Asked by Tammy
An object floats with half of its volume beneath the surface of the water. The weight of the displaced water is 2000N. What is the weight of the object?
I'm confused by this problem. How would I go about solving this? My guess would be that the weight of the object is 2000N. Is this right?
I'm confused by this problem. How would I go about solving this? My guess would be that the weight of the object is 2000N. Is this right?
Answers
Answered by
drwls
Yes. They are trying to confuse you with extra information.
No matter how much of the object is submerged, as long as it is floating, the weight of the object equals the weight of the displaced fluid. That is Archimides Law (or Principle).
It should be a Law by now.
No matter how much of the object is submerged, as long as it is floating, the weight of the object equals the weight of the displaced fluid. That is Archimides Law (or Principle).
It should be a Law by now.
Answered by
uday
v*d*g=200*gravity
v*d=200
volume * density = mass
somass remain 200 or 2000n as said above
v*d=200
volume * density = mass
somass remain 200 or 2000n as said above
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