Understanding the Mass and Density of Empty Aluminum Cans

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The mass of a pile of empty aluminum cans cannot be calculated using the formula m = ρV with the density of aluminum because the cans are not solid; they contain air. While the density of aluminum is 2700 kg/m3, the average density of the pile is significantly lower due to the air-filled space within the cans. This means that the actual mass of the pile will be much less than 2700 kg for a volume of 1.0 m3. Understanding that the volume includes both the cans and the air is crucial for accurate mass calculations. Therefore, the presence of air within the cans must be taken into account when determining the overall density and mass.
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Please correct me if wrong, but to explain why the mass of the pile is not pV = (2700 kg/m3)(1.0 m3) = 2700 kg, is because a pile of empty cans, its volume would be more of a dimension vs a weight. please explain concept if someone can. Thank you

Question: A pile of empty aluminum cans has a volume of 1.0 m3. The density of aluminum is 2700 kg/m3. Explain why the mass of the pile is not "p(alumin)*V "= (2700 kg/m3)(1.0 m3) = 2700 kg. Used from the formula p=m/V
 
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The mass of some volume of stuff is m = \rho V, where \rho is the density of whatever stuff fills the volume. Only if it were solid aluminum, would \rho equal the density of aluminum. An empty can is mostly air, so the average density is much less than that of pure aluminum.
 
Thanks for the assistance, I overlooked the cans as having space within. I guess sometime the obvious to some may be difficult to see for others.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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