Timeline for Why cant this situation is possible when we are considering a dimensional analysis approach in any problem?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Mar 16, 2022 at 0:22 | vote | accept | ProblemDestroyer | ||
| Mar 15, 2022 at 21:45 | comment | added | hdhondt | "full formula generation not just the dimensional formula" - You seem to be changing the question. The dimensional formula is not the same as the actual one. | |
| Mar 15, 2022 at 13:32 | comment | added | ProblemDestroyer | I am asking for actual full formula generation not just the dimensional formula | |
| Mar 15, 2022 at 13:31 | comment | added | ProblemDestroyer | Sir i am saying we know the dimensions (aT) from there we generate a formula for it suppose kaT , but this is not having the formula of kaT+alphaaT^2 ,,, kaT+alphaa^2T^2 ...any of them in cosideration for a actual formula | |
| Mar 15, 2022 at 11:11 | comment | added | hdhondt | It's all the same dimension. As I said in an earlier comment, $v=aT^2+bt^2+CT^2$ reduces to $v=pT^2$, where $p=a+b+c$ | |
| Mar 15, 2022 at 3:41 | comment | added | ProblemDestroyer | Yeah dimension is same , i mean this formula we cant get from dimensional analysis isnt? | |
| Mar 15, 2022 at 3:00 | comment | added | hdhondt | But, $T^2\times T^{-1}=T$ so the entire sum has dimension T | |
| Mar 14, 2022 at 23:32 | comment | added | ProblemDestroyer | ...this one depends on T^2 and T both while the dimensional analysis one gave depends on just T. | |
| Mar 14, 2022 at 23:31 | comment | added | ProblemDestroyer | Consider v(speed) is a just dependent on two quantities = tangential acc(a) and time(T) , by dimensional analysis we arrived at the form of it to be aT so it would be $v=kaT$ , where k is dimensionless quantity , but why this formula is not possible ? This formula : v = $kaT+(alpha)aT^2 + (beta)a^2T^2$ , where alpha and beta are constants having dimension of T^-1 , T^-1a^-1 respectively . Now we can observe that we can only factor out aT common and get aT[k+(alpha)T+(beta)aT] so its a function which is not told by dimensional analysis as... | |
| Mar 14, 2022 at 22:43 | comment | added | hdhondt | Not sure I understand... However, if "overall dimension of each term matches v", then you have something like $v=aT^2+bT^2+cT^2$, which reduces to $v=pT^2$ | |
| Mar 14, 2022 at 22:36 | comment | added | ProblemDestroyer | a,b,c have dimensions such that overall dimension of each term matches v | |
| Mar 14, 2022 at 22:35 | comment | added | ProblemDestroyer | I mean to say cant we have formulas like v= aT+bT^2 +cT^3 ? dimensional analysis will just tells its dimensional formula that is of form v=a'T but how exact like the above? | |
| Mar 14, 2022 at 22:29 | comment | added | hdhondt | That still doesn't alter the fact that a factored dimension is a different thing. $1/m$ is not a length and $1/m^2$ is not an area. | |
| Mar 14, 2022 at 11:52 | comment | added | ProblemDestroyer | a and b has dimensions of E^-1 , E^-2 respectively , i meant hdhondt | |
| Mar 14, 2022 at 9:13 | history | edited | hdhondt | CC BY-SA 4.0 |
added 37 characters in body
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| Mar 14, 2022 at 9:07 | history | answered | hdhondt | CC BY-SA 4.0 |