Recent content by EEristavi

Thanks all

It would be impossible to use first order approximation - so answer will be "my formula"?

##\Delta T## or ##\beta? ## Because as I think ##\Delta T## Can me "big" Also, Can you give me some direction: where I can look it up - Why I can discard that value in denominator? I know I will find it eventually, but I would appreciate it if you could help me spare some time P.S. I kind of...

##\Delta V = \beta V \Delta T####\rho _2 = \frac m {V + \Delta V} = \frac { \rho V} {V(1+\beta \Delta T)} = \frac \rho {1+\beta \Delta T}## ##\Delta \rho = \rho _2 - \rho = \frac \rho {1+\beta \Delta T} - \rho = \rho (\frac 1 {1+\beta \Delta T} - 1) = \rho \frac {-\beta \Delta T} {1+\beta \Delta...

when you pointed out in that way (step by step) - light bulb just flashed on :) Thanks mate! P.S. I was writing other question other day - forgot to change the title, sorry

I know that I have to calculate force that is generated by water. However, I do it wrong way. ## A = \pi r^2 ## ## v = \frac Q A = \frac Q {\pi r^2}## ## F = P t = mvt = ...? ## - have problem with mass. and is it even correct way? if I can calculate F - its easy afterwards... ##\tau = F l ##...

@kuruman @PhanthomJay Thank you for your answers and efforts. Your last comments really summarized whole idea and showed me my mistake. I can say - I understood the concept Thanks again very much to every person spending time and trying his best! I guess I have no answer and we can say -...

It is not Stated. However, my guess is that - "It's gently lowered by the had until it reaches the equilibrium"

Yes of course. However, As I understand, Bernouli is used when we have different densities (to "create" "upward" force).

Nothing helpful... Yes endeed, I need another equation/approach. However, I can't figure out which one :/

$$ H = \frac { V^2 - V_0^2 sin \Theta} {-2g} $$ $$ H = \frac {V_0^2 sin \Theta} {2g} $$ So, I need to calculate ## V_0 ## I'm thinking about pressure. $$ P = \rho g \Delta h $$ $$ \Delta h = h - L sin \Theta $$ $$ F_A = P S_A $$ $$ F_A = P S_B $$ Dead End here...

point A - where spring isn't stretched yet. Point B - where spring is stretched by ## \Delta x ## It moves because of the gravity. other force that is involved is force from spring F = k x if we take "starting point or 0 (zero)" to be point A: Mechanical Energy at point A - 0; Mechanical...

I have 2 Questions: 1. is mechanical energy conserved? 2. If it's not, why?Note: As I see it's not conserved. Am I right?

Guys I understood it mathematically. However, I can't figure it out as "Change of energy concept" - If the change of potential energy is ## mg \Delta x##, why the spring's potential energy isn't increased the same amount. As I understand we have a closed system (Or Isn't it?...). This is the...

Woow... now I get it! Very nice! Now I have to think what's the physics behind it (this I will manage on my own).. :D Thank you very much!