What is the density of a block 35% submerged in a liquid?

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To determine the density of a block that is 35% submerged in a liquid with a density of 2.5 g/cm³, the calculation involves using the principle of buoyancy. The assumption is that if the block were fully submerged, its density would equal that of the liquid. By setting up a proportion based on the submerged volume, the calculation yields a density of 0.875 g/cm³ for the block. This approach appears to be correct based on the provided information. The conclusion confirms that the calculated density aligns with the expected outcomes of buoyancy principles.
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Homework Statement



if I have a block of unknown density and if it is 35% submerged in a liquid of density 2.5g/cm^3
what is the density of the block

Homework Equations


d=m/v

The Attempt at a Solution


I thought that if the block was 100% submerged just under the surface, its density would equal that of the liquid.
Since it is 35% submerged, i thought i would approach it this way.
2.5/100=x/35
87.5=100x
x=0.875g/cm^3
Is that correct?
 
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That looks good to me :smile:
 
Thanks.
 
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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