# Tag Info

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Eventually the title forces will cause the object in this case the moon to have a circular orbit. The closer it gets to a perfectly circular orbit the less title forces there are or at least more constant.

-1

Wikipedia is totally unreliable for maths and science. You are right that it is the total differential that has to be zero, not the partial derivative. This is simply common sense, and has nothing to do with transport or anything specific, as Khwarazmi points out.

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Your principle ("you will loose half of the energy,... I feel that the principle is deeper and universal") does not work, and I will show that using a counterexample. Assume a similar example to the one you used with the water, but in this case the potential energy were $mgH^2$ instead of $mgh$. In such a case $U=mgH^3/3$, after splitting the water in ...

1

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 ...

1

The pressure at the bottom of the water tank is large. You have to push the balls in against this pressure - that is exactly as much work as you will get back. No free lunch in nature. Sorry.

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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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The energy required by the body will be given by: $$\Delta E = (U_2 + K_2) - (U_1 + K_1)$$ Since $U_2=0$, the energy required by the body is $$\Delta E = K_2 -U_1 -K_1$$ Since Gravitational potential energy is negative and Kinetic energy is positive, the $-K_1$ actually helps reduce the energy required to remove the satellite from its orbit. Also, it's ...

3

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 $\Delta V ... 1 Let me discuss a simpler version of your rocket-question: one where there is no gravity, so that we don't have to worry about gravitational potential energy. Consider a rocket in free space (vacuum), and consider that the rocket is at rest. Now the rocket fires it's engine for a short time. The engine accelerates the rocket. The rocket now has kinetic ... -2 Energy can neither be created and nor be destoryed it can changes from one form to another form. For daily life example just when we do any work as for example just when we play a football and kick in the football and take our energ to football to kick it then energy is going into football and it can move from KINETIC ENERGY INTO POTENTIAL ENERGY 1 For these kinds of system we often define a pair of quantities, one which is characteristic of objects or systems and one which is characteristic of interactions. Examples of these pairs are work (interaction) and energy (system) or impulse (interaction) and momentum (system). There is no commonly applied name for the interaction quantity that pairs with ... -1 Hamiltonian can be written as :$ dH=\frac{\partial H}{\partial q_i}dq_i+\frac{\partial H}{\partial p_i}dp_i+\frac{\partial H}{\partial t}dt $. or,$ \frac{dH}{dt}=\frac{\partial H}{\partial q_i}\dot{q_i}+\frac{\partial H}{\partial p_i}\dot{p_i}+\frac{\partial H}{\partial t} $. We lso know that$\frac{\partial H}{\partial p_i}=\dot{q_i}$and$\frac{\partial ...

0

The total energy as a function of time is a constant and it is always equal to the potential energy in the point where $x=x_{max}$ and $v = 0$. The total energy is the work done to bring the free end of the spring from $0$ to $x_{max}$. So $E(t) = E = Work = kx_{max}^2/2 + ax_{max}^4/4$ You simply integrate $f(x)$ from $0$ to $x_{max}$

1

Hint: Use $$m\ddot{x}=-kx-x^3 \\\ddot{x}=v\frac{dv}{dx} \\-\frac{kx^2}{2}-\frac{ax^4}{4}=\frac{m}{2}\left(\frac{dx}{dt}\right)^2$$ It will reduce to a form $$\frac{dx}{dt}=ix\sqrt{c^2+x^2}$$ This is a standard integral, and can be solved, then use $$U=-\int f(x) dx \\T=\frac{1}{2}m\dot{x}^2$$ Total energy $E=T+U\; .$

4

The Hamiltonian $H(\theta,p_\theta)$ needs to be formulated in terms of the coordinate $\theta$ and its canonically conjugate momentum $p_\theta = \frac{\partial L}{\partial \dot{\theta}} = R^2 \dot\theta$. The correct expression for the Hamiltonian is \begin{align} H(\theta,p_\theta) & = p_\theta \dot{\theta}(\theta,p_\theta) - ...

4

If you use a constant force along the path, the spring will move past the position where $F=kx$, because it will reach that point at some speed. Thus it is incorrect to use the force method in the way you used it, because at maximal extension $v=0$ but $a\neq0$. The energy method as you used it will give the correct answer. If, instead, the force is used to ...

4

The first method is giving the correct answer. In writing the work done by the force, you are assuming that the force $F$ itself is constant throughout the extension. However, this is not true. While extending the spring in a quasi-static way, the force $F$ must always match exactly the spring force at that time. This is needed so that at the end of the ...

3

You can do so. The energy put in is the integral of $VI$, the product of the voltage and current. It may be hard to calculate $I$ from $V$ because of the back emf and changing circuit resistance. The energy absorbed is the increasing magnetic field energy caused by the expansion of the current loop and the increasing kinetic energy of the wire and ...

1

Imaging the balls on a string. You are launching N balls per second, at a velocity $u$. This means the distance between the balls is $u/N$. And $N$ balls per second will pass a certain point in space. Now if the car is moving at a velocity $v$ (same direction as $u$), fewer balls per second can hit it - because subsequent balls on the string have further to ...

0

I did not quite understand your problem in solving a simple question. But i think i get your doubt. The friction when ever taken in the work energy theorem is always taken as negative and as the work is done against the direction of the gravity the work is negative. here do not worry about the potential energy just calculate the work done when you equate ...

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As the total work on the system equals the change in kinetic energy: $$W_{Friction}+W_{Potential} = \Delta K$$ Taking: $$v_{i}=v_{f}=0$$ we can write $$W_{Friction}+W_{Potential} = 0$$ $$W_{Friction}+\Delta V = 0$$ $$W_{Friction}-V_{i} + V_{f} = 0$$ $$W_{Friction} + V_{f}= V_{i}$$ $$E_{i} = W_{Friction} + E_{f}$$ obs: Work by definition is $$\int ... 0 Yes and no, but I think in your context the answer is yes: only account for friction work. This is a good place to consider that energy accounting depends on the choice of system. In introductory courses, potential energy is usually introduced as gravitational potential energy, mgh. IMO, this is dangerous because it can lead to confusion, as we see in ... 0 When you have a non conserving force in the system(such as friction) an energy only approach only gives you an estimate of the results. It's not a particularly good way to look at the situation. The initial energy is going to be the kinetic energy. The resulting energy will be the gravitational potential less the friction losses. The frictional losses ... 1 There are different forms of energy. Energy can be converted from one form to another but cannot be destroyed. In this case the kinetic energy of the hammer is driving the nail into the wood which is breaking the molecular bonds in the wood fiber. The energy is converted to heat energy as a result of the breaking of the bonds and the friction of the nail in ... 1 Your calculations are wrong, Hint: \frac{1}{2}I\omega^2 + \frac{1}{2}Mv^2 = Mgd\sin(t) wr=v 2 A physical system in GR is never isolated, in general, as it interacts with the curved metric, i.e., the gravitational background. (However a notion of isolated system can be given in the particular case of an asymptotically flat spacetime as discussed in auxsvr's answer.) Apparently this fact prevents the existence of conserved quantities because the ... 0 If the metric is asymptotically flat, it is straightforward to assign meaning to a quantity, such that it resembles the energy we know from special relativity. In particular, for the Kerr metric we may regard (\partial_t)^a as a vector representing the stationary observer at infinity, where space-time is Minkowski, and the rest appears to said observer as ... 0 The key problem is that heat is not an energy like the others: it's degenerated, unorganized, unoriented. Everything ends up as heat (not only friction, but also absorption of light by non-white surfaces, chocks when your feets touch the floor, activity of your cells, etc). But then, it's physically impossible to turn back 100% of this degenerate energy into ... 1 The city is not a perpetual motion machine, so it will not run forever. For example, all light will be absorbed by buildings, pavements, citizens, etc, and the resulting heat will not be recoverable. The same applies for any heat source including the metabolisms of occupants and pets. The final result will be a city which becomes slowly warmer, while usable ... 0 "In 1878, the physicist James Clerk Maxwell wrote in a book review for Nature: The truth of the second law is ... a statistical, not a mathematical, truth, for it depends on the fact that the bodies we deal with consist of millions of molecules... Hence the second law of thermodynamics is continually being violated, and that to a considerable extent, in any ... 0 Heat is not an intrinsic property; that is, we cannot say that a system "contains" a certain amount of heat. Heat is not the state function , instead we can say that a certain amount of energy can be transferred, into or out of the system in form of heat or work -Resnick, Halliday, Krane Another way of looking at temperature is as an indicator ... 1 No. Look up something called the Carnot efficiency, among many others. There is no extractable heat energy without a temperature difference. For your apparent level of physics, it's probably best to take this as a fundamental principle that just is. 0 The system you describe would not work because something has to prop the ball on the hill, using energy. Ideally a system of infinite energy would require no energy input. If it were that simple, someone would have done it long ago. That said, Gravity itself does violate the law of conservation of energy. Unlike electromagnetic and kinetic, Gravity has no ... 0 Yes, the strength of the Earth's gravity changes (slightly) from place to place. However, you can't just treat Earth's gravity as a varying number; it is a vector. It mostly points down towards the center of the Earth, but those same gravitational anomalies which cause the general strength of gravity to vary also cause it to deviate from what you'd generally ... 0 I think that the problem here is that U = mgh is an approximation that we can make by assuming a constant value of g. Even taking it a bit further and using$$U = -G\frac{m_1m_2}{r}$$will make some assumptions about the density and shape of earth if you consider one of the masses to be that of the whole earth. In reality, you'd need to sum the ... 0 Yeah I think that's right, put spins in a magnetic field and they fall into the lower energy state, (with a time given by T1) it's the first step in magnetic cooling. Edit: As Floris pointed out I got this backwards. As the spins go into a lower energy state the energy they had has to go somewhere.. into the lattice (or needle in this case.) 2 The total energy stored in the magnetic field goes down - that's where the energy to move the needle comes from. The "internal energy" of the atoms inside the needle has nothing to do with it. 1 In the first part you wrote E_{k}=E_{g} because kinetic energy is fully converted into potential energy. But in the second part, some of the initial kinetic energy (E_{f}) lost due to friction and part of energy left is E_{k}-E_{f} . Only this part is converted to potential energy E_{g} . Thus, E_{g}=E_{k}-E_{f} and this simplified as ... 1 The initial kinetic energy E_k gets partly dissipated as friction, E_f, and partly converted to gravitational potential energy, E_g. The sum of these two must equal the original energy input, so$$E_k = E_f + E_g

0

The EM drive is in the headlines again on Yahoo.com. In a new round of testing, NASA confirms yet again that the 'impossible' EMdrive thruster works There exists a web page and a "theory" by the proponent Suppose that an effect really exists. The main objection is momentum conservation, because the setup sounds like a variant of the bootstrap effect: ...

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This is actually quite simple. You measure the maximum no-load velocity of the Turbine and divide it with 2. At this speed you have the highest power. Practically the optimal value is between 1.8-2.3, depending on turbine type. This all is clearly explained in the book of Carl Pfleiderer, Strömungsmachinen, 1952. Page 248. The diagramm has the rpm on ...

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Refraction of Light is not a Thermodynamic process. If you study the QED basis of refraction, you notice that the difference happens in time. The speed of light is constant, and a photon which is refracted, doesn't actually travel any slower, it just travels a longer path, and needs thus more time. If it hit's somewhere, then it's not refracted. It's gone, ...

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