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October 23

Double variable including time

When I use variables like speed "v", power "P", frequency "f", etc... they have this particularity that they include a time value, a time variable "t". How could I replace these single variables by a double variable, in such a way that when I have a relation between two variables of this type I can demonstrate in an equation that this relation is independent of time or not ? How could I représente them, (distance, t) or distance/T, or what else ? Malypaet (talk) 04:30, 23 October 2022 (UTC)[reply]

Such variables may occur in formulas in mathematical models of physical phenomena, such as

E = ½ mv 2

, or

P = Mω

. These variables themselves are just that, variables. They do not "include" a time value or a time variable. A variable representing a time interval may occur in a formula for speed, such as

v = d/t

, but this is not a necessity. The dimensions of units used to express the values of some quantities, including speed, power and frequency, include the base dimension T (for time). The dimension of speed, using the base physical dimensions of the International System of Units, is T⁻¹M, that of power is T⁻³L²M, and that of frequency is just T⁻¹.

If you confuse the concepts of "variable", "quantity" and "dimension", you won't get very far with your investigations. One can replace anything by anything in any formula. For example, in Einstein's formula

E = mc 2

one can replace

E

by

E 2

and

c 2

by

ω

, and obtain

E 2 = mω

. This formula may be beautiful, but it does not make sense. It is physically meaningless. Your question also does not make physical sense to me. --Lambiam 16:54, 23 October 2022 (UTC)[reply]

What are you trying to accomplish (understand/express/solve) actually? 16:42, 23 October 2022 (UTC) — Preceding unsigned comment added by Bumptump (talk • contribs)

Science

How do plants make their own food — Preceding unsigned comment added by 197.231.239.18 (talk) 15:54, 23 October 2022 (UTC)[reply]

Plants are autotrophs. They use the process of photosynthesis to process water, sunlight, and carbon dioxide into oxygen, and simple sugars (The plant uses thist as "food"). Bumptump (talk) 16:39, 23 October 2022 (UTC)[reply]

And of course, plants need more than sugars, so they have to extract nitrogen compounds, phosphorous compounds, sulphur compounds, sodium, potassium and various spore elements from their surroundings. Most plants use their roots for that, but some digest animals. Most plants cannot directly extract nitrogen from the air, but a few can, with help from some bacteria. PiusImpavidus (talk) 09:34, 24 October 2022 (UTC)[reply]

Speed of temperature change

In thermodynamics is there a quantity used to represent the speed of temperature change in "°K/s", as for mechanics where the symbol "v" is used in "m/s"? If so, what symbol is used to represent it? Failing that, can we use "T/t" to represent a thermal power in watts in the equation "kT/t"? Malypaet (talk) 16:33, 23 October 2022 (UTC)[reply]

Thermodynamics as such deals with equilibrium states and does not consider time - state changes are considered to occur infinitely slowly, which is a good approximation in many cases. Non-equilibrium thermodynamics is what you're looking for. I don't think there's a special symbol. Usually, the derivative

{\frac {dT}{dt}}

or shorter

{\dot {T}}

("T dot") will be used. A measurement over a finite time

\Delta t

would be written as

{\frac {\Delta T}{\Delta t}}

. --Wrongfilter (talk) 16:54, 23 October 2022 (UTC)[reply]

Sorry, but the thermal equilibrium of a body which emits radiative power can only take place if this body receives an equivalent power in exchange. So we are indeed dependent on the time there and if one of the two powers increases or decreases, the other will have to be able to do the same so that we can recover an equilibrium. So for the transition from one equilibrium to another there will be a change in temperature which will take some time, hence my question on the speed of change "T/t". Malypaet (talk) 21:36, 23 October 2022 (UTC)[reply]

So apply the terms from non-equilibrium thermodynamics, what's the problem? Incidentally, do you know what a derivative is? --Wrongfilter (talk) 21:47, 23 October 2022 (UTC)[reply]