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0

consider ten kg substance .Take few kg substance and measuring mass density,the density is same as before substance. so we can say that from above explanation,density is an intensive property.


2

The obvious method is to burn a barrel of crude oil and measure how much energy is released. The only slightly less obvious method is to burn a small amount of oil and measure how much energy is released, and then mathematically figure how much energy a whole barrel would release, as @CuriousOne points out. The latter method is superior in both the ...


0

Unit problems can be tricky, especially if you start with SI definitions of unadorned fundamental constants. My technique is the think in energies and lengths whenever possible: \begin{align} [\hbar c] &= \rm eV\,nm \\ [\lambda] &= \rm nm \\ [k_B T] &= \rm eV \end{align} So the argument of the exponential, ${hc}/{\lambda kT}$, is dimensionless ...


2

Well, you could do it in any arbitrary units if you choose. Let's look at the perfect gas law: $P = \rho R T$. I can pick any consistent set of units that I want there, but it may involve offsets. For example, let's keep $P$ in Pascals, $\rho$ in kilograms per unit volume, $R$ in joules per kilogram per kelvin and let's use $T$ in Celsius. We know that $T_k ...


2

It should be obvious that the ideal gas law $$ pV=nRT$$ is not invariant under the transformation $T\mapsto T+T_0$. But that is exactly what the relation between Kelvin and Celsius is - they are not, like most other choices of units, a different scaling, but they choose a different zero point of temperature, so you may not use them interchangably in formulae ...


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 ...


1

Young's modulus of wire material is given by, \begin{align} Y=\frac{\text{stress}}{\text{strain}}=\frac{F/A}{l/L}=\frac{4FL}{\pi d^2 l}. \end{align} From given data, $F=mg=9.8\;\mathrm{N}$, $L=2.0\;\mathrm{m}$, $l=0.8\times{10}^{-3}\;\mathrm{m}$, and $d=0.4\times{10}^{-3}\;\mathrm{m}$. Substitute the values in equation to get $Y=1.95\times{10}^{11}\approx ...


2

Engineers created that problem. ;) (probably not) Many physics books use $Y$ for Young's modulus (Symon, Knight, Young & Freedman). Taylor's Classical Mechanics uses YM. Halliday, Resnick & -fill-in-the-blank- state that engineers use $E$. I suspect that physicists started using $Y$ for exactly this reason: to highlight a difference in the meanings ...


0

Coincidence, nothing deep I'd say. Note that the equation representing the electric field modulus depends on the units you've picked and as such putting so much emphasis on the exact characters appearing in the eq. is senseless. Note that it's possible to form many physics equalities and equations involving 3 characters. E, epsilon and sigma are quite used. ...


2

Yeah, that's just a coincidence. The easy way to see this is that $\epsilon$ is a relatively static property of a dielectric but a totally dynamic property of a stretching material.


4

Just a coincidence. There are too many quantities and not enough letters. It probably does make a difference that the fields in which these two equations exist (material science and electromagnetism) are well enough separated that you typically won't see them both in the same papers or textbooks; if that weren't the case, people would start using different ...


0

Units of temperature are K, degrees C, degrees F. Units of temperature change are K, C degrees, and F degrees.


1

Although the eV is defined with respect to a particle of unit charge (an electron) in an electric field, it is simply a unit of energy. There are simple relations between thr eV and everyday units of energy, such as the Joule or calorie. Thus energies of all sorts of things, in fact any energy, can be expressed in eV, even if it has nothing whatsoever to ...


1

One electron-volt $=1.6 \times 10^{-19} joules$ and is a unit of energy that is equal to the energy acquired by an electron falling across a 1 volt potential difference. The particle (neutrino) doesn't need a charge to have some energy. Instead of expressing the mass of a particle in kg, we can express it as $mc^2$ which is an energy (joules or eV... your ...


3

There is a smallest measurable time interval, known as Planck time, which is the time required for light to travel the smallest measurable length which is known as the Planck length, $$\ell_\mathrm{P} =\sqrt\frac{\hbar G}{c^3} \approx 1.616\;199 (97) \times 10^{-35}\ \mathrm{m} $$. So, the Planck time will be $ t_\mathrm{P} = ...


5

Planck time - the amount of time it would take a photon (or other particle travelling at the speed of light) to cross a planck length - the fastest known speed travelling over the shortest known distance. Time, as distance divided by speed, doesn't get much smaller than that. It is about 5.39 x 10-44 seconds Which can be expressed as ...


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

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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