Buoyance - pressure on brick immersed in water

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The pressure on a brick immersed in water is greatest at the bottom due to the increased depth, which results in higher pressure according to the principles of fluid mechanics. The pressure exerted on the brick is not uniform; it varies with depth, increasing as one moves deeper. Therefore, the correct answer to the homework question is that the pressure is greatest on the bottom of the brick. This understanding is crucial for solving related physics problems involving buoyancy and pressure in fluids. Overall, depth is the key factor influencing pressure on submerged objects.
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Homework Statement



Consider a brick that is totally immersed in water. The long edge of the brick is vertical. The pressure on the brick is

A)greatest on the sides of the brick.
B)greatest on the bottom of the brick.
C)greatest on the face with largest area.
D)greatest on the top of the brick.
E)the same on all surfaces of the brick.


Homework Equations



fb=pvg


The Attempt at a Solution



i am guessing that it would be the same on all surfaces of the brick.
 
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What determines the pressure at a given point under the water surface?
 


depth
 


bigboss said:
depth
Exactly! Pressure depends on the depth beneath the surface. Use that to revise your answer.
 


so it would be the bottom of the brick, because it has the most depth, thus the most pressure
 


bigboss said:
so it would be the bottom of the brick, because it has the most depth, thus the most pressure
Correct.
 
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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