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?
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?
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