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18

Temperature Amount of a substance Luminous intensity are pretty much bogus fundamental units. The unit temperature is just an expression of the Boltzmann constant (or you could say the converse, that the Boltzmann constant is not fundamental as it is merely an expression of the anthropocentric and arbitrary unit temperature). The unit energy will be ...


10

why have these been chosen as the fundamental units? Courtesy of the National Institute of Standards and Technology (NIST), we have some historical context. It basically boils down to wanting to have absolute measurements with respect to units of mass, length, and time. These few were originally chosen because these form a set of mutually independent ...


10

Any high slew rate (fast rate of change of power) stresses the grid. Lots of things cause high slew rates. People getting ready for work in the morning, having showers, turning lights and appliances on. Factories starting up at the same time. Faults on major international HVDC transmission lines. Safety shutdowns at nuclear reactors. Lots of people turning ...


5

The problem here is that you've neglected the effect of firing the bullet on the plane itself. It turns out we can account for the bullet's extra energy by examining the energy lost by the plane. To resolve this apparent contradiction, we need to examine the problem in terms of conservation of momentum: In the second case, where the bullet is fired from a ...


4

A few things: 1) it is in principle unknowable what's happening outside the cosmological horizon. Because notions of total energy depend on boundary conditions (or conditions at infinity), there are several different possible scenarios for "the total energy of the universe", all of which are completely consistent with observation 2) Assuming that ...


4

One Joule (the unit of enery) equals one kgm^2/s^2. So you see, a unit of energy can be expressed in terms of units of mass, distance, and time. The people who chose the SI units could, for example, have made the Joule an SI base unit and defined the unit of mass in terms of distance, time, and energy (kg=Js^2/m^2), but then we still have the same number of ...


3

Some rough estimates (you can dig up more accurate numbers): The oceans contain about 321 million cubic miles of water (source: http://oceanservice.noaa.gov/facts/oceanwater.html), or 3.5e20 U.S. gal. 1 gal seawater contains roughly enough deuterium to provide the same energy as 300 gal of gasoline (maybe slightly less - that's the part for your homework!), ...


3

Your answer : The bowlers and fielders rub the ball to make the ball smooth and shiny on one side and leave it rough on the other side. To generate reverse swing. There is a lot of science behind swinging the ball. The ball gets reverse swing when it is quite old and you see not only the bowlers but the fielders as well rubbing the ball before every ...


3

Purely historical and convenience reasons, people standardized measures that were more obvious from real life (remember, quantum mechanics and relativity didn't exist when SI was drafted). It's actually arbitrary how many quantities you define as fundamental (and associate them with base units), and it doesn't matter which ones are they. In quantum ...


2

Yes, absolutely. Potential energy is not a measurable physical quantity. What can be measured are differences in potential energy. So, if you compare the potential energies of a given mass at the Earth's surface and $100 \, \mathrm{m}$ above the surface, you cannot choose their difference, because that is governed by the laws of physics. However, you can ...


2

It shouldn't make any difference. The only force you're acting against is gravity if you don't take friction into consideration, and gravity is a conservative force, which means that the work it does (and hence the work that you do in this situation) doesn't depend on the path you take. So, the amount of energy you consume wouldn't depend on taking the steps ...


2

The relevant formula is Einstein's: $E^2 = p^2 c^2 + m^2 c^4$, where $E$ is energy, $p$ is momentum, $m$ is mass, $c$ is speed of light. If $p=0$ then the particle is at rest and we get the famous equation $E=mc^2$. For photons, $m=0$, and we get $E = pc = \hbar \omega$. Here $\hbar$ is Planck's (reduced) constant, and $\omega$ is the angular frequency. ...


2

There is a mistake in your working. The surface integral of the Poynting vector $$ \int \vec{\Pi} \cdot d\vec{S} = \Pi\ 2\pi a h = \frac{a I^2}{2\pi^2 \gamma a^4} 2\pi a h = \frac{I^2 h}{\pi \gamma a^2}$$ As pwf correctly points out, this power matches the power dissipated as Ohmic heating, $VI$, where $$V I = E h I = \frac{I^2 h}{\pi \gamma a^2}$$ This ...


2

I'd say part of the answer must be that whatever dynamic variable you use, like Enstrophy, Vorticity, their potential analogues, etc. those are always 'filtered' fields. Filtered in the sense, that you start with the velocity field $\vec v = \sum u_i \vec e_i$ that has full information over the dynamics and then apply some operators (integration and ...


2

If you want to do it in 1 million years then your basic problem is kinetic energy. The gravitational potential energy of the Earth around the Sun is $-GM_{\odot}M_{E}/(1au) = -5.3\times 10^{33}$ J. To get to Proxima Centauri within 1 million years requires a relative velocity of at least 1.27 km/s. So I'm not sure how exact an answer you need. The main ...


2

Dark energy is not negative energy. It causes a repulsion because of its unusual equation of state, which causes it to behave as if it has a negative pressure. There is some discussion of this in the answers to Have negative pressures any physical meaning? and possibly also 'Negative pressure' counteracting gravity?. When general relativists talk ...


1

The solar panels pictured convert light to electricity. If they were perfectly efficient then they would absorb all wavelenghts including available ultraviolet and infrared as well as visible and convert them all to electricity. In reality solar cells will be designed to absorb one set of wavelengths, though sometimes you might have sandwiches where one ...


1

The colour you're seeing is from the very small fraction of light that the panels are reflecting. The vast majority of light is being absorbed to generate electricity. Why some of the panels appear slightly blue while others don't I don't know. Presumably there must be small differences in the manufacturing process. The absorptance of solar panels does fall ...


1

Can someone resolve this conflict? Is a black body the only thing that exhibits the phenomenon of quantization or is quantization everywhere in nature? I am under the impression here that quantization ~ discreteness. I'll give an example where discreteness pops up in a classical system. Consider a simple stretched string with boundary conditions such ...


1

If you have current flowing one way through a resistor, then the electrons flow through the other way. Since current flows from the high voltage end of a resistor to the low voltage end, then the electrons come in at the low voltage end and come out at the high voltage end. When electrons (which are negatively charged) go from low voltage to high voltage, ...


1

One of many good reasons to know it is that you can also use the value you get for a metal, along with the dependence on the number of the energy level (usually "n") and the fact that electrons, as fermions, "pile up" in terms of their energy states, to get an approximation of the energy distribution of electrons in metals. I would second the notion given by ...


1

I think a good way to approach this question is with a Mach-Zehnder interferometer: The field landing on detector 1 is the interference between two waves, one from the lower path, and one from the upper path. Let's suppose the field in each arm is a collimated beam of coherent light, well-approximated as a plane wave, and the interferometer is well-aligned, ...


1

Natural frequency depends on the physical properties of a system. Some of the classical examples are masses attached to springs and pendula. For the former class, the basic model is based on Hooke's law, which translates into the differential equation (1D) $$m\ddot x + kx = 0,\qquad m,k > 0$$ where any sort of dissipative effect has been neglected. ...


1

I think of V^2 as the rate of change of the surface area of an expanding sphere , it's radius changing at constant V , but the surface area A = Pi r^2 expanding at v^2. Does that help ? Also in: E = MC^2 E = The total energy of a body at rest, V = 0 C^2 is a constant of proportionality, a real number, but not dimentionless - otherwise the formula would ...


1

Here's a simpler answer. Resonance is really all about the capture of energy into a system and its cyclic flow between potential and kinetic states. In mechanical systems we call these states potential energy and kinetic energy, but in electrical systems, as a another example, between magnetic and electrical fields. It's the rate of this cycling back and ...


1

The answer is negative. The gravitational acceleration is the same for all the atoms. Thus, the internal structure of the body is not affected. Using the terminology in your question, the average kinetic energy of atoms inside this object doesn't change. This is the reason for which given a body left free in the gravitational field, we usually consider the ...


1

$c$ is not only an invariant speed, $c$ is also a physical constant that factors in many well known formula, e.g., the electromagnetic fine structure constant $$\alpha = \frac{e^2}{4 \pi \epsilon_0 \hbar c}$$ In the case of the famous $$E = mc^2$$ the particle with mass $m$ has zero speed (in this frame of reference). If the particle has a speed $v$ in ...


1

There are two things. More photons means a brighter beam. power (Energy/sec) is proportional to the number of photons/sec. Photons with shorter wavelengths and higher frequencies have more energy. That is a bluer beam has more power. So $P = nh\nu$ where n is the number of photons/sec.


1

Since this question seems to be of the "homework and exercises" variety, I will, in keeping with the policy on this site, initially give just a pointer to the solution, rather than a fully worked example. See how far you get with this. Regardless of what the breakdown (not breakout) voltage is, there is an optimal solution for radius to maximize energy ...


1

First of all I recommend you to see in Internet Richard Feynman's WHY". It is exactly what he discusses, our questions about why. An electron in an atom has two major types of energy, kinetic and potential. The first one is due to the fact that the electron performs a motion, e.g. if we calculate the average of the absolute square of the linear momentum of ...



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