Buoyancy and Density Archimedes

AI Thread Summary
The discussion revolves around solving buoyancy and density problems related to a 300kg object placed on ice. For the first question, the volume of ice needed to keep the object above water is calculated as 3.75 m³. The second question remains unanswered, while the third question indicates that 6% of the object remains above water when the ice melts, based on the object's density of 0.94 g/cm³. The calculations confirm that the volume of the object above water is approximately 0.01914 m³. The final response reassures that the calculations for question three are indeed correct.
Koborl
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Homework Statement



1. a 300kg object is placed upon a block of ice what volume of ice is needed to keep the object fully above water.

2. if the object density is .94gcm ^-3 what volume of the object remains above water on ice half the size.

3. What volume of the object remains above water when the ice is fully melted.

p(water) = 1gcm^-3
p(ice) = 0.92 gcm^-3

Homework Equations

B = mfg
V = m/p

The Attempt at a Solution



1.
v= volume of water displaced

(denstiy of water)1000 kg/m^3* v = 300(weight of object)+ (denstiy ice)920 kg/m^3 *v
80 kg/m^3 v = 300
v = 3.75 m^3

2. ?

3. 6% of the object remains above water because buoyancy is .94 and water is 1?

V = m/p

p = 0.94 g/cm3
M = 300 kg = 300000g
V = 0.31914893617021 m^3

V above water = .01914m^3

I'm fairly sure I've got question 3 all wrong though :(
 
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Please respond.
 
Koborl said:
p = 0.94 g/cm3
M = 300 kg = 300000g
V = 0.31914893617021 m^3

V above water = .01914m^3

I'm fairly sure I've got question 3 all wrong though :(

No, it is correct. V(immersed)ρ(water)=mass of object

ehild
 
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