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| Question 1194859:  Assuming an alligator swallows stones with a mass 2.2 kg how much deeper, in millimeters, would it sink? Assume the alligator is initially floating and of 30% its body is above water and its body's surface area A= 1.2m^2; Use g=9.8 m/s^2, the p_water= 1,000 kg/m^3.
 *I'm not sure which section this truly belongs too, but I desperately need help because I don't understand this question at all.
 Answer by ikleyn(52879)
      (Show Source): 
You can put this solution on YOUR website! . Assuming an alligator swallows stones with a mass 2.2 kg how much deeper, in millimeters,
 would it sink? Assume the alligator is initially floating and of 30% its body
 is above water and its body's surface area A= 1.2m^2; Use g=9.8 m/s^2, the p_water= 1,000 kg/m^3.
 *I'm not sure which section this truly belongs too,
 but I desperately need help because I don't understand this question at all.
 ~~~~~~~~~~~~~~~~
 
 
 After reading this post,  my impression is
 that it is a  "joke"  entertainment problem.
 
 By supporting its joking attitude,  I will assume
 that the alligator is a creation of a cubic form.
 
 
 
 
Then from given surface area, I can find "the length" of the alligator: 
it is  L =  = 0.447 meters.
    +----------------------------------------------------------------------------------------------+
    |   From it, I conclude that                                                                   |
    |       1) the total volume of the alligator is  V =  =  = 0.0894 m^3,       |
    |      and                                                                                     |
    |       2) its body is  0.7*0.447 = 0.313 m under the water surface.                           |
    +----------------------------------------------------------------------------------------------+
0.7 of its volume is under water - - - hence, due to Archimedian law of floating, 
the mass of the alligator is  0.7*0.0894*1000 = 62.58 kilograms.
After swallowing stones with the mass 2.2 kg, the mass of the alligator becomes 62.58 + 2.2 = 64.78 kilograms.
Supporting the joking nature of the problem, we assume, naturally, that the dimensions of the cubical alligator
do not change: it is still a cube with the edge length of 0.447 meters.
Now its body is  = 0.324 meters under the water surface.
To find "how deep it would sink", take the difference 0.324 - 0.313 meters, and get the answer = 0.011 m = 11 millimeters. Solved.
 
 Yes, as I said at the very beginning, you may consider it as a joke problem and my post as a joke solution.
 
 The problem can be solved similarly by assuming that the alligator is a creature of spherical form (also as a joke).
 
 
 Amusingly ?
 
 
 - - - But,  who knows,  may be you will get a  Nobel prize or something equivalent for this solution.
 
 
 In such problems,  the achievement is not measured in accuracy of your calculations, but in the power of your imagination . . .
 
 
 
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