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The problem with this is that $$E = mc^2$$. Density, however, is given by $$density = \dfrac{mass}{volume}$$ Thus, if volume = 0, then density is infinite. Black holes have a finite mass. It is there density which is finite because all the mass is at a single point (singularity, volume = 0).


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No, the energy of a black hole is not infinite. It depends on its mass, angular momentum and charge. Infinite density at a point does not translate to infinite energy in the $E=mc^2$ sense. It is in fact possible to extract energy from black holes by exploiting certain properties of accretion disks or ergospheres, but this is a finite process.


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In nuclear fusion power plants, electromagnetic fields might be used to contain plasma. But I am not sure what would happen if you threw a solid at the electromagnetic field.


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Is it possible to make an electromagnetic shield to protect something (or ourselves) against everything which is attacking? It is possible to shield very well against some types of electromagnetic radiation. For example, putting a cell phone inside a "Faraday bag" will block enough radio waves from reaching the cell phone that it can not receive ...


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No, Chemical Processes and Mechanical Processes can also release more energy than is input. See Exothermic Process and Catalysis. Think of an explosion of, say, dynamite: for the low energy input of lighting a wick which can be done with a lighter, you can output a large explosion. Also, Palladium is not Fissile, though it is a Fission product (aka ...


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It will be total energy . But it can be change with respect to condition.Suppose if you consider this free electron then the electron have no potential energy .Only kinetic energy will remain.Then total energy becomes kinetic energy also.


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For a two cells series, you have to continue by repeating the same profile and putting Cu (from the 1st cell) and Zn (from the 2nd cell) at the same potential level. The e.m.f. will be doubled.


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$\newcommand{ket}[1]{\left| #1 \right>}$ $\newcommand{bra}[1]{\left< #1 \right|}$ $\newcommand{bk}[2]{\left< #1 \big| #2 \right> }$ In the level of QM you really don't have kinetic energy and potential energy. Some likes to call the expectation value of the operator $\frac{\hat{p}^2}{2m}$ on the state $\ket{\Psi}$ the kinetic energy and the ...


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Total Energy of the photon being considered for energy calculation


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The fact that the apparatus is using outer space as a heat sink or just using the atmosphere should not have any significant impact on radiative cooling properties. Is this reasoning correct? No, it is not. The cooling properties of the apparatus depend on the heat sink to which the heat is aimed, as the final result will be an equilibrium of ...


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When you pluck the string, you impart energy into it that's slowly radiated as sound. There are ways to radiate the energy faster, in which case the string loses energy faster. You're increasing the power and decreasing the time, so energy stays constant.


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The author confused things in section 3.1. In macroscopic EM theory of radiation (radiometry), technically irradiance at a given point of a plane is defined as time average of the normal component of the Poynting vector (normal to the plane): $$ I = \overline{\mathbf S \cdot \mathbf n} $$ As you have shown, this is function of $|\mathbf E|^2$ only in some ...


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I think this is off topic here and I'm not sure of a better site for it (maybe fitness.SE)... I'll try to put as much physics as I can to keep it on topic here though. It's all about an energy balance and understanding where the energy can go when an animal exerts effort. You can measure the basal metabolic rate (BMR) of an animal using respirometry. In ...


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A rope is the best example. Think of a rope as a chain of particles attached to each other. You now grab the particle in the end and lift it up. Your hand applies the force that causes the acceleration of the particle. As that particle is starting to move up, it will pull the next particle beside it by exerting the same force as the hand applies. This ...


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Is it possible to generate energy using gravity? Yes and no. It depends what you mean by generate. What gravity does, is convert energy from one form to another. When you lift a bucket of water you do work on it. You add energy to it - we call it potential energy. Then when you upend the bucket, the water pours out, and gravity converts potential energy ...


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This is the case with an elastic collision (kinetic energy conserved). Let $x$ be the initial velocity of the first neutron (the other neutron is stationery in the beginning) and let $m$ be the mass of the neutron. Then the intial momentum is given by $mx$ and the initial kinetic energy by $0.5mx^2$. Let $y$ be the velocity of the first neutron after ...


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Fundamentally there's a simple difference: When you are working with a perfect gas at constant volume you can take the variation of inside energy equal to the heat absorbed in the transformation. In this case you must use Cv, obviously. In the other side, where the pressure is constant you can't consider the equality defined previously. In fact you must ...


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The equation you need is that force is equal to the rate of change of momentum. The force is the weight of the rocket, $Mg$, and the rate of change of momentum is the mass ejected from the exhaust per second multiplied by the exhaust velocity. $$ Mg = v\frac{dm}{dt} $$ So choose your exhaust velocity $v$, and you can work out the required $dm/dt$. The ...


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If you take a point in the horizontal line in the middle,then the potential caused by the upper left charge will be the same as the lower left charge but with opposite signs.The same goes for the two charges on the right.Just take a random point on that line and let a be the distance from upper or lower left charges to the point and b the distance from the ...


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After asking lots of people, finally the right answer arrived to my inbox from Henry Reich, the creator of MinutePhysics. We can summarize the answer on: You can use lots of energy to deflect an asteroid, but The work done is zero because of the definition of work. Work="change of energy" According to Henry: All physicists mean by “work” is the ...


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It would, as you predicted, be neither, but it would be closer to the percentage than to the constant. Friction destroys lots of space debris which have a high energy than a dropped ball. The debris loses all of its energy while the ball loses a bit. This is far from a constant by not quite a percent. In the end, it relies on many factors.


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Sound is a pressure wave. Continuous sound is a continuous stream of high pressure fronts followed by low pressure fronts. The louder a given sound is, the higher the difference in pressure between a crest and trough; pressure is the amplitude of the sound wave. Furthermore, the volume of a sound is also dependent on frequency. Humans are not as sensitive to ...


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Along with your pounding heart, you're also experiencing the effects of resonance. In simple terms, there are certain frequencies of sound waves which correspond to the "natural" vibration frequency of your bones. At these frequencies, your bones vibrate with a greater amplitude than other frequencies, and you can think of it as "tapping" into the ...


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m includes all kinds of energies for the mass at rest, including the thermal energy, and those from the other degrees of freedom, such as the ones internal to the nucleus, as AnnaV mentioned. So you only have to compute mc^2, the HTM term is redundant.


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Eötvos-type experiments establish that QCD binding energy is exactly equivalent to mass as far as feeling the effects of gravity. Conservation of momentum is generically observed in all interactions in our universe. given 1. and 2., if energy was NOT active mass, then the force that a pound of gold exerted on the Earth would be different than the mass ...


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I don't want to spoil the beauty of 3D QM for you especially if you want to explore it by your own. However I thought I would give you some advise on how to approach to the transition from 1D to 3D. Considering 1D QM first of all think about what it means to be on the x-axis. Who is to decide which axis is x and which axis is the y or z axis for that ...


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The big arrow labelled energy is just an axis. It's supposed to indicate the upwards is positive, i.e. energy is being added to the system, and downwards is negative, i.e. energy is being emitted from the system. Re the extra question: where does the energy released in a nuclear reaction go, is it kinetic energy or heat? Heat is a collective ...


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You are overthinking, it looks like a simple inelastic collision. Before: Vertical (potential energy): $E_{plate} = 0$, $E_{object} = mgh$ Horizontal (kinetic energy): $E_{plate} = \frac{1}{2}Mv^2$, $E_{ball} = 0$ After: Vertical: The object smashes, not bounces, so its entire potential energy dissipates Horizontal: The plate now has a combined mass of ...


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It depends on the actual materials. If you have a big need and a big budget, you can make some mirrors that have very high reflectivity. 99.9% is achievable in some cases for certain energy ranges. As energies go up, (UV and higher) it gets harder to reflect cleanly. For sunlight, most of the energy is in the optical, so you can ignore the reflectivity ...


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I think the key thing missing in your thinking is that the energy drop across a resistor is not just determined by the properties of the resistor, but also by how much current flows. The cool thing is that no matter what resistors you put in, the current that flows is such that the potential will fall all the way back down. The reason for this is that ...


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Your post seems an awful lot like a homework question, so you should probably tag it as a homework-like question. If by "cup" you mean 250 mL on the dot, then we can say that the ratio of boiling water (100 Celsius) to room temperature water (22 Celsius) to achieve 80 degrees is $$ \frac {(T_1)(V_1) + (T_2)(V_2)}{V_1 + V_2}$$ In your case, we have: $$ ...


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The total energy can be expressed as the sum of the pairwise energys excluding the self-interaction energy. $$W=\frac{1}{2}\sum_{i\neq j}\frac{q_iq_j}{r_{ij}}=\frac{1}{2}\sum_iq_i\sum_{j\neq i}\frac{q_j}{r_{ij}}=\frac{1}{2}\sum_iq_iV_i,$$ where $r_{ij}$ is the distance between two point charge. The factor $\frac{1}{2}$ comes from the fact that the summation ...


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How come they can take energy without actually affecting the source? Well the source is the sun, so it just sends out light and it doesn't matter where it goes, the sun will keep out sending out light, so it will be sunny tomorrow, amd the next day and the next day.....it will keep on going for a long time Shouldn't reflected light (as a minimum) have a ...


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Good question! Solar cells made of silicon are naturally quite reflective (30%), and there is quite an art to making them less so - because any sunlight reflected is not available for power conversion. Antireflective coatings are not easy to make since they tend to work best at one wavelength and angle of incidence; you try to make them so the total power ...


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It is hard to make anything be truly black. At any frequency, some light will be absorbed, some reflected, and some transmitted. Eye response to light is logarithmic - We can see bright light, but also much dimmer light. So if 99% of the light is absorbed, that still may not appear black. Smooth surfaces produce a mirror like reflection. If illuminated by ...


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I'd like to offer a thought experiment in return: same scenario, but instead of hot gases coming from a rocket, we've got a device firing BB's in the same direction as the exhaust gasses were going. The results should be the same (gasses behave a lot like BB's in a vacuum), but BB's are a bit more tangible and have some heft, so the results will feel a bit ...


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There are multiple parts to your question. energy of a standing waves remains between the nodes When you look at the motion of a guitar string, it looks like this: The part between the nodes is moving, while the nodes remain stationary. When the string is flat (and moving fast) all the (kinetic) energy is indeed between the nodes. However, when the ...


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Energy transported by a standing wave gets thrown back by reflection. -> Consider a standing wave thrown back by total reflection - no energy transport takes place. A partly reflected wave sums up into a standing wave which gets overlapped by a running wave. In this case energy gets transported. If you want to understand why a guitar works, I would ...


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Let's think about sound waves. Sound waves is a pressure wave ie it propagates by the change of pressure in the air. If you have a standing wave then the places where there is high pressure and the places where there is low pressure remains the same. You can compare this fact with a swinging string fixed at both ends. The reason why this happens is that the ...


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Looking at a power cycle you want the largest possible change in volume at a certain pressure - and steam obliges. When water turns to steam its volume increase dramatically. How much depends on the temperature, obviously. The nice thing is that you don't have to do a lot of work to push the volume of water into the boiler because work =$P\Delta V$ and the ...


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http://www.launc.tased.edu.au/online/sciences/PhysSci/done/kinetics/wep/Work.htm The above link talks about WORK, ENERGY and POWER. Does this basic concept of physics applies all only to OBJECT and not to Human doing the work or applying power and using its own energy? Just confused here and I think I was wrong in linking with Human energy. If I pushed an ...


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Why steam? Because super-heated steam and above almost acts as polytropic, so our equations just work almost as expected. If you know chaos theory you would understand physicists & engineers just got lucky when it comes to all this fluid-mech. + thermodynamics scenarios I want to understand the mechanism of energy conversion in these equipment ...


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It takes more than kerosene to burn. You need kerosene and oxygen. The flame does not go back further into the cylinder because there is no oxygen there to allow it to burn.


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I suppose you might think of the ice cube "adding cold" to the system, but by convention, we typically discuss these matters in terms of heat. If cold was an inverse of heat, we might say that removing heat from a system is the same as adding cold, but in order to be parallel with the scientific community, it would be wisest to say you are "removing heat." ...


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I definitely disagree. Cold is a relative feeling, cant be added; Coffee will release temperature to equalize the temp.


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Assuming your question is about the concept of energy in physics: The muscle actually uses chemical energy. How this works in detail is not a physics but a biology question. The chemical reaction will create heat and cause your muscle to contract. Consequently, your body loses chemical energy, that's why you have to eat, drink and breath, to keep these ...


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Combustion is a chemical reaction and for this to happen you need the molecules to be sufficiently near to each other for the electrical forces to take over and cause the reaction. Thus you need to atomize the fuel and mix it well with air/oxygen in the presence of heat (= highly energetic molecules to bring collisions). This is why you have a long column of ...


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A flammable liquid in its liquid state will not burn, you can throw matches into a bucket of liquid petrol all day long without a problem. its only when you get gas/vapour coming off it that you gotta run ABSOLUTELY DO NOT TRY TO TEST THE ABOVE EXPERIMENT I know you know but just to be sure (not being smart with you) its LPG in the cylinder, Liquefied ...


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The cylinder is pressurised. It's pushing kerosene out quickly, so oxygen can't get in.


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Photon is a (one) quantum of electromagnetic radiation/light, ok? When it has low energy, means large wavelength, means low frequency, we know it as radio waves. Some higher energy, means higher frequency, means lower wavelength, we see it as light. At, even higher energy it is X-rays. Let us stop here. The formula for single photon's energy is ...



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