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4

So the BIPM has now released drafts for the mises en pratique of the new SI units, and it's rather more clear what the deal is. The drafts are in the New SI page at the BIPM, under the draft documents tab. These are drafts and they are liable to change until the new definitions are finalized at some point in 2018. At the present stage the mises en pratique ...


0

The units are not consistent. Or in less precise terms, wrong. Here's the only way I can think of for this to make some sense: just after equation (1), the paper says ...where $\rho$ is the density in $\mathrm{g\,cm^{-3}}$. My guess is that they intend you to take $\rho$ as a pure number. For example, if the density is $0.1\ \mathrm{g\,cm^{-3}}$, then ...


2

Each equation contain a different arbitrary constant: 1500, 2800, and 5 E20. It can be assumed that each arbitrary constant has exactly the right units to make everything come out right... It is sloppy to not specify the units of these constants... Edited for example: I could conduct experiments on the dynamics of falling objects, and publish that the ...


1

This is mostly to make an explicit connection with natural units - the unit system in which $\hbar$ and $c$ are both set to 1, which is the natural set of units for relativistic quantum theory. Because you've adimensionalized two units and you had three physical dimensions to start with (mass, length and time), natural units retain a single dimensional ...


0

The point is that luminous intensity is intensity as perceived by the human eye, and particularly taking into account the fact that the same amount of power will be perceived as brighter or dimmer depending on whether the wavelength is at a maximum of the eye's sensibility or at a minimum. This makes the candela ever so slightly washier than the other six ...


0

Personally, from a pure physics point of view, i don't think we need the candela as a base S.I. unit. it seems that all we need is: m,kg,s,A,K,mol These can all be related to basic physics constants (see Physics Today, July 2014, p. 37): c,h,delta nu(pick any atomic transition),e,k(Boltzmann),N_A(Avogadro's) Actually, even N_A is only necessary from a ...


-1

E=V/r=volt/meter as volt=Joule/coulomb E=J/C.m as J=N.m E=N.m/C.m as m and m cancel each other so,E=N/C



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