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PHYSICISTS OF NEWSCHOOLERS! I need your help...

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Edward.
Posts: 9366 - Karma: 445
An object is solid throughout. When the object is completely submerged in ethyl alcohol, its apparent weight is 15.9 N. When completely submerged in water, its apparent weight is 12.1 N. What is the volume of the object?

Here's a hint that my professor provides:
When and object is completely submerged in a fluid, its apparent weight is equal to its true weight minus the upward acting buoyant force: Click for larger image.
Write this equation for each fluid; once for the case that the object is submerged in ethyl alcohol and once for the case that it is submerged in water. The unknown volume V will appear in the expression for the buoyant force FB. Combine the two equations and solve for V.



Teach me how to solve it please!
Dec 2 2012 9:44PM
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Edward.
Posts: 9366 - Karma: 445
where it says "click for larger image" we are given this formula:
Apparent Weight = Actual Weight - Buoyant Force
Dec 2 2012 9:47PM
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Edward.
Posts: 9366 - Karma: 445
Oh and we are provided with the density of water and ethyl alcohol which is 1000 and 806 kg/m^3
Dec 2 2012 9:49PM
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lgski
Posts: 1832 - Karma: 2,291
i could've done this last year.. but i forgot everything i learned in physics. Im sorry to be no help
Dec 2 2012 9:51PM
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.Capn.
Posts: 1454 - Karma: 152
F_B = roh * V

F_B is the boyant force
roh is the density of the liquid the object is submerged in
V is the volume... What you're looking for.

Write out the two equations like he says. The "actual weight" is the same in each equation so solve each equation for it and set the two equal to each other. The only unknown left is the Volume. Solve for it!
Dec 2 2012 10:11PM
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Edward.
Posts: 9366 - Karma: 445
thank you!
Dec 2 2012 10:24PM
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JWoods
Posts: 2369 - Karma: 2,635
http://answers.yahoo.com/question/index?qid=20091124204820AA1xqZ2
Dec 2 2012 10:25PM
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tBatt
Posts: 9632 - Karma: 580
3/4 of my physics homework was done with ^.
The rest was with beer.


#college?
Dec 2 2012 10:26PM
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.Capn.
Posts: 1454 - Karma: 152
^Whoops, use that! I forgot gravity...
Dec 2 2012 10:28PM
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JWoods
Posts: 2369 - Karma: 2,635
You know it hahah. I <3 yahoo answers
Dec 2 2012 10:30PM
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MT_ripper
Posts: 85 - Karma: 9
I took physics 1 three years ago, but here is what I could find in my old notes.


Fb (water) = (density water)(volume)(gravity)

Fb (alcohol) = (density ethyl alcohol)(volume)(gravity)

SInce Wt = mg-Fb

we get: 15.9N = mg - (1000 kg/m^3)(V)(9.81m/s^2) - water

12.1 N = mg - (806kg/m^3)(V)(9.81m/s^2) - alcohol

subtract water - alcohol

we get: 3.8N = V(g)(Density of water - density of alcohol)

3.8N = V(9.81m/s^2)(1000 - 806)kg/m^3

V = 3.8N/(9.81m/s^2)(194kg/m^3)

V = 1.9967E-3 m^3






Dec 2 2012 10:43PM
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Edward.
Posts: 9366 - Karma: 445
Thanks everyone I just didn't know that the volume used to calculate the buoyant forces for each liquid was the same as the volume of the object
Dec 2 2012 10:47PM
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