# Tag Info

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The answer chosen as solution already explains everything but it is important to consider also the repair costs. If a car with rugged start/stop system can do 10 000 starts and the repair costs are about 400 Euro to replace the components of the starter system, it's already 4 cents per start (2.5 ml petrol at 1.6 €/liter). With the assumptions above, each ...

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The indigenous activist will travel in a circular arc of radius 10m. Using this image: And this formula from Wolfram: $r$ $=$ $1/2$ $\sqrt{4R^2 - a^2}$ Where $a$ $=$ $5m$ and $h +r = R$ You can find $h$, the height dropped and hence the change in gravitational potential energy.

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The hypotenuse of the right triangle formed by the vine and the gorge is 11.18 meters. In order to drop from the end of the vine on to the far edge of the gorge, he will have to propel the vine to carry him upward, against the Earth's gravitational acceleration, so that his ending position is at the far corner of a new right triangle with base of at least 5 ...

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The kinetic energy is defined as: $E_{k}=\frac{1}{2}m v^{2}$ Then the change of kinetic energy between, lets say, state 1 and 2 is: $\Delta E_{k}=\frac{1}{2}m v_{2}^{2} - \frac{1}{2}m v_{1}^{2}$ $\Delta E_{k}=\frac{1}{2}m (v_{2}^{2} - v_{1}^{2})$ as you correctly write. If you want to consider that the object starts from rest you just set $v_{1}=0$ and ...

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Answer: Electrical (or electromagnetic) The formula of the energy in magnetic fields can be written in two forms, as follows, $$W=\frac{1}{2}\int_V(\mathbf{A}\cdot\mathbf{J})d\tau \ ,$$ and $$W=\frac{1}{2\mu_0}\int_{all\ space}B^2d\tau\ ,$$ where $\mathbf{A}$ is the magnetic vector potential, $\mathbf{J}$ is the volume current density, and $B$ is the ...

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The prediction made by the cycle are quite accurate. In a real situation because of Air drag, rolling resistance and slope it takes a power output of around 300kWh for a 90kg cycle+rider. So at 250kWh, a less than 90kg rider+cycle can travel almost 40kms. The model you made is wrong because then energy spent is not equal to kinetic energy, it's rather equal ...

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Your formula for the first energy is incorrect. Instead use: $$W = \frac{1}{2}\sum_{i\neq j}q_{i}V_j(\vec r_{i}) \, .$$ Or even: $$W = \frac{1}{2}\sum_{i\neq j}\frac{q_iq_j}{4\pi\epsilon_0|\vec r_i-\vec r_j|} \, .$$ And now you see right away that you are avoiding the energy of the point charges themselves. Because naively it would be zero.

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What is the potential energy of the human body? 36 MJ $\simeq$ 8600 kcal. The basal metabolic state for a human is about 60 W. Let's assume (horrible) an immobile person dies after one week (604800 s) of starvation, he will have consumed 60 W * 604800 s = 36,288,000 J. Where does this energy come from? from the high energy chemical bonds of ...

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The difference is the zero point. When summing over charges, the reference is a state in which this charges are infinitely separated. Those are still distinct, localized charges, just separated from each other. When integrating $E^2$ over all space, the reference state has all charge separated. Even the individual charges from the first method are broken ...

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See for yourself. Just pour 2 cups of water into a (tall) glass. Then pour a third cup with some food coloring mixed in. You can see that only a small amount of the food coloring gets to the bottom before stopping.

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The hamiltonian is the mechanical energy of the system when the equations defining the generalized coordinates do not include explicitly the time and all the forces that do work can be derived from a conservative potential (Goldstein 8.1)

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It is inaccurate to say that we are "tearing matter apart" in accelerators. We are just changing particles into other particles. If we smash 2 protons, the collision excites various quantum fields and gives rise to multiple other particles. This happens because the energy of the incoming protons is used to create the outgoing particles. Both energies (of the ...

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Work is the change in kinetic energy. In both cases, the man starts on the ground at rest and end on the chair at rest. In both cases, the net work is zero. If you want to be more specific and describe the work due to his muscles, that positive work must exactly offset the negative work due to other sources. Gravity does negative work equal to minus the ...

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I'll answer the question for completeness. As discussed in the comments, if you think about all the posible ways of dissipating energy, maybe you will doubt of the veracity of the assertion made in the exercice. But, the clue is not to think in if the teacher wanted to be "captious"... but that in this case all these ways of dissipating are really small ...

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Their paper is inconsistent. They filled in $\omega = 264$ with the other quantities in SI units, so ω should be expressed in rad/s (often written $\mathrm{s}^{-1}$). So they assumed ω was already in rad/s. If they say they assumed $\omega = 264~\mathrm{rpm}$, that's not consistent with the values they plugged in. Your value of 69696 is hard to decipher ...

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For simplicity, look at the case where $m=\omega=1$ (where $m$ is mass and $\omega$ is frequency) so that the hamiltonian is $$H={P^2\over2}+{Q^2\over 2}$$ Put $$A={1\over\sqrt2}(P+iQ)\qquad B={1\over\sqrt2}(P-iQ)$$ so that $$AB=H-\hbar/2\qquad BA=H+\hbar/2$$ From this, check that if $H\phi=\lambda\phi$ ($\lambda$ a scalar) then ...

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Notice, the combustion is an exothermic chemical reaction with the oxygen that liberates the energy mainly in the form of heat, comparatively less amount of light, sound etc. Reactant molecules react with the oxygen that results in breaking of old bonds & formation of new bonds with comparatively less bonding energy. Thus, releasing the rest amount of ...

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An incompressible liquid is never completely incompressible, more like quasi-incompressible. So when you apply considerable force $F$ on the piston, pressure will wise by say $\Delta p$:: $$\Delta p=\frac{F}{A},$$ where $A$ is the cross-section of the piston (and assuming constant $F$). But the liquid will have decreased slightly in volume by $\Delta V ... 1$p=\frac{1}{3}\rho$is the well-known equation of state of a photon gas. It may be derived by looking at the ultra-relativistic limit of the energy momentum tensor for a bunch of particles.$^1p=-\rho$follows from the fact that the energy momentum tensor of$\Lambda$-style dark energy is proportional to the metric. Thus, at a point and in the proper ... 2 You're right that the unit "megawatt" is abbreviated MW. However, as Aniket comments, watt itself means "energy per unit time", so saying that the power plant produces 60 MW per hour doesn't make sense. In your comment, you question whether MW is a "basic unit". I'm not exactly sure what you mean by this, but the SI unit of power is watt, so if you want to ... 1$\mathrm{Watt}$(or Joule per second or$\mathrm{J/s}$) is the SI unit of power. So megawatt is a valid unit of power (expressing power with order of magnitude$10^6$) and is used mainly in commercial statements. Definition: Power means the quantity of energy consumed or produced per unit time. So$\mathrm{MW/hr}$actually makes no sense since it ... 0 Notice, the unit of power is$~\mathrm{J/s}$or$~\mathrm{W(Watt)}$. The unit$~\mathrm{MW}$indicates the energy (in$~\mathrm{mega\ joules}$) produced by power-plant per unit time (in$~\mathrm{seconds}$) The unit$~\mathrm{60 \ MW\ per \ hour}$doesn't represent a physical quantity. 3 Expanding on what Couchyam said: To some extent, this is actually the definition of what it is to be a "particle." Intuitively, a particle should have some particular definite energy, and be stable enough that it can exist in its own right. These two requirements are linked by energy-time uncertainty. The natural "clock" to compare against when asking if ... -2 Energy can neither be created and nor be destoryed it can changes from one form to another form. For daily life example just when we do any work as for example just when we play a football and kick in the football and take our energ to football to kick it then energy is going into football and it can move from KINETIC ENERGY INTO POTENTIAL ENERGY 1 No it is separate issue to total internal reflection. The important thing here is frequency dependent emissivity of the materials. Very simply speaking, the absorber needs to have high emissivity in the optical and NIR region (so it strongly absorbs sunlight) and low emissivity in IR, so that the thermal energy is not radiated away. It is like a green ... 1 The formula that WhatRoughBeast uses is not correct. It should be $$KE = \left(\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}} - 1\right) mc^2 = \left(\frac{1}{\sqrt{1-0.22^2}} - 1\right) (22,000)(3\cdot10^8)^2 = 5\cdot10^{19} J$$ Your calculation wasn't far off since relativistic effects aren't very large at 22% of the speed of light, so$KE = ...

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You've underestimated the effect, although your math is correct as far as it goes. At 22% of c, relativistic effects do rear their ugly heads, and the proper equation is $$KE = \frac{\frac{mv^2}{2}}{\sqrt{1-\frac{v^2}{c^2}}} = 7.6\times10^{19}\text{ J}$$ Divide by $63\times10^{12}$ and the ratio is 1,200,000 (1.2 million). And yes, this is an unreasonably ...

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If the WiFi antenna is emitting at 2.4 GHz, you could detect a slight improvement of the signal, but unless the door is solid and very thick, I doubt it will make much difference. If it is emitting at 5 GHz the improvement could be bigger, as the wavelength is reduced and the door appears "bigger" to the electromagnetic wave. Finally, if your antenna uses ...

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For part (1.) think about this: just as the spring comes to rest momentarily all of the energy in this system is stored in potential energy of the spring itself. First find the spring constant $k$ using Hooke's law. From there you can find the potential energy stored in the spring. You then know that just as the mass comes in contact with the spring the ...

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You can draw a schematic diagram to analyze it and suppose that elastic force is $F=-kx$ by Hooke's law and the gravity is $mg$.

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A material of high specific heat, coupled with another material of high thermal conductivity, might just do the trick. I would suggest water (high specific heat) taken in a metal tank(high thermal conductivity). Though this can be the best option, scientifically(talking about efficiency), it might not be the same aesthetically

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Its not that tough. You can work it out by using just two equations. But the one thing you should keep in mind is that when the comet is at the minimum distance from the sun, its velocity must be perpendicular to the radial vector (sun to comet). So the minimum distance is itself the minimum perpendicular distance used in the angular momentum formula at ...

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What happens is not a transfer of kinetic energy from the weight to the pulley. Rather, the pulley accelerates slower and then reach a lower speed because part of the work done by gravity is directly transformed into rotational energy of the pulleys. Yes, both translational and rotational motion energy are forms of kinetic energy.

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from the figure: liquid that cannot be compressed... so forget compression of springs. Molecularly, particles in a liquid have much less space to move around (short mean free paths) compared to particles in a gas. The result is that any applied pressure to these particles is instantaneously transferred through the fluid. The mechanism by which this happens ...

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In the general case the frequency of a wave and its kinetic energy are not related. As you can derive from Energizer 777's answer one can increase the frequency and this time decrease the amplitude of the wave generator and you approximately need the same amount of energy to support the damping of the wave from dissipation processes. The point is that an ...

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Work and kinetic energy are interchangeable, so they are the same, ie. they have the same units. Your work can give kinetic energy to a body and a body with kinetic energy can produce work.

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The Newton unit is not a fundamental unit but consists of: $$\mathrm{[N]=[kg\cdot m/s^2]}$$ which you can convince yourself of from Newton's second law $F=ma$. Plug it into $\mathrm{[N\cdot m]}$ and you'll see.

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Suppose the maximum angle of the pendulum to the vertical direction is $\theta$, where the pendulum has a zero-velocity. According to the conservation of mechanical energy, the total energy of the system, $E$, should be: $$E=mgl(1-\cos\theta )$$ If $\theta$ is small enough, we can expand $\cos\theta$ up to the second order: \cos(\theta)\approx ...

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There are a fairly large number of replications of both the Nickel-Hydrogen variant and the original Fleischmann-Pons Palladium-Deuterium electrolytic cell variant. See this for a recent survey: https://www.academia.edu/17964553/Condensed_Matter_Nuclear_Science_October_2015 The results are becoming more mainstreamed (Wired, Forbes, Huffington Post), and thus ...

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The acceleration of gravity is the weight of the object on the surface. If the structure, either of the object or the surface, is deformed by the weight, then part of the gravitational potential energy of the object will turn into kinetic energy of surface moleucles and therefore heat, and this will be shared between the object and the surface. In this ...

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Checkout John Williamson's work, for example being discussed here: https://www.physicsforums.com/threads/is-the-electron-a-photon-with-toroidal-topology-what-is-that.614799/ He derives Maxwell's equations for the electron with the Lorenz transformation taken account for relativity. It's quite good IMHO; carrying on with the work that Dirac was doing with ...

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With respect to current theory my answer may prove quite unpopular (down votes expected), so please take this answer with a grain of salt if you are still in school. Your approach in this 'thought experiment' that you propose is very much like some of my own research conclusions. In direct answer to your question, yes, General Relativity should require an ...

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I) What does it exactly mean when we say that energy has moved from one body to another, what has physically been transferred? Contrary to the view expressed above by Mark H < http://physics.stackexchange.com/a/217505/76263 >, I think energy is a fundamental physical property and not just “an abstract [useful] mathematical quantity”. In fact, the ...

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First two answers to your questions must have gave you some good understanding of Energy, Spectra & some history of early Quantum Mechanics. Please read them again before you read my analogy. I'm directly moving onto analogy without wasting "Space" & "Time". In Classical systems, the movement of energy is like an movement upwards on a Ramp. In ...

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Mass is energy, not being able to go at the speed of light in a direction. If the photons creating the electrons and positrons are going at the speed of light in a tight loop as John Williamson describes in a half dozen papers, then it can't also go at the speed of light in the perpendicular direction.

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Energy does not require force, as a moving object will have kinetic energy even if no force is acting on it. To give energy to an object is using work, so the second question is really asking if force requires work. The definition of work is the integral of the force function from the inital position to the final position. Even if there is a force, if the ...

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