What is the length of the can above the water level?

AI Thread Summary
The discussion revolves around calculating the height of a soda can above water when it is floating half full. The can's dimensions and mass are provided, leading to the use of buoyancy equations to determine the height of fluid displaced. After performing calculations, the height above the water level is found to be approximately 5.88 cm, though an error is noted, and the final answer is stated as 5.2 cm. Additionally, another user seeks help with a fluid dynamics problem involving a cube immersed in fluid, asking for assistance with density and resultant up-thrust calculations. The conversation highlights the importance of understanding buoyancy and fluid properties for solving related physics problems.
rhodium
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Hi everyone,

I hope you can help me out with this question as my exam is tomorow.

Homework Statement



A 355 mL soda can with diameter=6.2 cm has a mass of 20g. If it is half full with water, and it is floating upright on water, what length of the can is abover the water level?


Homework Equations


\rho = m/V

F of buoyancy = \rho of fluid*V of fluid displaced*g = \rho of object*V of object*g

Since V= Area*height, and the area is the same, then the Area cancel out and we are left with:
\rho of fluid*h of fluid displaced = \rho of object*h of object

Isolating for height of fluid displaced

height of fluid displaced = \rho of object*h of object/\rho of fluid


\rho is density


so height above water level = (H of object) - (H of fluid displaced)

The Attempt at a Solution



m of object = 0.02 kg + (355 mL/2)(1 g/1 mL)
= 177.52 g

density of object = m of object/ volume
= 0.17752 kg / 0.000355 m^3
= 500 kg/m^3

h =V / (\pi * radius^2)
= 11.76 cm
= 0.1176 m

Now, subbing into the buoyancy eqation:

height of fluid displaced = (\rho of object) (h of object) / (\rho of fluid)
= (50)(0.1176 )/1000
= 0.0588 m

height above water level = (H of object) - (H of fluid displaced)
= 0.1176 -0.0588
= 5.88 cm

The answer is 5.2 cm.
 
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rhodium said:
m of object = 0.02 kg + (355 mL/2)(1 g/1 mL)
= 177.52 g
Recheck your arithmetic.
 
hey,

God...Thank you sooo much. It took me a while to notice it actually. :)
 
i got a similar problem with fluid dynamics , i got a test later in the week haven't been prepped for it at all just been handed some sheets wondering if anyone could help with an example i have got , as it will help me see what i need to do for future reference
thanks in advance




QUESTION

A cube of 0.25m length sides which has a weight of 50N is immersed in a tank of fluid .

If the specific gravity of teh fluid is 0.8, calculate
a) the density of the cube material
b) the density of the fluid
c) the resultant up-thrust of the cube


ATTEMPTS

w=mg
50=mg
50/0.8 = m
m = 62.5 kg

p=m/v
p=62.5/0.25x0.25x0.25
p=4000 kg/m3




Thats all i can manage i haven't really been taught this yet so any help would be appreciated
 
Hey,

I'll try to help with what I know.

It might help you to know this:

specific gravity of a fluid = density of fluid/density of water.
 
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