Tag Info

Hot answers tagged

27

In general, the sun's light (particularly the UV that causes sunburn) has to pass through a lot more atmosphere (or a greater amount of air mass) in the morning and evening to get to a vertical surface than it does when it is at zenith to a horizontal surface. An example is shown in the generalised image below (all graphs are obviously generalised): The ...


11

The frequency is determined by the energy spacing between two configurations of the caesium atom. Caesium has a single electron in the outermost $6s$ orbital, and this electron can be aligned with or against the nuclear spin. These two configurations differ in energy by about 0.000038 eV, and transitions between them produce/absorb light with a ...


5

What causes sunburns is UV radiation, which damages our skin cells. Heat on the other hand is the same as the one felt when near an incandescent light bulb (doesn't cause sunburns). Most of the UV radiation coming from the sun is absorbed by the earth's atmosphere. During sunset and sunrise the radiation emitted by the sun passes through more air until ...


5

The sun is an extended source. This means that it occupies a definite solid angle in the sky $\omega = 6.8\times 10^{-5} Sr$. To visualise this (not to scale), let say that the black area in the following diagram is the angular extend of the sun as seen from the surface of the Earth (ignore the other labels), What happens when we concentrate sunlight ...


4

Wikipedia has a good article on hybrid photovoltaic-thermal systems. As you proposed they consist of a solar cell with a thermal collector at the rear. Solar energy conversion is a fascinating topic from a thermodynamic perspective and has been summarised beautifully by the work of De Vos, The Thermodynamics of Solar Energy Conversion, ISBN: ...


3

But what experiment showed that cesium-beam's period was so terribly consistent? They compared it with other clocks. That frequency is terribly consistent, because it isn't actually constant, because it varies with gravitational potential. Moreover this frequency is quoted in Hertz, which is cycles per second. And it's used to define the second. Spot the ...


3

The refractive index of air is about 1.0003, so the speed of light in air is about 0.9997$c$. You can work out the energy of the proton at this speed using: $$ E^2 = p^2c^2 + m^2c^4 $$ where the momentum is: $$ p = \frac{m_p v}{\sqrt{1 - v^2/c^2}} $$ I get the energy to be a shade over 38GeV.


3

The ionisation energy of a nitrogen atom is about 14.5eV but in $^{14}$C beta decay the electron is emitted with an energy of 156keV, which is far higher than the ionisation energy. So beta decay will normally produce an N$^+$ ion and a free electron.


3

The rest mass of the intermediate particle is not relevant. Even if the W boson was real and not only a virtual particle, the energy-momentum is conserved during beta decay and the whole energy-momentum couples to the gravitational field in the Einstein equations $$ G_{\mu\nu}=\frac{8\pi G}{c^4}T_{\mu\nu} $$ where $T_{\mu\nu}$ is the energy-momentum tensor. ...


2

Planck's law of black-body radiation can be stated in many different ways, depending on whether one is interested in the spectral energy density per volume or per area. It can also be expressed in terms of radiation wavelength or frequency. The energy of a photon is $$ \epsilon = h\nu = \frac{hc}{\lambda}$$ I will not derive Planck's law here. It can be ...


1

There are very few people in the world besides physicists and nuclear engineers who know the difference between radiation and radioactivity. Radiation is energy that radiates (i.e., travels in straight lines) from a source. Usually, when we say "radiation" we are talking about ionizing radiation---either photons or massive particles that carry enough ...


1

There are two distinct ways for a previously non-radioactive material to become radioactive. Contamination refers to any behavior where existing radioactive material is sticks to or is incorporated in an previously non-radioactive object or body (this can include a tract of land). This requires you to go to where there is mobile radioactive material or for ...


1

For example aluminum, even though in the Thorlabs mirror it has been coated to prolong the life of the mirror, the base is the reflective aluminum. Check also Refractiveindex.info for reflectance of materials.


1

You can actually do this a bit more simply (or at least without integrations). Luminance is invariant in geometrical optics. That is, the brightness of an image cannot be brighter than the source. The radius of the sun is 0.6958 x 10^6 meters. The radius of the earth's orbit is a mean of 149.6 x 10^6. Then the brightness at the surface of the sun is the ...


1

Jezstarski is mostly correct, The para-positronium (p-PS) state ends up being the main mode of annihilation of positronium (PS). Positrons can annihilate in at least eight different ways but once ortho-positronium (o-PS) forms in a void/vacuum, it has additional time to undergo another mode of annihilation. P-PS annihilates in under 125 picoseconds. O-PS ...


1

Stoves and other hot objects heat up, but don't burn. Burning is very different. Burning is a chemical reaction. In the example of stoves, they work by conduction.


1

Here's a simple mental picture to have of a how a burner on a stove heats up water in a pot (which is sitting on the burner). (In what follows, I will use the term "molecules" for both molecules and atoms.) Also, keep in mind that thermal conduction is different than electrical conduction. One (electrical conduction) concerns the flow of charge, so in this ...


1

When you put the pot on the stove, the heat from the stove is somehow getting to the pot, which gets hot. The pot and the stove are obviously in contact with each other. Therefore conduction plays a role here. If you have an old pot, with a warped bottom, it will heat up slower, because the contact surface between pot and stove is smaller. When you hold ...


1

The radiation of a body is given by the Stefan-Boltzmann equation: $$E= \epsilon \sigma T^4$$ Which means that if you can measure the radiated power $E$, the temperature is $$T=\sqrt[4]{\frac{E}{\epsilon \sigma}}$$ If the emissivity is lower than you think, it follows that you see less emission that you would expect for a given temperature and you would ...



Only top voted, non community-wiki answers of a minimum length are eligible