# Tag Info

0

Yes, the amount of radiation (i.e. power) would decrease $N^4$ times, since the whole geometry of the system does not change, only the power emitted by the sun. But the temperature of the earth would decrease only $N$ times, since the ratio of the temperatures remains equal - again because of the unchanged geometry. The earth radiates a power proportional ...

1

Background The specific intensity or brightness, $I_{\nu}$, is defined as: $$I_{\nu} = \frac{ dE }{ dA \ dt \ d\Omega \ d\nu } \tag{1}$$ where $\nu$ is the frequency, $dE$ the differential energy, $dA$ the differential area, $dt$ the differential time, $d\Omega$ the differential solid angle, and $d\nu$ the differential frequency. We can define a net flux ...

1

International Agency for Research on Cancer groups cell phone in group 2B, which is for possibly carcinogenic to humans. Not even probably. So, it is not that harmful . But, you can protect your by Electromagnetic shielding. You can figure out a way to connect conductive or magnetic materials either to your or the phone, but if you connect it to your ...

0

background radiation is very dependent on where you live it's different depending on altitude rocks around where you live ect. but generally comes to around 0.1 - 2 micro sieverts per hour the "ionizing radiation regulations 1999" states: Anyone under 18 should not be exposed to more than 6 milli sieverts per year above background radiation ~ 1 micro ...

0

The answers already given explain the differences between the processes by which heat is transferred from one body to another but there are also differences in their relative importance in particular situations. A central heating radiator does radiate heat but convection is a more important process once the air is heated and before that conduction through ...

2

I don't fully comprehend why the positive (and therefore quicker) parts appear to be retarding and eventually making a N-like sawtooth wave? I wrote a more detailed answer at http://physics.stackexchange.com/a/139436/59023, but the basic idea is that the larger amplitude parts of the wave have a higher phase velocity than the lower amplitude parts. ...

1

Neutrons and protons are hadrons and are identified as an isotopic spin 1/2 hadron called baryon. There are many hadrons, pions, kaons, detas, etc , and one of the quantum numbers that separates them is the baryon number. Protons and neutrons each have baryon number 1. Thus a neutron turning into a proton does not change the baryon number. Isobaric means ...

1

I think the figure of 33eV comes from the definition of ionising radiation i.e. electromagnetic radiation with an energy above around 10 to 33 eV (the boundary is somewhat vaguely defined). However gamma radiation from radioactive sources can be much, much higher energy than this. The mechanism is that a gamma photon is absorbed by an atom and produces an ...

1

I think the term "released" is applied a bit too liberally here: The underlying idea should be the following: As per your Chem 101 course, you need energy to ionize an atom. But the atom will not stay ionized, it will immediately attract an electron again. This electron will then release energy - the ionization energy. Since the original energy to ionize ...

1

Ionization is not just to do with knocking electrons out of atoms and molecules. If there is enough energy available the bonds between molecules can be broken to form ions. Energies of 33 eV are equivalent to the extreme UV band of the electromagnetic spectrum. Energies of this magnitude can break up water molecules and produce free radicals (reactive ...

3

First of all the Earth's radius is about $6.4\times 10^6\mathrm{m}$, while its orbital radius is about $1.5\times 10^{11}\mathrm{m}$: about $2.3\times 10^4$ times its radius: yes, it is a good approximation that the whole Earth is at the same distance from the Sun. Secondly, this is the flux passing per unit area of the imaginary shell: the flux per unit ...

0

This is the periodic table of elements, showing nuclei at the center of atoms: Each nucleus except of the hydrogen atom, contains a mixture of protons and neutrons held together by forces that can only be described using quantum mechanics. Stable nuclei are the ones where the forces binding protons and neutrons allow fixed energies and no probability of ...

1

I've worked out a formula that assumes no absorption in the atmosphere, and that light is only backscattered. It takes into account the backscattering of the light reflected off the surface. In the diagram below: TOA is Top Of Atmosphere $\tau_a$ and $A_a$ are the transmittance and albedo of the atmosphere such that $\tau_a+A_a=1$ $A_s$ is the albedo of ...

1

The Albedo describes the fraction of radiation reflected for a given area of the surface. If you are combining albedos for different parts of the surface of a body, you need to weight them by their areas, for example consider albedos $\alpha_1$ and $\alpha_2$ corresponding to areas $A_1$ and $A_2$, then the effective albedo for the total area is: ...

0

The force exerted by the magnetic field on diamagnetics is not given by its actual value, but the gradient of its magnitude. So in truly homogeneous 45T field, no net force would be generated; on the contrary, you could use a strongly varying field to achieve levitation also with weaker field. The health risks from strong static magnetic fields are ...

0

When you say things like My question is just about the typical amplitude value in an experiment with lasers. you run into trouble, because "typical experiments" use laser powers which cover many orders of magnitude. In general, you find the electric field from the intensity $I$ of the light, which is equal to the amount of power per unit area ...

4

The electric field strength is related to the power of the laser by the Poynting vector. This is given by: $$\mathbf{S} = \mathbf{E} \times \mathbf{H}$$ and the magnitude of $\mathbf{S}$ is the power. Assuming we can treat your laser as a plane wave (which seems reasonable) then $\mathbf{E}$ and $\mathbf{H}$ are at right angles so the power is simply: ...

2

Yes, it is essentially just the Coloumb interaction. Sometimes, this will strip electrons from atoms. However, those electrons might have been responsible for a bond in a molecule, so this will have effectively destroyed the molecule. Other mechanisms include having a nucleus recoil from a proton (perhaps imparting enough energy to break a bond), or ...

Top 50 recent answers are included