# Tag Info

## Hot answers tagged energy

3

Expanding on what Couchyam said: To some extent, this is actually the definition of what it is to be a "particle." Intuitively, a particle should have some particular definite energy, and be stable enough that it can exist in its own right. These two requirements are linked by energy-time uncertainty. The natural "clock" to compare against when asking if ...

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 $\Delta V ... 2 You're right that the unit "megawatt" is abbreviated MW. However, as Aniket comments, watt itself means "energy per unit time", so saying that the power plant produces 60 MW per hour doesn't make sense. In your comment, you question whether MW is a "basic unit". I'm not exactly sure what you mean by this, but the SI unit of power is watt, so if you want to ... 1$p=\frac{1}{3}\rho$is the well-known equation of state of a photon gas. It may be derived by looking at the ultra-relativistic limit of the energy momentum tensor for a bunch of particles.$^1p=-\rho$follows from the fact that the energy momentum tensor of$\Lambda$-style dark energy is proportional to the metric. Thus, at a point and in the proper ... 1 For simplicity, look at the case where$m=\omega=1$(where$m$is mass and$\omega$is frequency) so that the hamiltonian is $$H={P^2\over2}+{Q^2\over 2}$$ Put $$A={1\over\sqrt2}(P+iQ)\qquad B={1\over\sqrt2}(P-iQ)$$ so that $$AB=H-\hbar/2\qquad BA=H+\hbar/2$$ From this, check that if$H\phi=\lambda\phi$($\lambda$a scalar) then ... 1 Their paper is inconsistent. They filled in$\omega = 264$with the other quantities in SI units, so ω should be expressed in rad/s (often written$\mathrm{s}^{-1}$). So they assumed ω was already in rad/s. If they say they assumed$\omega = 264~\mathrm{rpm}$, that's not consistent with the values they plugged in. Your value of 69696 is hard to decipher ... 1$\mathrm{Watt}$(or Joule per second or$\mathrm{J/s}$) is the SI unit of power. So megawatt is a valid unit of power (expressing power with order of magnitude$10^6$) and is used mainly in commercial statements. Definition: Power means the quantity of energy consumed or produced per unit time. So$\mathrm{MW/hr}$actually makes no sense since it ... 1 You've underestimated the effect, although your math is correct as far as it goes. At 22% of c, relativistic effects do rear their ugly heads, and the proper equation is $$KE = \frac{\frac{mv^2}{2}}{\sqrt{1-\frac{v^2}{c^2}}} = 7.6\times10^{19}\text{ J}$$ Divide by$63\times10^{12}$and the ratio is 1,200,000 (1.2 million). And yes, this is an unreasonably ... 1 The formula that WhatRoughBeast uses is not correct. It should be $$KE = \left(\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}} - 1\right) mc^2 = \left(\frac{1}{\sqrt{1-0.22^2}} - 1\right) (22,000)(3\cdot10^8)^2 = 5\cdot10^{19} J$$ Your calculation wasn't far off since relativistic effects aren't very large at 22% of the speed of light, so$KE = ...

1

No it is separate issue to total internal reflection. The important thing here is frequency dependent emissivity of the materials. Very simply speaking, the absorber needs to have high emissivity in the optical and NIR region (so it strongly absorbs sunlight) and low emissivity in IR, so that the thermal energy is not radiated away. It is like a green ...

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