal energy, but in practice, most often it is not. I know of two reasons. because for systems that are discussed in thermodynamics, it is believed that gravitational energy is negligible compared to electromagnetic potential energy of the constituting particles; because it ...

1

Your separation into potential energy of the system as a whole due to external force fields and energy contained within the system known as internal energy seems a bit arbitrary. Still, if you want to split the PE up this way gravitational interactions within the system would have to go into internal energy. Take the Solar System as an example. Everything ...

1

The gravitational potential for particle 1 is $V_1(r_{12}) = -Gm_2/r_{12}$ and for particle 2 it is $V_2(r_{12}) = -Gm_1/r_{12}$. $m_2$ is in $V_1$ and vice versa because the gravitational potential energy is the potential energy in a gravitational field per unit mass, and therefore only depends on the mass that is generating that gravitational field. When ...

4

When a cup of coffee is hot, the air molecules directly above it get hot as well. After some time, they reach equilibrium and no heat transfer (or maybe very little transfer) occurs. By blowing, you disturb that equilibrium and replace the hot air molecules directly above the cup with colder air and therefore create once again a steeper temperature gradient. ...

1

The experiment is correct. Well, what is really correct is the final interpretation of the experiment. Radiation gets redshifted in the gravitational fields. It's one of the well-known consequences of general relativity. It's been known theoretical since 1915 or so and directly and safely experimentally verified in the 1960s. All the "magic" terminology on ...

0

If the masses of the charged particles are the same then I think that you can only apply an electric field to separate them, thus doing work on the plasma. However if the masses are different, as it probably is, you can try to exploit some thermodynamic-statistical properties of the species, here are just a couple of ideas. Lightest particle will travel ...

2

Shouldn't the energy gain be greater than this formula describes since the energy from the electric field is applied for so long? The electron gains energy and accelerates until it encounters a collision. This is a statistical process and there's a distribution for the energy loss for many electrons. Then, it can accelerate again from that point on, ...

1

The continuity equation (without sources) is usually written as follows $$\partial_t \rho + \nabla \cdot \mathbf{j} = 0$$ If you identify $\rho$ as the mass density, integrate over some volume $V$ and use the divergence theorem you get the result that you mention in your question. Namely, the change in mass in $V$ equals the amount of mass flowing through ...

0

Reading in the disclaimer section: The authors of this website are not responsible for ridicule, which can cause yourself by an effort to reproduce the devices described here. The authors of this website are not responsible for the time you may waste reading this website. They are also not responsible for the time you may waste by trying ...

1

As Wikipedia clearly lays out, the concept of a free energy generator - a machine that will perform work on external systems eternally and without needing external intervention - is inconsistent with either the first or the second law of thermodynamics. There exist, as yet, no credible and reproducible experiments that shed any type of doubt on either of ...

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You should read the (rather funny) disclaimer on that site: This website presents a serious risk of damaging your self-confidence in case you decide to take any content of this website seriously and try unsuccesfully to utilize it any usefull way. "Free energy" is not possible due to energy conservation.

2

Work. Potential energy exists because of some force that exists, and moving an object relative to that force causes work to be done. And by the work-energy theorem, the work done on an object is equal to the change in kinetic energy of that object.

0

This is not specific to converting potential energy to kinetic, but the term you're looking for might be "transduction" or "to transduce"- the process of converting energy from one form to another.

2

You'll want a much bigger heatsink!! (and maybe just one TEC) If it's being cooled only by convection then maybe a heat sink area* that is 10 times that of the TEC. (maybe bigger) The classic mistake with a TEC is to make the heat sink too small. With too small a heatsink the hot side of the TEC gets hotter, more thermal leakage through the TEC, it has ...

2

If efficiency is the issue, then definitely parallel TECs (or use a single unit rated for twice the power, same thing). The only reason for stacking TECs is to get a lower temperature. However that comes at great expense to efficiency and overall power consumption. Another point is that paralleling TECs is actually more efficient overall. The reason is ...

3

Many students confuse the term work in physics with the conventional term of work. Your body wastes energy when you push something, and when that something doesn't move... 100% is wasted in the biological efficiency. 1st step: forget the concept of how hard it would be for you to do it. How much work is a table doing by holding up a 1kg weight? zero. It ...

0

It seems to me the obvious answer is that it agrees with all the experiments. I am an experimentalist after all, (the neutrino is perhaps the best "missing energy" story.)

1

With Regards to the Energy Conservation To turn a turbine requires energy. A turbine will convert the energy that is required to turn it (mechanical energy), into another form of energy (electricity). This is why there is resistance in turning a turbine; if there was no resistance and it produced energy, it would defy the law of conservation of energy. ...

0

eVs= hv + work function of the metal. So putting e= 1.6*10^-19 and h stands for planck's constant=6.62606957 × 10-34 m2 kg / s v= frequency and simply hv as the energy of the photon.

0

You want to put in joules and take out watts? Watts = joules/second. Done. As for friction, if it is turning in space, why is there any?

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From its potential energy $V=mgh$. If you have 1Kg of water falling from 10meters you can produce at most 100Joules of energy. ( I used $g\approx10m/s^2$).

3

Can someone explain the atomic process, if it even exists, of how this would work to convert energy in to matter, and what form of energy was initially present, and what is required to cause this change? It is not an atomic process, it is an elementary particle process, atoms are made up of elementary particles in a non trivial way. At the level of ...

1

They are actually trying to do so in the lab. They need a very potent laser (I believe they didn't reach the critical power yet). The idea is simple, inside the lase cavity you generate a very powerful electromagnetic field, powerful enough such that the photos have enough energy to transform a virtual electron-positron pair into a real one. See this link ...

0

Only half of the total number of nuclear reactions release energy. The other half are the reversed processes that absorb energy. However almost the totality of the spontaneous processes are the ones that release energy. Processes that requires some external energy to be inserted into the system are much more rare as can take place only when this energy is ...

0

The atomic hypothesis describes processes, and so we shall look at vaporization from an atomic standpoint. We shall picture the molecules of water forming a body of liquid water (and the surface). Above the surface we see a number of things. First of all there are water molecules, as in steam. This is water vapor, which is always found above liquid water. ...

0

![Might orbitals appear as depicted here][1] [1]: http://i.stack.imgur.com/0MYZ6.jpg This image appeared during a test and impressed me as possibly displaying orbitals or energy fluxes generated by them.

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Concerning the factor $\frac{1}{2}$: It seems that OP in his classical reasoning only accounted for the Coulomb potential energy $$\tag{1}\langle U\rangle ~=~-k_e e^2 \langle \frac{1}{r} \rangle ~=~-\frac{k_e e^2}{a_0} ~<~0.$$ Here $k_e$ is Coulomb's constant and $a_0$ is the Bohr radius.$^1$ However we should also take the kinetic energy $\langle ... 3 Very interesting question! In chemistry you spend lots of time discussing exothermic and endothermic reactions: when you put your reagents together, sometimes the reaction heats things up, and sometimes the reaction cools things down. Nuclear reactions are very different, in that essentially all spontaneous reactions studied in laboratories are exothermic. ... 10 If they didn't release energy, they wouldn't happen. The alternative, nuclear reactions that require energy, clearly need said amount of energy, which has to come from somewhere, e.g. kinetic energy involved in the collision of two nuclei (even ones that release energy usually have a "barrier" and some amount of initial kinetic energy is needed to overcome ... 0 Yes jim is right wikipedia article will answer your most of the questions. Still ill list few facts about nuclear reactions here hope those will help you Nuclear reaction basically has two major types fision reaction and fusion reaction (There are other types also i am just listing only two to answer your quenstion in simple words) In fusion reaction two ... 0 Your intituition is totally different because ennumerous forces change the situation from the ideal situation predicted by work enrgy theorem, some of them are: Air Drag/Friction, Rotational Friction due to differences in size of tyres, different aerodynamic effects due to different body design.. etc. 1 I know which book you are referring to. It is the book "Finite elements in Engineering" by Chandrupatla and Balegundu.I also have the same question. We learnt in Physics that the Work done by the force is stored as Potential Energy. There was no mention of Work Potential. 0 There, in fact, is such a way to convert heat produced from radioactive decay into electrical energy. Many systems doing just this have already been designed and used. The most straightforward device is a Radioisotope Thermoelectric Generator, which does exactly what you are asking. However, in principle, almost any system could be implemented that uses ... 1 By considering the variation in the potential energy$V=mg\Delta y$due to the vertical displacement$\Delta y$. A displacement of an angle$\alpha$would move the mass of the pendulum horizontally by$\Delta x=L \sin\alpha$and vertically by$\Delta y=L(1- \cos\alpha)$. Therefore the energy required for a displacement of an angle$\alpha$amounts to$mgL( ...

1

You seem to be interested in the concept of dualities. Dualities are incredibly informative in Physics in that every time we've come across one, it's led to unification of the two dual entities. You've already stated the most common one of spacetime. This was of course unified by relativity. You mentioned the Mass-Energy duality. This arises right from ...

0

From a thermodynamic point of view, when you have no current in the wire, the free electrons in the wire are, for the most part, not moving around very much in a random fashion. Temperature is the sum of the random, vibrational motion of the particles in a system. Once you put a current through the wire, ie introduce an electric potential, you set the free ...

1

Try showing that the kinetic energy of a particle of mass $m$ and momentum $\vec { p }$ can be written as $$K=\frac{p^2}{2m}$$ The solution to your problem should be clear from there. This expression for kinetic energy is actually a pretty handy formula to have memorized.

0

Your particles have the same momentum, so one has twice the velocity of the other. Their KE is in the ratio $$\frac{\left( \frac{m}{2}v^{2} \right)}{m\left( \frac{v}{2} \right)^{2}}=2$$ It gets confusing because energy is not conserved as a particular form. Once you get a handle on the work-energy theorem or the relationship of work and energy many things ...

2

You can write kinetic energy as $$K=\frac{1}{2}mv^2=\frac{mv^2}{2}=\frac{(mv)^2}{2m}.$$ You can take the next few steps of rewriting this in terms of momentum $p=mv$.

2

I can think of two reasons why we need a lower bound, one statistical, one inuititive. First, the intuitive: The annihiliation/creation operators represent adding/removing particles (or excitations, or whatever). The vacuum state as lowest energy state represents obviously the empty state from which no further particles can be removed. It is also clear (in ...

1

It's obvious that the kinetic energy must be compute with respect to the inertial frame. Using Lagrangian method for your problem is very similar to applying this method in order to derive the equations of motion of a gyroscope that can be found anywhere (for example here). The only difference between your case and a gyroscope is that in your case, the ...

1

Hint: If the transformation equations $$\mathbf{r}_i~=~\mathbf{r}_i(q_1,\ldots, q_n ,t) \tag{1.38}$$ do not contain the time explicitly, then the explicit time derivative $\frac{\partial \mathbf{r}_i}{\partial t}=0$ is zero. Note that the position $\mathbf{r}_i$ of the $i$'th point particle also depends implicitly on time $t$ through the generalized ...

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