# Tag Info

1

If you call $\chi$ the exergy (availability) then $\chi = U + p_o V - T_o S$ where $p_o, T_o$ are the pressure and temperature of the environment (and are assumed to be constant). To find the maximum amount of useful work that can be extracted form the system it is sufficient to analyze reversible processes only so that $dU=TdS-pdV$ and then the exergy ...

0

According to our current scientific knowledge we know that we don't know what the 70% of the energy of the Universe is. Also, a comprehensive description of the thermodynamics of the Universe is impossible with the current standard Cosmological model and Einstein's General Relativity. In particular it's very complicated, and incomplete, as I said, to ...

0

On the internet there is plenty of talk of how the continuity equation applies to conservation of charge, fluid dynamics, and so forth, but I can't find any mention of how it applies to the conservation of energy. Why? Is it because it is problematic to talk about energy current density (j)? The continuity equation is fine for energy, and sometimes ...

0

A perfectly inelastic collision (also known as a plastic collision) occurs when the maximum amount of kinetic energy of a system is lost. In a perfectly inelastic collision, i.e., a zero coefficient of restitution, the colliding particles stick together. In such a collision, kinetic energy is lost by bonding the two bodies together. This bonding energy ...

0

The fundamental reason why energy is conserved, is invariance of the physical laws by a time translation $t \to t + t_0$, in a Lagrangian formulation. This is a particular case of the Noether theorem, which states, that if a Lagrangian has a continuous symmetry, there is a corresponding conserved quantity. Now, if we look in detail, conservation of a ...

0

Usually, when physicists talk about energy being conserved, they mean Energy being a Noether charge on the fundamental level, c.f. wikipedia. As a very general result, one can derive that the time derivative of Energy is zero $$\frac{d}{dt} E = 0.$$ This result is only true if the Lagrangian description of your system does not explicitly depend on time ...

1

This is an experimentalist's answer and yes, accelerated charged particles either in stable circular orbits or in linear acceleration do radiate. Classically, any charged particle which moves in a curved path or is accelerated in a straight-line path will emit electromagnetic radiation. Various names are given to this radiation in different contexts. For ...

0

Neglecting air resistance (whether this is a good idea or not is besides the point), suppose you drop the egg from height 10.0m, and it fell to a height of 0.10m, and then rebounded to a height of 9.0m. What can you deduce from this fact? Edit: Since you want something a bit more along the modelling side, if you know how the tension changes with ...

0

Mainly because you need a lot of speed to go into space, and to each that speed, you need to accelerate. If you need a high speed, you will need to accelerate for a long time, thus the need for a large quantity of fuel. You also need to compensate for gravity the whole lift. There are ways to reduce that fuel requirement, like a horizontal takeoff, you ...

-4

$E = mc^2$ The larger the mass, the more energy can be produced. And we still haven't found any fuel which in small quantities gives the needed amount of energy. I know you will be thinking of nuclear energy; we cannot fit a nuclear reactor inside a rocket with current technology, and even if we can fit it I don't think our existing knowledge of nuclear ...

19

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 ...

50

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 $$... 3 Because most payloads are quite heavy. I am not sure what kind of payloads you had in mind, I am no expert on this, but I think that most launches contain satellites, which might be heavier then you think, for instance the satellite in this BBC Documentary weighs 6000 kg. And according to Wikipedia, miniaturized satellites weigh less than 500 kg (so heavier ... 0 Consider the problem in the from of a ratio, what is the ratio of mass used to lift the rocket(fuel), to the mass finally put into orbit(cockpit). That proportion will be much the same regarding smaller objects that must be put into orbit. If you use the same ratio or proportion to calculate the needed fuel mass for a small craft, you will find you can't ... 0 The work done by a force along a path \gamma is defined as$$ W = \int_\gamma \vec{F}(\vec{r})\cdot d\vec{r} = \int_a^b \vec{F}(\gamma(t))\cdot\dot{\gamma}(t)\,dt$$where the last equality is the actual definition of the line-integral of a vectorfield along a curve. Note that the force is explicitly depending on position. As you state correctly, a force ... 0 Does the work done by the force remain 0 even if it varies at all points on the loop ? Yes. For example, the gravitational force. Note, that in general fields are not conservative. So if you write an arbitrary force, the work will not be zero. 1 In the standard formula given in the question posed, the potential energy is zero. The formula applies to a free particle only. For a charged particle of charge Q in an electromagnetic field, the correct formula for the total (kinetic plus potential) energy is$$E= c\sqrt{(mc)^2 + (p+QA)^2} -QA_0, (e.g., $Q=-e$ for an electron) where $A$ and $A_0$ are ...

1

This should get you going in the right direction. I am using different variables so you can understand the concepts behind impacts and collisions (or explosions). Two rigid bodies are attached at common point A located a distance $\vec{r}_{A1}$ and $\vec{r}_{A2}$ from their respective centers of mass. With the CM velocities $\vec{v}_{1}$ and ...

1

The answer is simple. The optimum strategy to slow (accelerate) the system is to remove equal amounts of mass from each small mass at the same time and in the same (opposite) direction as the current motion of the pair of point masses connected by the rigid rod. By symmetry the momentum change is then all in translational momentum in the direction ...

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