Pressure, Density, and Atmospheric Pressure

AI Thread Summary
To find the absolute pressure at a depth of 15 m in seawater, use the formula P = ρgh, where ρ is the density (1.03 g/cm³ converted to kg/m³), g is the acceleration due to gravity (10 m/s²), and h is the depth (15 m). Calculate the pressure due to the water column first, then add the atmospheric pressure of 101 kPa. The pressure from the seawater alone is approximately 153.5 kPa, leading to a total absolute pressure of about 254.5 kPa. Understanding that pressure depends on density, gravity, and height, rather than volume, is crucial for solving this problem. The final answer combines both the hydrostatic pressure and atmospheric pressure.
kallisti
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Homework Statement



What is the absolute (total) pressure at the depth of 15 m below the surface of the sea. Assume density of seawater is 1.03 g/cm3 and g=10 m/s2. Give answer in kPa. Use atmospheric pressure is 101 kPa.


Homework Equations



P = F/A
P1V1=P2V2


The Attempt at a Solution



i don't know how to solve it because neither of the formulae my teacher gave me have atmostpheric pressure or density and the problem doesn't have volume please help!
 
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Pressure in this case doesn't depend on volume, it just depends on gravity, density and the height of the fluid column above you. You've probably seen the equation before- p=d*g*h.

The total, then, is just that plus atmospheric pressure.
 
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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