What is the change in density of water at a depth of 400 m?

AI Thread Summary
To determine the change in density of water at a depth of 400 m, the initial density is given as 1030 kg/m³ and the bulk modulus is 2 x 10⁹ N/m². The relevant equation for new density involves the change in pressure (dP) and the bulk modulus (B). There is uncertainty about whether to use the surface density for calculating dP or if integration is necessary due to varying density with depth. The pressure can be calculated using the formula P = h × surface density × g. Understanding these calculations is crucial for accurately determining the density change at that depth.
Abhishekdas
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Bulk modulus question...please help

Homework Statement


What is the change in density of water in a lake at a depth of 400 m below the surface? The density of water at the surface is 1030 kg/m3 and bulk modulus of water is 2*109 Nm2


Homework Equations


For fluid
New Density= Initial density*(1/1-dP/B)
dP=change in pressure and B=bulk modulus


The Attempt at a Solution



My doubt is can we apply dp=h*density*g here taking the initial density (1030) because density is not constant throughout...Do we have to integrate or something...? What do do?
 
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ANy replies here...please...
 


Abhishekdas said:
ANy replies here...please...

You can take P = h X surface density X g
 
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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