Recent content by fer Mnaj

I don´t get that part, does it mean that water is "pushing" the air throughout tube 4? Because continuity equations have worked when we were talking about the same fluid

So i decided to add $$Q_2$$ and $$Q_3$$, so I would substract them from $$Q_1$$ later. I thought it would be like a $$Qnet$$, and that would be equal to $$(Atank) (vrise)$$. Does it make sense?

Indeed: $$A_1v_1=π(1.5m)^2(20m/s)$$ $$A_2v_2=π(1)^2(8m/s)$$ $$A_3v_3=π(1.25)^2(10m/s)$$ Then we can see that $$Q_1= 141.37 m^3/s$$ is way higher than $$Q_2= 25.13 m^3/s$$ and $$Q_3=49.08 m^3/s$$ So more water enters than exits. In that aspect makes sense. Now I was thinking to treat tube 2 and...

Hi, I´m quite lost and would appreciate guidance I have solved for 2 tubes using Bernoulli´s equation before, but now how does it change? Is it really going to rise water level inside? Why?

why?

##m (v^2/r)##

Not really, can you explain it?

Any idea? I am pretty lost

changed the statement, and you are right in the first thing you said. In this case we are talking about a central force, so no angular acceleration as you say. But we know that ##\vec{F} = d \vec{p}/dt ## so i hoped we could get the velocity by integrating

Im studyin mechanics, so I don't think is related directly with electromagnetism The torque in central force is zero. Ive corrected m, I was thinking in order to get velocity from that force I need to integrate it

Is it correct to say that that τ=0 since r has the same directacion as F?? and for \vec{L} que need to find \vec{p} So I thought solving this dif equation ## \int dp/dt =−kq/r^2 +β^4/r^5## Do you agree in the path I am going?

HI τ= r ˆr x - ##k / r ^ 2## ˆr= 0 right? since ˆr x ˆr is zero What about L?

I missed the integration constants... how to aply them with the initial conditions i gave aa the begining?

So I left it in: ##-\int1/w^2 dv=C_1\int \cos (\omega t) dt+ C_2\int\sin (\omega t) dt## now let ##\omega t=u## so ##du/dt=\omega## and ##dt=du/\omega## ##-1/w^2\int dv=C_1\int \cos (u) du/\omega## + ##C_2\int\sin (u) du/\omega## ##-1/w^2\int dv=C_1/\omega\int \cos (u) du/## +...

I also noticed it was the simple harmonic oscillator, but usually its solved from position to acceleration. Its taking me too much time to figure out in the opposite way