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May 17

Mass and Weight

What is the difference between mass and weight?

Inkonvin (talk) 09:07, 17 May 2020 (UTC)[reply]

Mass is a property of the body. Whether you stand on the equator of the Earth or on the pole, whether you standard on the Moon or visit the Space Station, your mass remains the same (ignoring the size of you r last several meals). Weight is a force the body pushes the ground below it, equal to the force of gravity. That depends on how big and how massive the planet is, if you stand on it - hence your weight on the Moon would be about a sixth part of your weight on the Earth, and your weight on an orbiter would be zero. --CiaPan (talk) 13:43, 17 May 2020 (UTC)[reply]

Erm, this isn't really a math question. –Deacon Vorbis (carbon • videos) 13:56, 17 May 2020 (UTC)[reply]

Our articles: Mass, Weight, Mass versus weight. -- ToE 14:39, 17 May 2020 (UTC)[reply]

Mass is a measure of how much stuff there is, weight is a measure of how much the local gravitational field interacts with that stuff. 2A01:E34:EF5E:4640:F50A:AD79:5008:239B (talk) 15:32, 19 May 2020 (UTC)[reply]

number of unordered items in a permutation

let's say I have an almost ordered set,and I count the number of pairs of items that are ordered correctly. And I find out that it's n(n-1)/2-5n. what's number of misplaced items? --Exx8 (talk) 13:49, 17 May 2020 (UTC)[reply]

Unless I'm misunderstanding the question, knowing the number of pairs of items that are correct relative to each other isn't enough information to determine the number in the right overall place. For example, even in a list of just 3 items, consider the two possibilities: (1,3,2) and (3,2,1). In both, there's exactly one item in the correct place, but the first has 2 pairs correct relative to each other ((1,3) and (1,2)), while the second has no pairs correct relative to each other. –Deacon Vorbis (carbon • videos) 14:34, 17 May 2020 (UTC)[reply]

The pairs that aren't ordered correctly are called inversions. For n≤10, n(n-1)/2-5n is negative, n would have to be at least 11. If n=11 then the number of inversions is 0 n(n-1)/2 and all the items are in ~~the correct place~~ reversed order, but, as DV pointed out, there's not much else you can say. --RDBury (talk) 17:13, 17 May 2020 (UTC)[reply]

RDB, I think you worded that backwards. n(n-1)/2-5n (=0 for n=11) was the number of correctly ordered pairs, not the number of inversions. -- ToE 17:26, 17 May 2020 (UTC)[reply]

Thanks, I made the appropriate (hopefully) corrections. The actual number of inversions is then 5n. --RDBury (talk) 00:48, 18 May 2020 (UTC)[reply]

Exx8, how are *you* defining misplaced? Are you calling every element in (5,1,2,3,4) misplaced? Or are you saying only one is, since if you discard the 5 the remaining terms are ordered? -- ToE 17:26, 17 May 2020 (UTC)[reply]

for example. in (1,5,2,3,4), 5 is misplaced regarding to 2,3,4. hence is misplaced by 3. thanks.--Exx8 (talk) 19:38, 19 May 2020 (UTC)[reply]

There are three inversions in that example. So out of the ten pairs, seven are correctly ordered. By the general understanding of "misplaced item",

\pi (i)\neq i

, there are four misplaced items. Does that correspond to *your* usage of "misplaced item"? -- ToE 20:04, 19 May 2020 (UTC)[reply]