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For steady upward bubbles in a stationary liquid, assume fixed bubble mass, quadratic drag force, negligible gas weight and inertia, and negligible wall force. Then $a\propto {\rho }_g^{-1/3}$. Hydrostatic mixture support and terminal drag force balance give $v^2\propto a(1-\alpha )$. The gas mass flux $J=\alpha {\rho }_{g}v$ is constant, hence $J^2\propto {\alpha }^2(1-\alpha ){\rho }_g^{5/3}$. Comparing at a reference point proves the displayed relation. The approximation requires a small gas/liquid mass density ratio and macroscopic variation scales; it need not hold near a gas-dominated endpoint or rapid acceleration region.

Terminal velocity is a constant limiting velocity approached when a velocity-dependent resistance balances the other applied forces. A falling body may keep moving while its net force and acceleration tend to zero. For downward motion in a stationary medium, its limiting speed is set by balancing gravitational weight against drag force, including buoyancy if appreciable. The balance, rather than a universal numerical value, is the defining idea. ^[ref]

The terminal velocity under linear drag gives a simple example with an explicit approach to the limit. Real falling objects need an appropriate resistance model: linear drag is useful for slow motion of a small sphere in Stokes flow, whereas many faster falling objects require quadratic drag. Changing mass also changes the question: the existing accreting raindrop with quadratic drag has a speed-radius attractor of an accreting raindrop, rather than the fixed-mass balance used below.

Consider a point mass of fixed mass $M>0$ released from rest in a stationary medium. Choose downward as positive, assume constant gravitational acceleration $g>0$, neglect buoyancy, and impose linear drag $F_d=-bv$ with constant linear friction coefficient $b>0$. Here $v$ is the signed downward velocity component; it equals speed during this descent. Newton's second law becomesThe terminal velocity balances the two forces, and the time constant $\tau$ controls how quickly that balance is approached:For the illustrative parameters $M=1\mathrm{kg}$, $b=2kg/s$ and $g=9.8m/{\mathrm{s}}^2$, the time constant is $0.5\mathrm{s}$ and the terminal velocity is $4.9m/s$ downward. At $t=1\mathrm{s}=2\tau$, the downward velocity is $4.9(1-e^{-2})m/s\simeq 4.23686m/s$ and the downward acceleration is $9.8e^{-2}m/{\mathrm{s}}^2\simeq 1.32629m/{\mathrm{s}}^2$. This is a toy linear drag model, not a claim that every $1\mathrm{kg}$ falling object has this resistance law.

For a proof outline, subtract the equilibrium value: $u=v-v_T$ obeys $du/dt=-u/\tau$. The exponential decay solution and the initial condition give $u=-v_T{e}^{-t/\tau }$; one derivative gives the acceleration. Every time constant multiplies the remaining gap $v_T-v$ by $e^{-1}$. The exact limit is reached only as $t\to \infty$. This makes the model useful for estimating a settling duration without mistaking vanishing acceleration for vanishing velocity.

Newton already analyzed vertical motion under uniform gravity and resistance proportional to velocity in Book II, Section I, Proposition III of the Principia. His geometric construction includes a greatest downward velocity, illustrating that a limiting rate can emerge from balancing a driving force and resistance. ^[ref]

Mass per unit length of a filament or yarn. The tex unit equals one gram per kilometre, or $10^{-6}\mathrm{kg}{\mathrm{m}}^{-1}$.

A unit of mass per length, equal to $1g/km$. Force divided by a linear density stated in tex gives tensile tenacity in newtons per tex.

The original study combines optical light curves, companion-star radial velocities, rotational broadening and a distance constraint to infer the compact object's mass. The result $14.8\pm 1.0$ solar masses is the estimate from this 2011 analysis, rather than a claim about the latest mass determination.

Charge released per specified active or composite mass under complete stoichiometric utilization. For $n$ electrons per formula unit, active-material capacity is $nF/(3.6M)$ in mAh per gram.

A geometrically thin, optically thick disk has a radially varying effective temperature. Relativistic transport and an atmosphere model convert its thermal spectrum into an inferred emitting area and inner radius. Black-hole mass, distance, inclination and spectral hardening are important external constraints. Selecting thermal-state observations with moderate luminosity limits contamination and the breakdown of the thin-disk approximation. Relating the fitted inner radius to the innermost stable circular orbit remains a physical modeling assumption.

The three-dimensional translational momentum integral is $1/{\lambda }^3$ per unit volume, where

If nonrelativistic emitters of mass $m$ have rest-frame angular frequency ${\omega }_0$, then the one-dimensional Maxwell-Boltzmann velocity distribution produces a Gaussian line with standard deviationMeasuring the line width therefore determines the gas temperature when nonthermal broadening is negligible or separately modeled.

The thermomechanical effect couples a temperature difference to a pressure difference across a superleak. For the closed-chamber model in the apparatus figure, consider neighboring equilibrium states with small velocities, negligible height difference and pure helium-4. Let ${\mu }_m$ be chemical potential per unit mass, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. The Gibbs free energy differential per unit mass givesThe superleak allows the superfluid to redistribute until the two sides have equal ${\mu }_m$. Setting $d{\mu }_m=0$ therefore givesThe sign matters: heating one side at initially equal pressure lowers its ${\mu }_m$; inflow raises its pressure until chemical balance is restored. The hotter side has the higher pressure. The integral follows the constant-${\mu }_m$ path; it need not be evaluated at fixed pressure when material properties vary appreciably.

For an original local example, stipulate constant $\rho =150\mathrm{kg}{\mathrm{m}}^{-3}$, $s=100\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $T_h-T_c=0.020\mathrm{K}$. The predicted pressure difference is $300\mathrm{Pa}$. These are teaching assumptions, not helium measurements. This static thermodynamics relation predicts neither the required heater power nor the filling time: those require heat leakage, channel conductance and chamber heat capacity. An outlet would introduce kinetic energy and a flowing balance, so the static $300\mathrm{Pa}$ is not a predicted fountain height.

For a thick rectangular barrier with equal exterior potentials, constant mass and $0<E<U$, the exact matched transmission probability has the displayed limit when $\kappa a\gg 1$. The exponential is fixed by the evanescent action; the interface prefactor remains energy-dependent. A leading-exponential estimate can therefore differ by an order-one multiplier from the matched result.

The thin-film mass flux is the layer thickness times its depth-averaged velocity. In a quasi-steady transition region, mass conservation makes this flux approximately constant along the layer.

An ideal circular ring whose mass lies at a single radius $R$ from its symmetry axis. About that axis, the moment of inertia is $\int R^{2}dm=M{R}^2$. The result assumes negligible radial and axial thickness; it is different from the $M{R}^2/2$ of a uniform solid disk.

For the geometrical diffusion-capacitance analogy approximation $C=2r/\pi$, a mass law $m=\alpha r^3$ and constant ice supersaturation $s>0$, the growth law $dm/dt=4\pi Cs/F$ gives $d(r^2)/dt=16s/(3\alpha F)$. Integration yields the displayed time. Comparing to a liquid sphere must use equal mass rather than equal radius; evolving supersaturation, ventilation and surface attachment invalidate the constant-coefficient estimate.

For unit particle mass, this symmetric third-order tensor transports the pressure tensor in a kinetic fluid description. Its contraction gives the heat flux $q_i=Q_{ijj}/2$. It vanishes for a local Maxwellian, whose odd central velocity moments vanish, but it need not vanish for a general anisotropic or skewed phase-space distribution function.

A third-law interaction pair consists of the two forces belonging to the same mutual interaction, with their agent and recipient exchanged. The names action-reaction pair do not imply a delay: in the instantaneous model, the two forces occur together. Test a proposed pair by naming both bodies and the interaction, rather than merely noticing equal magnitudes and opposite directions.

For a book resting on a table, the table's upward normal force and Earth's downward gravitational weight both act on the book. They can balance in the book's free-body diagram, but they are not a third-law interaction pair. The partner of the table's normal force on the book is the book's downward contact force on the table. The partner of Earth's gravitational force on the book is the book's gravitational force on Earth. Their equality does not require equal accelerations: Earth's much greater mass makes its response correspondingly smaller for a given net force.

Similarly, friction on a sliding block from a surface is paired with friction on that surface from the block. For the collection containing only the block, that friction is an external force; including the contacting surface makes both members internal forces. Changing the collection changes the bookkeeping of momentum, not the physical interaction.

A physicist.^[ref]

The Thomson charge-to-mass experiment infers the magnitude of an electron's charge-to-mass ratio by combining electric and magnetic deflection of cathode rays. The central idea is to determine speed separately from magnetic curvature: a known speed turns the observed gyroradius into a measurement of $e/m$.

In 1897, J. J. Thomson used two grounded slits in an evacuated glass tube, then aluminium plates $5\mathrm{cm}$ long, $2\mathrm{cm}$ wide and $1.5\mathrm{cm}$ apart. A battery supplied their electric potential difference. Deflected cathode rays produced a phosphorescence patch on the glass; an external scale and movable luminous pointer recorded its position in darkness. Greater evacuation reduced gas electrical conductivity that screened the electric field; reversing plate polarity reversed deflection. External coils supplied the magnetic field, calibrated by induction against larger coils. He adjusted equal electric and magnetic deflections. ^[ref]

His comparison included air, hydrogen, carbon dioxide, and aluminium versus platinum cathodes, supporting a common carrier. He explicitly noted uncorrected magnetic bending outside the plates. His tables give $m/e$, the reciprocal of the magnitude used here. The inferred ratio alone could not separate electric charge from mass. ^[ref]

To connect this readout with the Lorentz force, derive a simple nonrelativistic model in modern SI notation. Let the uniform deflecting region have length $\ell$, followed by a field-free drift of length $L$. Assume a narrow speed distribution, small deflection angles, negligible collisions, and matching electric and magnetic field profiles. For an electron with speed $v$, the electric acceleration has magnitude $eE/m$, the transit time is approximately $\ell /v$, and the subsequent drift adds exit transverse velocity times $L/v$. Thus the displacement magnitudes measured separately with only electric field or only magnetic field areThe electric expression follows from constant transverse acceleration; the magnetic expression uses its small-deflection approximation. Equal displacement magnitudes then giveThis is an ideal derivation of the comparison principle, not a numerical reanalysis of the historical tube. Different exterior magnetic field profiles invalidate cancellation of the common geometrical factor. The comparison determines $e/m$ through deflection and the known apparatus dimensions; it does not require a downstream semicircle.

For a second, deliberately different teaching geometry, combine a velocity selector with a separate region containing only a uniform magnetic field $B_a$ perpendicular to the selected velocity. Here $B_s$ belongs to the velocity selector, while $B_a$ belongs to the analyzer: they need not be equal. The magnetic Lorentz force supplies the centripetal acceleration, soIf the analyzer turns the beam through a semicircle before it hits a screen in the entry plane, the separation $D$ between entrance and impact is the diameter: $r=D/2$. For a shorter observed arc, the screen displacement is not generally a radius; fit the appropriate circle geometry. This two-stage velocity selector and curvature apparatus is an original ideal teaching model, not a claim that Thomson's Fig. 2 used a separate selector and semicircular analyzer.

For a worked example, stipulate $E=2.00\times 10^3\mathrm{V}{\mathrm{m}}^{-1}$, $B_s=1.00\times 10^{-3}\mathrm{T}$, $B_a=2.00\times 10^{-4}\mathrm{T}$, and $D=0.1136\mathrm{m}$. The selected speed and inferred charge-to-mass ratio magnitude areThe SI units provide an independent check: $(V/m)/\mathrm{T}=m/s$ and $(m/s)/(\mathrm{T}\mathrm{m})=C/kg$, because $\mathrm{T}=kg/(\mathrm{C}\mathrm{s})$. The charge-to-mass ratio of the electron is negative; this calculation gives its magnitude. The bending direction gives the sign of the electric charge once the magnetic field direction is known. Scaling both $e$ and $m$ by the same positive factor preserves every displayed trajectory, so an independent electric charge measurement is required to determine mass separately.

The curvature formula assumes nonrelativistic momentum $p=mv$. At higher speeds, the relativistic cyclotron orbit gives $p=\gamma mv=e{B}_{a}r$, hence $e/m=\gamma v/(B_{a}r)$, where $\gamma$ is the Lorentz factor; the ideal straight-through selection $v=E/B_s$ still follows from vanishing Lorentz force. The stipulated speed is about $0.0067$ times the speed of light, making $\gamma -1$ about $2.2\times 10^{-5}$. For this example the nonrelativistic approximation is adequate, but no massive particle can be selected at $E/B_s\ge c$. Finite slit widths, collisions with residual gas, uncertain magnetic field calibration, and fields outside the ideal boundaries limit a real inference; repeat at several selected speeds and analyzer strengths to test consistency of $e/m$ rather than trusting one spot position.

A trajectory whose dynamics require gravitational forces from more than one attracting body. Around a synchronously rotating moon, a maintained trajectory involving the planet–moon system is distinct from a simple moon-centred stationary Kepler orbit; the moon's Hill sphere limits the latter approximation.