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2

According to the SI Brochure as well as ISO 80000, plane angle is an ISQ derived quantity. The SI coherent derived unit of the plane angle expressed in terms of SI base units is ${\text{m/m}}$. Thus, expressed in terms of other SI units, the SI coherent derived unit is simply $1$. The radian is a special name for the number one that may be used to ...


11

The angle $\theta$ might have units of degrees, yes. It could also have been in e.g. radians. But putting it through the sine function removes the unit. The $\sin(\theta)$ is unitless. The sine and cosine functions are defined as the "distance" vertically and horizontally, respectively, to the point on the unit circle. That is, a distance per unit lenght. ...


0

These equations are often done in units where c=1 to make things easier, in this case: $$ c^2=1 $$


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Light years help give an idea of distances through space and time. When we look at a star 100 light years away, that 100 light years not only gives an idea of the immense distance to the object but will also tell us that what we see is light from 100 years in the past.


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Ultimately, the answer boils down to convenience. When we want to describe the distance between here and, for instance, the star Sirius (the brightest star in our night sky), it would be a little cumbersome to write $\ell=8.13\times10^{18}\,{\rm cm}$ any time we want to write its distance from us. And really this goes for any astronomical object: they're ...


1

Whenever you see things at a distance, your perceptions are of things that happened in the past. You can see it at a football game, for example, where someone kicking the ball is seen very much before it is heard (sound is slower than light). Light only has a finite speed, too; and light turns out to be the fastest thing there is. A light year is the ...


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This is a more complicated question than you think. Superficially, if you have a volumetric flow rate of $V$ through a pipe with constant area $A$, then the average velocity of the fluid is given by $$v = \frac{V}{A}$$ If you use cubic feet per second and feet squared, the result will be in feet per second. HOWEVER The actual velocity of a fluid in a ...


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Any suggestions? Summarizing the answers in the comments: Divide the volumetric flow rate (in cm^3/s) by the cross-sectional area of the pipe (in cm^2) to get the velocity in cm/s.


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If you introduce any unit $v$ in terms of which you express the numerical values of $V_1,V_2$, then you see $\ln\left(\frac{V_2}{V_1}\right)=\ln\left(\frac{V_2/v}{V_1/v}\right)=\ln\left(V_2/v\right)-\ln\left(V_1/v\right)$ So if $V_1$ and $V_2$ were the integral bounds of a single integral sign, it's only important that make sure you express the numerical ...


6

How can you account for this apparent discrepancy? There is no discrepancy in the value of the difference of the logs $$ \log(V_1)-\log(V_2) $$ and the value of the log of the ratio $$ \log{V_1/V_2}\;. $$ And neither of these depends on the choice of units. This is because, for both the difference and the ratio, the units cancel. For example, suppose I ...


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Given the way that you've presented your table, I would personally put a "-" rather than a 1 in the units column. This to me would signify that units such as "g, km, s, A" etc. do not apply here. In terms of your symbols, in many branches of physics it is common to use a "hat", "tilde" or "star" notation above a symbol to indicate that it is a ...


5

Conventionally we use $1$ for dimensionless quantities, although it may cause some confusions. In additon, The International Committee for Weights and Measures contemplated defining the unit of 1 as the 'uno', but the idea was dropped. --https://en.wikipedia.org/wiki/Dimensionless_quantity


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The only convention I am familiar with is to put a $1$ for dimensionless quantities. If you feel that this could give rise to confusion, you could explain the convention somewhere above or below the table.


3

Yes. In the unit column, put $1$.


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The entire problem would disappear if temperature is measured in energy units, which is quite natural. Then we can write simply S=logW, entropy is dimensionless and TS has the dimensionality of energy.


2

There are many of such formulas. Start with Snell's law that relates sines of angles (which are just numbers) to refractive incides (which are also just numbers) but obviously has physical content. What's even more astonishing (but might also be a bit confusing) is that in so-called "natural units" most equations have no units. All you have to do is to ...


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In oscillators, we have a fourier transform of the green's function that have complex poles $\omega_0 + i\Gamma$ . The real part of this pole is a frequency $\omega$ (have units too) and the imaginary part is the inverse of the mean life $\tau$ of the oscillation $\omega_0$. This pole is a complex numeber that has physical dimensions. (more here) Complex ...


4

Some quantities are complexified for mathematical convenience and only the real part retains a physical meaning. When you have a general phasor, like an oscillating potential or current, you can think that the amplitude is rotating on the complex plane, so that both the real and imaginary part have the same physical dimension, and the actual phenomenon is ...


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The system of equations belong to the ESU, and presumably gaussian, where charge^2 = force*length^2. Putting this in, you get $F^2 * L^4 / F * L * L^3$ gives F = E/L. Hint, the correcting factor for SI is to replace $e^2$ with $e^2/4\pi$. If you really are going to read a lot of older physics, it is best to be versed in the gaussian units: their units ...


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Without having access to the book, the best guess I can come up with is that the formula is written in a natural unit system where $c=1$, $\hbar=1$, and $e=1$. In this system, charge is unitless, and energy has the same unit as inverse length. So, using $\equiv$ to mean "congruent to" (i.e. having the same units), ...


2

The units for $u$ should be $K$, $m$ and $s$ aren't involved here. The partial derivatives introduce the factor of $T$ and $L$, so it should read, in terms of the units, $$ \frac Ks+\frac{m^2}{s}\cdot\frac{1}{m}\cdot\left(\frac{1}{m} \cdot K\right)=\frac Ks\tag{1} $$ which all the terms in the middle reduce to $K/s$. Moving the $\alpha$ to inside the ...


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Imagine you are given a large number of different springs, and a large number different objects. You attached different objects to different springs (with the same extension for simplicity). Repeat the experiments many many times with different combination, and measure the accelerations $a$ of the different objects when they are attached to different ...


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Distance travelled can be measured as a length (meters) [$m$]. Now if we differentiate this function with respect to time ie. get the change in distance divided by the change in time we get the average speed which can be measured in distance divided by time (meters per second) [$m s^{-1}$]. If we repeat this process again we can get the acceleration which ...


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Actually $$\tag{1}\sum_i\vec{F_i} = \frac{\mathrm{d}}{\mathrm{d}t}\vec{p}$$ is the so called "Newton's second law". It's an axiom if you will, the validity of which must be tested by experiments. The law sometimes (if the mass of the body under consideration is constant etc) simplifies to (say in the $x$-direction) $$\tag{2}F_x = ma_x. $$ Now, to avoid ...


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Mathematicians usually are not worried about the conecpt of physical units. As such, a mathematician probably would argue that $\mathbf M_O$, $\vec{OP}$ and $\mathbf F$ belong to $\mathbb{R}^3$, as MyUserIsThis did in his comment. If this is not satisfactory to you, you could consider three distinct fields of numbers, say $\mathbb{R}_F$ for forces, ...



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