Calculating the Volume of Water Displaced by a Kayak

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The problem involves calculating the volume of water displaced by a 25kg kayak with a total volume of 0.17m^3. According to Archimedes' principle, the buoyant force must equal the weight of the displaced water for the kayak to float. The weight of the displaced water is 170kg, which corresponds to a volume of 0.025m^3. The calculations confirm that 0.025m^3 is indeed the correct volume of water displaced. Therefore, the answer is validated as accurate.
psruler
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Hi, I need help on this problem:

You own a 25kg plastic ocean kayak with a sealed interior filled with air. THe kayak has a total volume of 0.17m^3. The mass of water the kayak displace is 170kg. WHile floating, what volume of water does the kayak displace?

Thanks!
 
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For the kayak to float, the buoyant force must equal its weight. From Archimedes' principle, the buoyant force equals the weight of the displaced water.
 
I got 0.025m^3 as the answer. Is that correct?
 
psruler said:
I got 0.025m^3 as the answer. Is that correct?
Looks right to me.
 
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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