What Is the Value of g in the Pressure Calculation Formula?

AI Thread Summary
In the discussion about calculating pressure in a cylindrical tank, the formula P = P_a + pgh is highlighted, where P_a is atmospheric pressure, p is the density of water, g is the acceleration due to gravity, and h is the height of water above the point of measurement. The user is attempting to find the absolute pressure 6 meters above the bottom of a 20-meter tall tank filled with water, questioning the value of g, which is indeed the gravitational constant typically approximated as 9.81 m/s². The conversation emphasizes the need to determine the pressure exerted by the water column above the measurement point. It is clarified that the measurement point is 14 meters below the water surface, which affects the pressure calculation. Understanding these factors is crucial for accurately determining the absolute pressure at the specified height.
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Homework Statement



Cylindrical tank of water is 20m tall, and full. atmospheric pressure is 1.01x10^5 Pa and the density of water is 1000Kg/m^3. Find absolute pressure 6m above the bottom.

Homework Equations



P=P*sub*a + pgh

The Attempt at a Solution



I got so far... P=(1.01 x 10^5) + (1000)g(6)
How do i find g?what is g??
 
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Isn't g gravity? Which is a given constant most of the time.
 
P=(1.01 x 10^5) + (1000)g(6)

u're finding the pressure exerted by 6m of water?
 
xiankai said:
u're finding the pressure exerted by 6m of water?
Good point. The water tank is 20 m high and the gauge is 6 m above the base so how many meters below the water line is it?
 
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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