What is the volume of the bubble

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The volume of an air bubble released from a submarine at a depth of 115 m is affected by pressure changes as it rises to the surface. At this depth, the pressure is calculated to be approximately 12.4 atm, while the pressure at the surface is 1 atm. Using Boyle's Law, which states that p1v1 = p2v2, the volume of the bubble can be determined. As the bubble ascends, its volume increases due to the decrease in pressure. The final volume of the bubble when it reaches the surface can be calculated accordingly.
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An air bubble has a volume of 1.50 cm3 when it is released by a submarine 115 m below the surface of a lake. What is the volume of the bubble when it reaches the surface? Assume that the temperature and the number of air molecules in the bubble remains constant during its ascent.

I'm guessing that the depth of the bubble effects the pressure (that being p1) and then the pressure at the surface is one atm? How can I calculate the pressure based on its depth?

Thanks in advance for any hints??
 
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Boyle's Law
 
so p1v1=p2v2? i figured out p1 to be 12.4 atm based on the 115 m depth. would p2 be 1 atm?
 
uhm remember the pressure at the bottom has an increment of row g h i think... been a while since i last did this, but just look into that ;-p
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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