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

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Homework Help Overview

The discussion revolves around a physics problem involving buoyancy and fluid dynamics, specifically focusing on a floating soda can and the calculations related to the height of the can above the water level. The problem includes parameters such as the volume, mass, and density of the can and the fluid.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of buoyancy equations and the relationship between the densities of the object and the fluid. There are attempts to calculate the height of the fluid displaced and the height of the can above water level. Some participants also question arithmetic accuracy in the calculations.

Discussion Status

The discussion is ongoing, with some participants providing guidance and corrections to arithmetic errors. There is an exploration of similar problems in fluid dynamics, indicating a broader context of understanding needed for upcoming tests.

Contextual Notes

One participant mentions a lack of preparation for a related test, indicating that the discussion may also serve as a study aid for future reference. There are references to specific gravity and density calculations that are relevant to the problems being discussed.

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 the 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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