# Tag Info

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1-When you put $v=c$ the Lorentz factor gets to $\infty$ and this is impossible for the $e^-$

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$E = pc$ is only true for massless particles. For massive particles you have the mass-shell relation: $E^2 = m^2c^4+p^2c^2$ After you use $E=T+mc^2$ and you can find $p$

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Short: Using the assumptions that you provide, you should be able to comfortably bring the water to boiling point in 1 hour, starting with water at 5 degrees C (as you specify). If necessary and practical the pot can be wrapped in a towel or other insulator to reduce thermal losses. (Remove towel before using pot or electric or gas stove :-) ). A rolling ...

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In today's understanding of Nature, there is nothing completely isolated. So technically there will always be interaction with the surrounding, at least from a quantum physical perspective. Here vacuum is not empty i.e. it does allow for electromagnetic interaction and there will be heat loss due to these vacuum effects. Furthermore also the other concepts ...

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Although it has been said in other comments and answers, it bears repeating succinctly: photons (as far as any experiment can tell) are massless and therefore always move at the universal, invariant speed of light. There is NO non-relativistic description of the photon. Even the "classical" description of light - Maxwell's equations - can be interpreted as ...

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People are addressing the speed question, but just to be clear: a photon can be very low energy. For instance, radio waves are much lower energy than gamma rays, even though both are made of photons (and, in vacuum, both travel at the speed $c$). What determines the energy of a photon is the frequency of the excitation (frequency of the corresponding light ...

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Refractive Index is when light travels more slowly in a medium. Here is an example of light being slowed down to 38 miles per hour. The speed of a photon does not affect its energy. It has zero mass, therefore zero kinetic energy. The energy it has is due to its frequency (color), and nothing else. (However, it does have momentum!)

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Is this the correct way to find the derivative of kinetic energy? $$K=\frac{1}{2}m v^2 \\$$ So: $$\frac{dK}{dt} = \frac{1}{2} \left(\frac{dm}{dt} v^2 + 2mv \frac{dv}{dt} \right)$$ If the mass does not change over the time, then $$\frac{dm}{dt}=0$$ And finally $$\frac{dK}{dt} = \frac{1}{2} \left(2mv \frac{dv}{dt} \right)$$ So simplifying: $$... 2 Here is the procedure: KE = 0.5mv^2 \frac{d}{dt}KE = 0.5m\frac{d}{dt}v^2 So the question becomes,how do we find the derivative of v^2 with respect to time? One can easily see that \frac{d}{dt} = \frac{dv}{dt}\frac{d}{dv} (Notice how the dv cancels top and bottom) Therefore, \frac{d}{dt}v^2 = \frac{dv}{dt}\frac{d}{dv}v^2 = \frac{dv}{dt}\times ... 2 The time derivative of v^2 is 2v \frac{dv}{dt} not 2v. You must use the chain rule. 2 The black machine is a weight lifting machine. It is self contained with no power source. If it can lift an external weight and return to its original state as shown below, it is a perpetual motion machine. Suppose the blue weight is water. We could add a water wheel and generator on the right. You start at the top and work your way to the bottom. Then you ... 0 Use the definition of work done over a period of time t , and you have :$$ E_k = \int_0^{t} \vec{F} \vec{dx} = \int_0^{t} \vec{v} d (m\vec{v}) = \int_0^{v} d\bigg{(} \frac{mv^2}{2} \bigg{)} = \frac{mv^2}{2}$$1 Single particle energy eigenstates for a system of particles in a box are given by$$ E_n=\frac{\hbar^2 \pi^2}{2mL^2}\,n^2 + E_0. $$The Fermi energy for a single particle is, by definition, the value of its energy that exhausts all the possible states given by N indistinguishable particles; in the case at hand, for fermions (electrons), this is given by ... 1 You might find the wiki article on this topic helpful. Summarizing: When you have a 1-D box, the energy states of an electron can be given by$$E_n = E_0 + \frac{\hbar^2 \pi ^2}{2 m L^2} n^2$$Now the things to note are this: Two electrons (with opposite spin) can occupy the same level The Fermi level is the energy of the last electron After each pair ... 1 According to the phase diagram of diamond (see for example http://files.umwblogs.org/blogs.dir/6093/files/2011/10/carbon_phase_diagram2.jpg) there is a region where diamond is stable and graphite is metastable, at pressures and temperatures in the range you are asking about: Given the slope on the diagram, you would actually expect that this reaction is ... 1 Chemical vapor deposition is used to produce diamonds at low pressure https://en.wikipedia.org/wiki/Synthetic_diamond#Chemical_vapor_deposition 0 "If you double the mass of an object, its kinetic energy doubles" is simply wrong. A true statement would be: "If you double the mass of a point object, its kinetic energy doubles." Let's take for example a ball, moving with a rototraslational motion. Its kinetic energy would not be$$ E_k = \frac{1}{2} mv^2$$, but it would be$$ E_k = \frac{1}{2} mv^2 + ...

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Zeldredge's answer is great, answering the question from a mathematical point of view. Since you asked this question in a physics forum, I'll just add to that by answering from a physical point of view, clarifying why the energy is linear in mass and not quadratic: Suppose you have two objects of equal mass moving parallel to each other with velocity $v$, ...

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Yes, doubling the mass doubles the energy, while doubling the velocity quadruples it. Your question is basically about order of operations; the exponent only applies to the variable it's immediately on. As you note, you'd have to put parentheses to make it cover the $m$ as well. We say that energy is linear in mass, but quadratic in velocity.

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You correctly identified there are two angles of interest, labeled $\theta$ and $\phi$. I actually want to pick two different angles for my analysis - see this diagram: First thing to note is that if the cylinder rolls without sliding, the length of the green arc and the red arc must be the same. Length of green arc: $(\phi + \theta) a$ Length of red ...

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You ask: If the clock is running slowly compared to a distant clock is this equivalent to the clock having a lower energy compared to a distant clock? but you have to very careful what you mean by energy in general relativity. As it stands your question too vague to be usefully answered. However in the weak field limit there is a sense in which time ...

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Okay, that was the physics part. If your answer was that the windmill should be built as large as possible, then what are the common engineering problems that occur as we scale up? As large as possible is definitively the correct answer. The common engineering problems which comes, is that you don't have a big-enough crane to build it. It's practically ...

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My question is, are we really must say that the energy of the tide and the loss of the kinetic energy of the moon are equal? The answer is obviously "YES." It must be so. I would refer at this point these words; Nature is relentless and unchangeable, and it is indifferent as to whether its hidden reasons and actions are understandable to man or not.- ...

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You are right that the actual axis of rotation is goes through the contact point with the ground. To see this, roll your bike back and forth. You'll see that the contact point is always standing still: the rest of the wheel turns around it and as it turns, the axis of rotation shift to the next point that makes contact with the ground (it's the static ...

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By heating the liquid the heat energy absorbed by the molecules and then it tends to vibrate more compared to its ground state. Due to the heavier vibrations the atoms moves more far apart from its equilibrium position. Now the phenomenon called "phase change" occures. Then the liquid is transformed into gaseous state.

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The bond that holds water as a liquid is a simple static electricity bond. it has a strength and will 'break' with sufficient energy. this happens all the time. water evaporates, when a random chance of circumstances through thermal agitation and exterior pressure are at the right amount the molecule leaves the liquid and goes flying off as a gas. the higher ...

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Mythbusters ran an episode on this one. When you learn of relative velocity, you learn that from the perspective of one of the drivers, the other driver will appear to come at you at 80 m/s. So the thought is that it should be the same as hitting a wall at 80 m/s. The energy doesn't add up though. Essentially, from the perspective of the driver, they are ...

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The total kinetic power of the system will be $\frac{1}{2}mv^2_1 +\frac{1}{2}mv^2_2$. The first equation that you mention is wrong, because this equation says that you have an object of mass $m$ with speed $v_1+v_2$. If you expand the squared term you will see that it is different. Now, what do you mean hit each other? Do they have opposite velocities? In ...

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I may be over-simplifying the problem a bit, but the laws of motion of a flexible sheet are very similar to that of a classical field. And the classical field is described by a Lagrangian density of $$\mathcal{L}(\phi;x,y,t) = \frac{1}{2} \left[\left(\frac{\partial \phi}{\partial t}\right)^2 - \left( \frac{\partial \phi}{\partial x}\right)^2 - \left( ... 1 See for example this table which contains the excess energy for each nuclide. You can take this table to compute the number you are interested in. The answer depends not only on the atomic number, but on the number of neutrons as well. This is why you need to think about how you want to represent this. I recommend you study that table and then figure out ... 0 The force is just the gradient of the potential energy. So it's not true that the energy is just force multiplied by r. It is the integral of the force w.r.t. r, which gives you the 1/r dependence. Edit: here "potential energy" and "interaction energy" are used interchangeably. 1 break earth into 2 parts in which each part is nearly equal to half of volume of the earth Yes and no. No, as in you can't neatly split the planet in half. Most of the earth is liquid and will re-form once the cutting device has passed through it. Much like cutting pudding. Yes, if you want 2 half-earth balls when you are done and don't care about the ... 4 One can answer this question by calculating the energy needed to shift half the Earth's mass so that it is infinitely far from the other half. Let's calculate the gravitational potential energy released as we create a planet: assuming a constant density \rho, when the planet is growing and of radius r and thus of mass M(r)=\frac{4}{3}\pi\,r^3\,\rho, ... 0 No, because the vast majority of the planet has a molten interior and where it is not in the liquid phase it is held in solid phase by the internal pressure. You could maybe disperse it into space with a big enough bomb, but not actually break it into two parts. 1 In the nuclear shell model, you can assign "orbitals" to each nucleon analogous to the orbitals assigned to electrons. Like the electrons, you can promote a nucleon into an empty orbital above the Fermi energy by exciting the nucleus with a photon — a gamma ray rather than a visible photon. The structure of the energy levels is pretty complicated: much more ... 0 There is no fundamental physics reason why carrying a bag should use more energy than wheeling it. Whether you be carrying a bag or wheeling it, you are essentially sliding it along a line of almost constant gravitational potential (aside from a little jiggling up and down with your stride), so the bag's total energy isn't changing and in theory does not ... 0 "Sound energy" must mean the vibration of the air (or other medium in contact with your eardrum). If water molecules collide with each other and with air, then there is a transfer of vibrational kinetic energy. Vibrations in air propagate to your ear as a pressure wave. For your second question, the more disturbance of the water flow and the more ... 1 Internal energy is a concept from thermodynamics, defined with help of the 1st law of thermodynamics. One formulation of this law, based on experience: If a system is in a state of thermodynamic equilibrium X_i, addition of definite amount of heat to it with no work being done results in certain equilibrium state X_f. The same result may be ... 0 The energy difference between two arbitrary speeds v and v' cannot be found by \frac{m(v'-v)^2}{2} except for the case where v=0. So your calculation about the energy change between 50m/s and 25m/s is incorrect. You instead have to calculate the initial and final energy states and subtract directly.$$E(50) = \frac{(20kg)(50m/s)^2}{2} ...

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The formula for kinetic energy is $\frac12mv^2$. If your initial velocity is $v_i$ and your final velocity is $v_f$, then your initial kinetic energy is $KE_i = \frac12 m v_i^2$ and your final kinetic energy is $KE_f = \frac12 mv_f^2$. The difference is $\Delta KE = KE_f - KE_i = \frac12 m(v_f^2 - v_i^2)$ It appears you're thinking that you can define ...

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A quantum well is a much more general concept that applies to any potential... not just to the energy bands in solids. For example, it could refer to a space between two charged metal plates. The real potential profile in a crystal is actually rather complicated because it consists of an enormous number of atoms. However, you can use a simple quantum well ...

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Part of the problem here is that you want two things: regardless of where you hit the large surface, you want the small piezo to record the "event" you would like the force on the piezo to scale with the force of the hit The obvious material to give you both these properties would be an incompressible fluid - preferably a non-conductive one that doesn't ...

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If you "push the pedal" (presumably the accelerator) and the car isn't moving, then 2 things are happening. 1) The engine is burning fuel, the pistons are going up and down, and the crankshaft is turning. All of this takes energy. 2) But the wheels aren't moving. This says that something is interrupting the transfer of motion from the crankshaft to the ...

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dmckee's comment is correct, this problem is hard (impossible?) to solve with pen and paper. But it's not too hard to set up, which hopefully will answer your question. There will advanced calculus by high school standards (including differential equations), no way around that, but I'll try to explain the equations at a high school level, I hope... I'll ...

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(Wheels will not work on a frictionless surface... But let's assume you mean that surface under the box is frictionless and not the surface under the car) then I will have to keep pressing on the pedal of the car so as to apply a force that continuously counteract Something is wrong here. For equilibrium and a non-moving box, you don't want the box or ...

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I'm sure this isn't what you had in mind, but... Reflected optical light (I can't think of a reason that UV wouldn't be just as good, provided there is enough UV radiation around to be reflected, e.g. sunlight) is 'radiation from a human'. A picture is a pretty good way of identifying humans. Provided the picture is sufficiently detailed (for instance, ...

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here is what I think he is trying to explain. You have two machines A and B. A is reversible B is not necessarily reversible. Both these machines are placed side by side. let both machines A and B be initially horizontal. (Left , right pans of machine A will hold 3 unit and 1 unit mass respectively. Similarly, the left and right pans of machine B will hold ...

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You're exactly in trouble! Wh is a symbol of ENERGY, not the POWER!

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You should look at an electric circuit a like a river or a water slide attraction in an amusement park (see my little artist impression below). The resistors are the steep parts: that's where the potential energy is lost. The wires are the horizontal parts, so there no potential energy is lost. But as the water is already moving, it doesn't stop moving in ...

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