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Tom59593
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First off, thanks in advance to anyone who can help at all. I've attempted to do a few things, and none of them seem to work...let me know what you think!
THANK YOU!
A sperical bubble of radius r has a volume v and a pressure p when just below the surface of a liquid of density d (top of bubble is tangent to the water level).
Let F1 be the magnitude of the buoyant force on the bubble when it is at this location.. Assume P is the atmospheric pressure and that the temperatureof the liquid and bubble do not change with depth.
Let the y-axis be vertical. Let the x-axis be the water level. The bubble (though it doesn't matter) is on the positive side of the x-axis).
a) Show that the magnitude of the buoyant force (F2) on the bubble as a function of its vertical position y in the liquid is: F2 = F1 [ p / (p - d*g*y) ]
b) If the liquid is water, show that the buoyant force is half the surface value at a depth of 10.3m.
Buoyant Force = (density of liquid)(volume displaced)(gravity) = d*v*g
a) We know that as the bubble descends in depth, it is moving in the negative-y direction. We also know that:
Buoyant Force = d*v*g = d*(cross-sectional area)(depth)*g = d*A*(-y)*g
What I don't understand, is why you would need to go any further, given the information that we have. However, the question has been asked so I must solve it, right : P
If it helps at all, this question is in a chapter about the Ideal Gas Law (PV = nRT)
My next best attempt was the following:
we know that F2 = d*A*(-y)*g and that (by Ideal Gas Law) v = nRT/p
so,
F2 = d*A*(-y)*g = d*v*g
F2 = d*A*(-y)*g = d*(nRT/p)*g (this gets P in the denominator)
F2 = A*(-y) = nRT/p
F2 = (nRT/p)/(A*(-y)
F2 = (nRT)/[A*(-y)*(p)]
F2 = (nRT)/[(pi*r^2)(-y)(p)]
And that's as far as I get before I realize I'm really not getting anywhere!
There must be an equation or formula or form that I am missing somewhere, because this is really not even close to the form we are asked to show...and I am positive if I had this "missing equation" I could finish this...
THANKS IN ADVANCE!
-Tom-
THANK YOU!
Homework Statement
A sperical bubble of radius r has a volume v and a pressure p when just below the surface of a liquid of density d (top of bubble is tangent to the water level).
Let F1 be the magnitude of the buoyant force on the bubble when it is at this location.. Assume P is the atmospheric pressure and that the temperatureof the liquid and bubble do not change with depth.
Let the y-axis be vertical. Let the x-axis be the water level. The bubble (though it doesn't matter) is on the positive side of the x-axis).
a) Show that the magnitude of the buoyant force (F2) on the bubble as a function of its vertical position y in the liquid is: F2 = F1 [ p / (p - d*g*y) ]
b) If the liquid is water, show that the buoyant force is half the surface value at a depth of 10.3m.
Homework Equations
Buoyant Force = (density of liquid)(volume displaced)(gravity) = d*v*g
The Attempt at a Solution
a) We know that as the bubble descends in depth, it is moving in the negative-y direction. We also know that:
Buoyant Force = d*v*g = d*(cross-sectional area)(depth)*g = d*A*(-y)*g
What I don't understand, is why you would need to go any further, given the information that we have. However, the question has been asked so I must solve it, right : P
If it helps at all, this question is in a chapter about the Ideal Gas Law (PV = nRT)
My next best attempt was the following:
we know that F2 = d*A*(-y)*g and that (by Ideal Gas Law) v = nRT/p
so,
F2 = d*A*(-y)*g = d*v*g
F2 = d*A*(-y)*g = d*(nRT/p)*g (this gets P in the denominator)
F2 = A*(-y) = nRT/p
F2 = (nRT/p)/(A*(-y)
F2 = (nRT)/[A*(-y)*(p)]
F2 = (nRT)/[(pi*r^2)(-y)(p)]
And that's as far as I get before I realize I'm really not getting anywhere!
There must be an equation or formula or form that I am missing somewhere, because this is really not even close to the form we are asked to show...and I am positive if I had this "missing equation" I could finish this...
THANKS IN ADVANCE!
-Tom-