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rhodium
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Hi everyone,
I hope you can help me out with this question as my exam is tomorow.
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?
[tex]\rho[/tex] = m/V
F of buoyancy = [tex]\rho[/tex] of fluid*V of fluid displaced*g = [tex]\rho[/tex] 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:
[tex]\rho[/tex] of fluid*h of fluid displaced = [tex]\rho[/tex] of object*h of object
Isolating for height of fluid displaced
height of fluid displaced = [tex]\rho[/tex] of object*h of object/[tex]\rho[/tex] of fluid
[tex]\rho[/tex] is density
so height above water level = (H of object) - (H of fluid displaced)
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 / ([tex]\pi[/tex] * radius^2)
= 11.76 cm
= 0.1176 m
Now, subbing into the buoyancy eqation:
height of fluid displaced = ([tex]\rho[/tex] of object) (h of object) / ([tex]\rho[/tex] 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.
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
[tex]\rho[/tex] = m/V
F of buoyancy = [tex]\rho[/tex] of fluid*V of fluid displaced*g = [tex]\rho[/tex] 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:
[tex]\rho[/tex] of fluid*h of fluid displaced = [tex]\rho[/tex] of object*h of object
Isolating for height of fluid displaced
height of fluid displaced = [tex]\rho[/tex] of object*h of object/[tex]\rho[/tex] of fluid
[tex]\rho[/tex] 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 / ([tex]\pi[/tex] * radius^2)
= 11.76 cm
= 0.1176 m
Now, subbing into the buoyancy eqation:
height of fluid displaced = ([tex]\rho[/tex] of object) (h of object) / ([tex]\rho[/tex] 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.