Recent content by Gavran

https://en.wikipedia.org/wiki/Siphon#Maximum_height

True. P2-P1=ρgh holds when v1=v2, and as you said in the original post, v1 can not be equal to v2, except in the case when v1=0 and v2=0.

Bernoulli's equation P+ρgh+(1/2)ρv2=const is the logic for analyzing statements (ii) i (iii). The statement (ii): P1+ρgh1+(1/2)ρv12=P2+ρgh2+(1/2)ρv22 Is it possible to have the next situation P2-P1=ρgh1-ρgh2=ρgh? The statement (iii): P1+ρgh1+(1/2)ρv12=P2+ρgh2+(1/2)ρv22...

If b, p and q are integers then from the post #9 we have: 70=qa0 ⇒ q∈{1,2,5,7,10,14,35,70} 27=pa9 ⇒ p∈{1,3,9,27}. 1/27 is the smallest value of q/p and 70/1 is the biggest value of q/p among all possible values of q/p. The function f(y)=27y+70(1/y) from the post #3 strictly increases for...

You are right. There are the local maximum and the local minimum. In my opinion, the question only holds for q/p<0.

Try to express b as a function f(y) where y=q/p and f(y)=27y+70(1/y). We can speak about the greatest value of b for q/p<0. If q/p>0, then we can speak about the smallest value of b.

Generally, yes. But based on the original post, it is obvious that the vector space does not vary, so here we have only a single vector space which is enough for modelling forces.

The only important thing is that every vector (force) must belong to the same vector space, so every vector in the vector space can be expressed as a linear combination of the vectors in the basis of the vector space.

For more complicated problems than this one in the original post the formula is $$ N=\frac1n\sum_{d|gcd(n_1,n_2,...,n_k)}^{}\phi(d)\binom{n/d}{n_1/d,n_2/d,...,n_k/d} $$ where: ## n ## - the number of balls, ## k ## - the number of colors, ## n_i ## - the number of balls of the i-th color, ##...

You do not need any formula to solve this problem. With two blue balls and two red balls you have two cases BBRR and BRBR. How many different options do you have when you insert the yellow ball into BBRR? How many different options do you have when you insert the yellow ball into BRBR?

The first statement of the theorem was published in 1887 by Georg Cantor. If M and N are two such sets that components M' and N' can be separated from them, of which it can be shown that |M|=|N'| and |M'|=|N|, then M and N are equivalent sets.

The answer to the post #8: Yes, it does.

The Axiom of Choice and the existence of a Hamel basis imply each other.

A ⇔ B The exclusion of A does not produce B is false.

It can be shown that if the Axiom of Choice is false, then there are vector spaces with no Hamel basis. The exclusion of the Axiom of Choice means the exclusion of the proof based on the Axiom of Choice and this does not mean there are vector spaces with no Hamel basis. Every vector space has a...