Hot answers tagged energy
42
Individual photons are very small and don't have much energy.
If you put a lot of them together in one place you can hurt somebody - by simply supplying enough power to melt an object (ask any spy on a table underneath a laser beam).
There is another very odd feature of photons. Although lots of them can provide a lot of energy and heat an object, it takes ...
16
I have a somewhat non-physics answer for you. If you allow me to broaden your question a bit to "why doesn't light kill or otherwise make all life on Earth impossible" the answer is that the Earth is in what we call "the habitable zone".
If the Sun produced so much light or light at such high energies that it would kill you, it also would heat the planet ...
11
The horizontal component of running is believed to be fairly negligible for humans. Some research suggests that the limit isn't strength related at all, but design --- in particular, based solely on power, humans could theoretically run up to almost 40 mph. The issue is two fold: first, our limbs are actually too heavy, for big strength (e.g. climbing in ...
8
This question is more interesting than I thought at first. I like it. There are several different parts to an answer to this question; I'll just contribute a couple that have something in common: our bodies (and everything else, it has nothing to do with bodies) also emit photons about as fast as they absorb them.
On the macroscopic/thermal scale, we have ...
8
A general photon isn't too dangerous. Most photons that we encounter have the power to heat our bodies and not much else. The heat we absorb from photons daily isn't that much, so this is rarely a problem.
Now, an interesting thing about photons is that two photons of a lower energy do not make a single photon of higher energy (frequency). So a million ...
6
Imagine that you have just two particles with the same mass and same speed, but going in opposite directions. They have opposite momenta, so the total momentum is zero. But they each have energy, and the total energy is not zero. The reason is because kinetic energy is just $\frac{1}{2} m v^2$. That square means that the kinetic energy can never be ...
5
It is just easier, i.e. less expensive, to build and maintain them that way. There exist alternative designs that are more efficient but also more difficult (= more expensive) to build, put up and maintain. You can check those out via this link.
5
The work in the first law is exactly the usual work $W=\int Fdx\rightarrow\int PdV$. For point particles, this is enough to completely specify the behavior of the system using Newton's first law, or energy methods. However, for macroscopic objects, the motion of the internal components (in thermodynamics these would be particles) have some additional degrees ...
3
You have to take into account the differentials. The actual equation is
$$
f_\text{MB}(\mathbf{v})\,\text{d}v_x\text{d}v_y\text{d}v_z =
n\left(\frac{m}{2\pi k_BT}\right)^{3/2}e^{-mv^2/2k_BT}\,\text{d}v_x\text{d}v_y\text{d}v_z.
$$
Changing to spherical coordinates, we get
$$
\text{d}v_x\text{d}v_y\text{d}v_z = ...
3
Large systems with many degrees of freedom (e.g. a ball consisting of many molecules) tend to settle into low energy states. This is a direct consequence of two fundamental laws, the first and second laws of thermodynamics: energy conservation and entropy increase.
A system with many degrees of freedom can be in many different microscopic states (think ...
3
The first law of thermodynamics says "the increase in internal energy of a body is equal to the heat supplied to the body minus work done by the body". Assuming there is no heat flow (for simplicity), this says "the increase in internal energy of a body is equal to the work done on the body". Since you are doing work on the gas, the internal energy ...
3
You are talking about relativity and gravity together so the question can only be answered in the context of general relativity, but concepts like gravitational potential energy and gravitational force acting over a distance are Newtonian and do not really carry over to general relativity.
However, the gravitational field does contribute to total energy and ...
3
Depends on what you're doing. General relativity handles it for you, in the sense that the Einstein field equation links geometry to the non-gravitational stress-energy tensor. That general relativity is non-linear can be interpreted in part as gravity itself contributing to gravity, but it's generally not even possible to localize gravitational energy in a ...
2
Energy is a scalar quantity, a single number, that doesn't change in physical systems where all interactions are inside the system.
For a number of particles that only interact with one another at the same location, kinetic energy is this single number that's conserved. When they interact at a distance via the electromagnetic field, the total kinetic energy ...
2
It sounds like the way you're imagining this is the source of your confusion. Electrons in one atom are not attracted to the electrons in another atom. What actually happens is that it requires less energy for two atoms to come together and share some electrons in a covalent bond. How much can be saved and the configuration of the bonded atoms depends on ...
2
You have a few different but related questions so I will try to explain them in a simple, no-math way.
If a radio tunes to a specific frequency, where does the excess energy go?
Almost every object that has radio waves (electromagnetic waves) around it absorbs some of the radio energy. When the radio waves hit the electrons in the atoms and transfers ...
2
Energy and momentum has to be conserved. That the electron / photon has to have enough energy for the excitation is obvious. What is interesting is what happens when they have too much energy.
For radiative transitions between bound states the orbital anuglar momentum has to change by 1. This means that the photon has to be absorbed which in turns means ...
2
The $W$ term in the first law expression exclusively refers to the mechanical work done by a system and all other things , all other possible exchanges of energy are clubbed together in $Q$.
Suppose I am the system under consideration , and I apply a force on a block and that does some mechanical work (that is the point of application moves a distance) ...
2
As far as we can tell (up to energy scales we've measured so far), spacetime is a nice and smooth manifold. It might happen that the smoothness is approximate and spacetime is discrete at a much more microscopic scale, or it could turn out that spacetime is smooth all the way through. Short answer: We don't know.
About the notion of energy quantization: ...
1
This kind of exponential decay toward "equilibrium" can be derived when one looks at a Markov process.
In this case, if we call $S_t$ the state of the system at time $t$ and $S_{t+1}$ the state at time $t+1$, one has for the evolution:
$S_{t+1} = T S_t $
where $T$ is called the transition matrix. This implies that $S_t = T^t S_0$. The idea is then to ...
1
The other answers tackle the statistical/thermodynamic aspect. I will tackle the "falling apple " aspect.
Why does the apple fall?
From this observation onwards nature was modeled mathematically as interactions between masses, in this case, charges in the electromagnetic case etc.
The observations of gravitational interactions led to a mathematical model ...
1
I will address such sample systems as a point (or a small metal ball) rolling or bouncing on some hard surface with hills and pits, and an atom which can be either in excited or in basic state.
I. If we consider an ideal closed system, then the enegly is conserved. But real systems do not (exactly) behave this way. For a macroscopic mechanic motion we can ...
1
The probability of finding a system in a state with energy $E$ is $P(E) = \exp(-\beta E)/Z$, where $\beta = (kT)^{-1}$, $k$ being the Boltzmann constant and $T$ being the absolute temperature. $Z$ in the formula for $P(E)$ is the canonical partition function. For our purpose, we can consider it as a factor introduced to ensure that $0 \le P(E) \le 1$. The ...
1
When radio waves hit the antenna it creates an electric potential difference between the antenna and ground. An electric current flows from antenna to ground, through the radio receiver. The radio receiver is able to extract information (the signal) from this current and amplifies it. Virtually all the electromagnetic energy collected by the antenna flows to ...
1
energy is always positive or 0. And these are just numbers we associate with a body due to its motion according to different set of mathematical rules so that we can study these particles . As such they have no physical meaning . For example momentum is just $m$ x $\vec v$ . It is just a number we associated with a body by the quantities we defined ourselves ...
1
Use an equal-arm balance beam, with a 250 kg counterweight suspended 5 cm in the air on one end, and the mass to be moved just touching the ground on the other end.
Energy storage is in gravitational energy. Losses can be made insignificant. Problem was originally solved by elevator engineers...
1
I apologize "basics foundations of thermodynamics" still does not make a lot of sense to me.
Steve B already provided some answer associated to one way to interprete the word "foundation" that is from statistical mechanics. I will kinda play here devil's advocate and assume that you are refering to axiomatic thermodynamics.
As far as I am concerned, the ...
1
There's a group I call "thermodynamic purists" who think that thermodynamics is a self-contained system based on semi-mathematical "axioms". I disagree! I think that thermodynamics is fundamentally a consequence of statistical mechanics, and that this is the best way to think about it and understand it. I acknowledge that reasonable people can differ on ...
1
You can think of it this way -
The field isn't being created 'constantly'. That's like saying our hands and legs are being 'created constantly'. The field is coupled to an electric charge, and that's how it is.
When a charge is being accelerated, it is gaining energy (since work is being done on the charge) and it also looses energy (through the ...
1
I) Here is at least a partial answer. Assume the following set-up. Let there be given a classical Lagrangian field theory in $d+1$ spacetime dimensions, with dynamical field variables $\phi^{\alpha}(x,t)$, and with no explicit time dependence.
Action $S[\phi]:=\int \! dt~ L[\phi(t,\cdot)]$.
Lagrangian functional $L:=T-V$.
Energy functional $E=T+V$.
...
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