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Pressure, Buoyant Force problem 
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#1
Nov1311, 04:07 PM

P: 35

1. The problem statement, all variables and given/known data
A cube of metal(density=6.00 kg/dm^3) has a cavity inside it. It weighs 2.40 times as much in air as it does when completely submerged in water. What fraction of the cube's volume is the cavity? 2. Relevant equations Density=m/v Fb=Density of fluid times volume of fluid times g F=mg 3. The attempt at a solution I first made two equations: Density=m/v (for water) Density=m/v 6.00 kg/dm^3=m/1000 kg 6.00 kg/dm^3/2.40m/V Idk what 2 do from there because I think I set the equations up wrong could someone please explain 2 me what I should do cuz what's really bugging me is the amount of information that is given to me? 


#2
Nov1411, 07:11 AM

HW Helper
P: 3,440

The information is enough to work out the answer (as long as you also know the density of water). And you could either assume the density of air is approximately zero or you could use the actual value, to get a more accurate answer.
To start this question, you should use what they give you. They are saying that the apparent weight of the cube is 2.4 times as much in air than in water. And you know the equation for the apparent weight. So you can use this to find the actual weight of the cube in terms of the volume of the cube. And then you can use this along with the equation for the actual weight of the cube due to the masses contained, to find the fraction of the cube that is hollow. 


#3
Nov1411, 08:48 PM

P: 35

I still kinda dont get what ur saying can u show me ur work because I dervived Fw=Density of object times g times V and I made this equation equal to Fa plus 9800 times volume of the fluid and I really need help please



#4
Nov1511, 09:07 AM

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P: 3,440

Pressure, Buoyant Force problem
So you wrote:
[tex]Weight_{object} = F_{apparent} + 9800 V[/tex] Is this the apparent weight when its underwater? what does the 9800 mean? Maybe I should go through it stepbystep, since there are a few steps which could get confusing when written all in one paragraph. So first, you need the equation for the actual weight of the object. the object has two parts. an air cavity and a metal shell. I think you should write an equation for the total weight of the object in terms of the volumes of the metal part and cavity part and the density of air and metal. 


#5
Nov1611, 05:12 PM

P: 35




#6
Nov1611, 07:30 PM

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P: 3,440

And just use symbols for now. (I find it is easier to use symbols in calculations until you get to the end of the problem). So you could use something like [itex]V_m[/itex] for the volume taken up by the metal (for example). 


#7
Nov1911, 12:58 PM

P: 35

Fa=FwFb Fa=m(object)g(density of fluid)(volume of fluid)(g) Fa=(Density of the object)(Volume of the object)(g)(Density of fluid)(volume of the fluid)(g) Fa=g((density of object)(volume of the object)(density of fluid)(volume of fluid)) m(a)=(density of object)(volume of object) minus (density of fluid)(volume of fluid) m(a)=(6000 kg/m^3)(V of object) minus (1000 kg/m^3)(Volume of fluid) Idk what 2 do from there but the only other thing Ik is that since its submerged the volume of the fluid is equal to the volume of the object and the apparent mass=m/2.40 but idk what 2 do from there please show me the steps and the math out :) 


#8
Nov2111, 05:42 PM

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P: 3,440




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