An air bubble arises from the bottom of the sea. Find its acceleration if the resistance force is proportional to $\rho$*A*$v$ where $\rho$ is density of water, A is cross section area and $v$ is velocity of the bubble. Density and temperature of water are constant.
There was also a part a) of this exercise, where there was no resistance force, and i used Newton's law:
$$ma=\rho Vg -mg$$
and
$$pV=nRT$$
and obtained $a= g(\dfrac{\rho RT}{M(p_a+\rho gh)}-1)$, which means acceleration depends only of the depth $h$. But in the case with the resistance force I write:
$$ma=\rho Vg -mg-k\rho Av$$
and after I plug in $a=\frac{dv}{dt}$ and solve the differential equation i get
$$a=\dfrac{e^{\dfrac{-t}{p_a+\rho gh}}}{M(p_a+\rho gh)kART}$$
So, my question is, can I have $t$ and A in the final solution, am I doing this correct? Every advice would be helpful
I think the same as the other person who replied to you. You cannot solve this exercise since the value of the cross section is unknown. (You can't relate it to volume since the height of the bubble is unknown.) Anyway, a negative exponential is the typical behavior of viscous friction, but it didn't take into account the fact that v = reh / dt.