What is the Tension in the String Holding a Submerged Cork?

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SUMMARY

The tension in the string holding a submerged cork at a depth of 0.713786 m is calculated using the principles of buoyancy and weight displacement. Given the cork's density of 157 kg/m³ and volume of 3 cm³, the weight of the displaced water is determined to be 294 N, while the weight of the cork is 46.158 N. The resulting tension in the string is 247.842 N, highlighting a common error in unit conversion where 3 cm³ is misinterpreted as 0.03 m³ instead of the correct 0.000003 m³.

PREREQUISITES
  • Understanding of buoyancy principles and Archimedes' principle
  • Knowledge of density calculations and unit conversions
  • Familiarity with basic physics equations, particularly d = m/v
  • Ability to perform calculations involving gravitational force (weight = mass × gravity)
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  • Review buoyancy and Archimedes' principle in fluid mechanics
  • Study unit conversion techniques, especially for volume and density
  • Learn about the implications of density in submerged objects
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Students studying physics, particularly those focusing on fluid mechanics, as well as educators looking for examples of buoyancy and tension calculations in submerged objects.

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Homework Statement


A cork is held at the bottom of a bucket of
water by a piece of string. The actual depth
of the cork is 0.713786 m below the surface of
the water.

If the density of the cork is 157 kg/m3
and the volume of the cork is 3 cm3
, then what is the tension in the string? The acceleration
of gravity is 9.8 m/s
2
. Assume the density of
water is 1000 kg/m3
.


Homework Equations


d= m/v Fb= weight of displaced fluid


The Attempt at a Solution


The tension should be the force required to keep it in equilibrium...
using d=m/v, into dv = m (1000)(0.03) = 30 kg (mass of displaced water)(9.8)
weight of displaced water = Fb = 294 N
using dv = m again for the cork. (157)(0.03) = 4.71 kg (9.8) = 46.158 N

294 - 46.158 = 247.842 N .
Its a multiple choice answer and the choices are either that number but the decimals in the wrong place (all the answers are smaller), or 0.0123921 N.
maybe I'm supposed to use the depth and P=pgh somehow if that other numbers right, but I'm not sure how it would apply to this problem.
 
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Units, Units, Units!

3 cm^3 is not the same as 0.03 m^3

You are essentially saying that 3 cc of water has a mass of 30 kg!
 
that makes sense... so it would be like (0.03)^3 then i guess
 
Looks like the problem setter didn't think of this particular mistake, or there would have been a choice to match.
 

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