# Tag Info

## New answers tagged units

1

The value of the gravitational constant (aka Newton's constant and big G) is a man-made convention - it's dimensionful and depends upon the our definition of units. In common SI units, $$G = 6.674\times 10^{-11}\,\text{Nm^2/kg^2},$$ but clearly the numerical value would differ in e.g. units of $\text{N miles$^2$/stones$^2$}$. With that in mind, it ...

-1

Ultimately, a mystery of sorts, but see https://en.wikipedia.org/wiki/Dirac_large_numbers_hypothesis and google "Dirac large number hypothesis" for what might (or might not) be a step towards explanation. For example, the relation $GM/Rc^2\approx1$, where $M\sim\mbox{"mass of universe"}$ and $R\approx13.7\times10^9\mbox{l.y.}\sim\mbox{"size of universe"}$, ...

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Newton just find that the gravitational force is proportional to the product of mass and the inverse of the square of the distance. Some years after, Cavendish mesure the value of that constant G. That constan could be 1, or 1000000. But G=6.674×10−11 N · (m/kg)2. Why?, nature works that way.

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I can provide only "classically newtonian" answer cause my orientation in general relativity is only very limited. As it's written in the wikipedia article, the $G$ is empirical constant. How to understand that? Experimental scientists are usually aware of dependency on parameters, but the absolute values are corrected using various additive or (in this ...

4

This video by Richard P. Feynman might explain how hard it is to answer such a 'why' question. An excerpt: But the problem, you see, when you ask why something happens, how does a person answer why something happens? ... When you explain a why, you have to be in some framework that you allow something to be true. You have to know what it is that you’re ...

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 ...

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$\mathrm{Watt}$ (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 ...

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Notice, the unit of power is $~\mathrm{J/s}$ or $~\mathrm{W(Watt)}$. The unit $~\mathrm{MW}$ indicates the energy (in $~\mathrm{mega\ joules}$) produced by power-plant per unit time (in $~\mathrm{seconds}$) The unit $~\mathrm{60 \ MW\ per \ hour}$ doesn't represent a physical quantity.

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Physics processes are independent on the measurement units that you use. Your question is not about physics but about math. If you square a number less than one it will result in a smaller number, if you square a number larger than one you will get a larger number. But changing units will not change anything. For instance, suppose you have a square that is ...

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I'm not sure about your rain comment and how you were taught rain formed, but I can give you a bit of info on units. It doesn't matter what length you choose to be the base unit, as long as you are consistent. Right now our definition of the meter is the length light travels in $1/c$ seconds, where $c$ is the defined speed of light. You have to ...

2

You cannot convert "watts per meter squared", $\rm W/m^2$, to "watts per meter cubed," $\rm W/m^3$. Square meters measure area, cubed meters measure volume; they are quite different. You might as well say, "I have an acre of sod; how tall is it?" The extra factor of length changes things completely.

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Writing $(Q_m)$ (and so on) for the numerical value of the constants in the units stated, \begin{align} \frac{Q_m}{\omega_b \rho_b c_b} &= \frac{(Q_m)\mathrm W/\mathrm m^3}{(\omega_b)\frac{\mathrm{ml}}{\mathrm{ml\: s}} \times (\rho_b) \frac{\mathrm {kg}}{\mathrm m^3} \times (c_b)\frac{\mathrm J}{\mathrm{kg}\:^{\circ}\mathrm C}} \\ & = ...

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Since mL cancels with mL, J/s and N-m are compatible and therefore you should need no scaling.

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This is nothing but a choice of units. Let me make that (hopefully) more clear by explaining more about how choosing units in electromagnetism works: Coulomb's Law is $\vec F = k \frac{q_1 q_2}{r^2} \hat{\vec r}$ for the force between two charges $q_1$ and $q_2$. $k$ is different in the various systems of units - essentially, it depends on how the unit of ...

0

It's valid, but in physics it is usually safer that each partial term keeps some physical meaning (and physical consistency). At any point you can temporally "leave physics for maths": calculations are equivalent, at your own risk of mistake ! (e.g. if you start identifying terms with similar-looking terms coming from other sources). It might happen also ...

4

As you said $\log(n_1/n_2)$ is perfectly valid because even though $n_1$ and $n_2$ are not individually dimensionless but their ratio is dimensionless. The relation $$\log(a/b) = \log(a) - \log(b)$$ is only true if $a$ and $b$ are real positive numbers. Since $n_1$ and $n_2$ are not real positive numbers (they are quantities with dimensions), you can't ...

0

You have to take the log of a dimensionless quantity. I assume you have a model for the relationhship between $D$ and $V$ that looks like $D=m \log V + b$, but since you can only take the log of a dimensionless quantity, this should really be $D=m \log \left(V/V_0\right) + b$. Typically $V_0$ is just $1$ in some choice of units, so for example, \$V_0 = 1 ...

0

You cannot meaningfully evaluate the logarithm of a dimensionful number. The reason for this should become clear if you attempt to series expand a logarithm with a dimensionful argument. The usual way to get around this is to write something like: $$\frac{\frac{D}{1\,{\rm Gray}}}{\log_{10}(\frac{V}{{1\,\rm cm}^3})} = 1.23$$ Some people prefer to omit the ...

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Physicists also use g/kg in many fields (chemistery, atmosphere dynamics), and Hubble constant is in (km/s)/Mpc. Light transport and scattering have many very close formula that might only differs by having or not a cosine inside, or be integrated or not with a unitary weight function. Of course you would be right to say g/kg is dimensionless and Hubble ...

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