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58

The problem is what Konstantin Tsiolkovsky discovered 100 years ago: as speed increases, the mass required (in fuel) increases exponentially. This relation, specifically, is $$ \Delta v=v_e\ln\left(\frac{m_i}{m_f}\right) $$ where $v_e$ is the exhaust velocity, $m_i$ the initial mass and $m_f$ the final mass. The above can be rearranged to get $$ ...


20

TL;DR: This answer arrives at roughly the same conclusion as Kyle Kanos', i.e. in addition to payload considerations, the difficulty lies in stuffing a small rocket with a mass of fuel exceeding to the mass of the rocket itself. This answer, however, is more rigorous in how the $\Delta v$ budget is treated. Developing a relationship between rocket and ...


14

Repeat lifting books to height h several times per minute for an hour or two, and observe who gets tired. You do work lifting the book, burning up biological energy stored in ATP molecules in your cells, to activate muscles, which apply some force upward on the book (as well as suffer some internal friction). The biological energy used up is converted ...


11

Note: This question was cross posted by the OP on the Mathematics Stack Exchange. Here is a copy of my answer for it there. This is it.  The perfectly centered billiards break.  Behold. Setup This break was computed in Mathematica using a numerical differential equations model. Here are a few details of the model: All balls are assumed ...


10

The best intuition is a calculation but in this simple case, the calculation is really intuitive so you shouldn't turn off when you hear the word "calculation". The height reached by initial velocity $v$ is the height of the object after the initial velocity $v$ drops to $0$ (and then reverts the sign) because of the downward acceleration $g$. How much ...


10

As the magnet approaches the solenoid, a current is induced. The current generates a magnetic field. The field repels the magnet, slowing it's approach. The amplitude of the oscillations diminish. If there was no resistance, this would work in reverse as the magnet receded from the solenoid. The magnetic field would accelerate the magnet. The magnet would ...


8

Gravity is doing that work! If you observe, the domino is in a position of unstable equilibrium. Edit: as pointed out in the comments, this position is of a metastable and not unstable equilibrium. This means that the domino is in a state where it hasn't achieved the minimum possible energy state yet. The energy I'm talking about here is the ...


8

No. Angular momentum and energy are two different quantities. In fact, angular momentum can be conserved when energy is not. See, for instance, the famous case of the figure skater pulling their arms in and spinning more quickly. Let's say our figure skater reduces their moment of inertia to some fraction $n$ of their original moment of inertia. Then: ...


8

You do not need to invoke friction. The magnetic forces are in equilibrium by themselves so if you place the magnets in that configuration, they will not spontaneously begin to move. The reason is that there is a corresponding force on the magnets when they are vertical that matches the ones you've already drawn. Let me make a simple model. First of all, ...


7

Let $E$ denote a quantity that does not change over time (from the first principle). Consider a ball with mass $m$ dropped from a height $h$. As the ball drops, its speed changes due to the gravitational acceleration $g$, reaching a final value $v$ at impact. Thus, we can infer that the quantity $E$ depends on these 4 parameters: $$E(m,H,g,V)$$ where $H$ ...


7

Your first assumption, that there would be a weakness in the material of the spring and it would suddenly break if corroded enough is what realistically would happen. In your idealized situation each atom dissolved in the liquid had a proportional part of the potential energy of the stressed spring. As it looses its bonds with the surface, bonding with the ...


7

The amplitudes do become arbitrarily small, and there's nothing at all wrong with this. In fact the exact same thing happens with electromagnetic waves. Sure we have a quantum theory with photons that places limits on how small a packet of energy can be detected, but light can travel across the universe just fine and become as dim as it wants. The intensity ...


7

This is an experimental physicist's answer: The linked article is careful to state: That means that conservation of energy can appear to be violated, but only for small values of t (time) Italics mine. Conservation of energy is an experimental fact that has been validated in innumerable experiments. This means, as far as the correspondence with ...


6

What you say is correct in principle, but ignores the important fact that practical car engines are horribly inefficient, and their effeciency changes quite a bit over the range of speed and power required to move the car. Note that this is the point of transmissions. At best they don't loose any power, but they make the overall process more efficient by ...


6

How is it proved to be always true? It's a fundamental principle in Physics, that is based on all of our currents observations of multiple systems in the universe, is it always true to all systems? Because we haven't tested or observed them all. Could it possible that we discover/create a system that could lead to a different result? A physical theory, ...


6

Courtesy of the book Carl found we have an answer! Consider the element of the liquid helium at a height $h$ above the fluid surface and distance $y$ from the wall. To raise that element above the fluid surface costs an energy $mgh$, but because there is a Van der Waals attraction between the helium atoms and the wall you get back an energy $E_{VdW}$. ...


5

I believe the key point here is that the electron is already excited. So yes, a photon enters and two leave, but the electron goes from an excited state to a lower energy level. In a way, you could say that in a photon and an excited electron changes to 2 photons and an electron with lower energy than before. Hence, conservation of energy!


5

The reason for the asseveration If time $t$, does not appear in Lagrangian $\mathcal{L}$, then the Hamiltonian $\mathcal{H}$ is conserved. This is the energy conservation unless the potential energy depends on velocity. is that, from the definition of the hamiltonian as the Legendre transformation, ...


5

If you are cremated after death, all your fat will get burned and convert to heat energy. If you were buried, your body would decompose, turn into some other form of chemical energy and then get used up by the organisms in the soil, somehow eventually turning up as heat. So energy conservation is still valid, and no your energy does not need to go to some ...


5

The conservation of angular momentum and energy in a mechanical system are distinct, independent conservation laws. The deep reason for this is that, because of Noether's theorem, conservation laws are in correspondence with the symmetries of the system. Thus if a system is time-translation invariant - if experiments behave the same no matter what the ...


5

"They" are probably talking about symplectic integrators. Most numerical integrators for (partial) differential equations do not specifically consider the energy of the system; they are generic integrators capable of solving any set of DEs, and not all DE's have a concept like "energy". When these are applied to a classical dynamics problem concerning ...


5

yes, the waves fall off in intensity as they get farther from the source. This does not violate conservation of energy, because you'll just be spreading the same amount of energy out over an ever larger volume, but the (energy density)*volume will be constant, minus energy transferred from the waves to matter.


5

He's basically saying assume you have some complicated system of weights connect by pulleys, and each weight can be in only one of two states: up or down. But you can trade off which ones are up and down, for example you can make 3 light weights go up by having one heavy one go down, and there are many other moves like this you can do. Now his point is ...


5

I apologsise for this being only a partial answer. I can answer the questions in your title and in the final paragraph, but the first paragraph seems like quite a different question, and for the moment I'm not sure of the answer. (Some more details would help, e.g. a link to the discussion you mention.) To answer the question in the title, consider unitary ...


5

Suppose the ramp wasn't there, then the trajectory of the object would the same as if it fell off a cliff: To get the equation of motion you simply note that the horizontal and vertical coordinates are given by (neglecting air resistance): $$ x = ut $$ $$ y = \tfrac{1}{2} g t^2 $$ So you can get the trajectory by substituting for $t$ to get: $$ y = ...


5

When you compute the energy of a rotating object you either consider the tangential velocity or the angular velocity. The two expressions are two different ways to look at the same observable, they both lead exactly to the same result and summing them is a mistake. For instance for a point like mass on a straight line we have: $$E_l = \frac{1}{2}m v^2$$ ...


5

See Anand's answer: I'm not sure whether this one is simply misguided or instead subtle fraud (as calls for money to fund research are involved). Actually the claims made in the article are true in one sense, which gives the idea the whiff of sophisticated fraud. In the linked article, it is claimed that the device is powered by a 1kW source and then ...


4

In an alpha decay no electrons are created or destroyed. There is a small correction needed for the Coulomb term when the alpha escapes without carrying two electrons with it, but that is at chemical, not nuclear energy scales and is (usually1) sorted out by chemical means in fairly short time scales. So, no you do not figure the mass of any electrons into ...


4

The factor $2$ comes from the equation $v_f^2 = v_i^2 - \color{#C00}2gh$. That's how we derive the equation for max height.$$v_f^2 = v_i^2 - 2gh$$Now, we want $v_f$ to be $0$ (because at the maximum height, the speed is always zero). Calculating $h$, we have:$$ 0 = v^2 - 2gh \Rightarrow -v^2 = -2gh \Rightarrow v^2 = 2gh \Rightarrow h = \dfrac{v^2}{2g} $$You ...


4

If you apply the same force for the same period of time, the linear velocity of the body will be the same in both cases, assuming the body is unconstrained. However, having applied the same force for the same amount of time does not mean that the same amount of energy has been transferred. The energy, or the work done by the force, is the force times the ...



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