What is the mass of the sphere?

  • Thread starter lollypop
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hello everybody:

A hollow, plastic sphere is held below the surface of a freshwater lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.700 M^3 and the tension in the cord is 930 N.

Calculate the buoyant force exerted by the water on the sphere. Take the density of water to be 1000 kg/m^3 and the free fall acceleration to be 9.80 m/s^2.
**for this I set up Bouyant = density *Volume*gravity = 6860 N

What is the mass of the sphere? Take the density of water to be 1000kg/m^3 and the free fall acceleration to be 9.80m/s^2 .
**here I used Buoyant = mg-T and solved for m, so answer for m= 605 kg

The cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged? Express your answer as a percentage.
** here is my problem :eek: , so far all I have is that the sphere is at rest so there is no external force acting on it, so F=B+T+(-mg) =0 right?? I don't know what else to use for this part. Any clues??
 

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lollypop said:
**for this I set up Bouyant = density *Volume*gravity = 6860 N
Correct.

lollypop said:
**here I used Buoyant = mg-T and solved for m, so answer for m= 605 kg
Not so sure about that one. Shouldn't be it:
[tex]B = mg + T[/tex]
Since both mg and T act in the same direction, downward? The cord isn't pushing the sphere up, it is pulling it down.

lollypop said:
The cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged? Express your answer as a percentage.
** here is my problem :eek: , so far all I have is that the sphere is at rest so there is no external force acting on it, so F=B+T+(-mg) =0 right?? I don't know what else to use for this part. Any clues??
For the sphere to be at rest, mg must equal B. There is no tension anymore, since the cord was broken. You need to express B as a function of the volume of the sphere that is still submerged, and find it.

[tex]mg = B = \rho V' g[/tex]
[tex]V' = \frac{m}{\rho }[/tex]
 
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