Maximum Depth for Organisms Under the Sea

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The maximum pressure most organisms can withstand is approximately 1000 times atmospheric pressure, equating to around 100 million Pascals. Small organisms like tadpoles and bacteria are capable of surviving these high pressures. Calculations suggest that the maximum depth for these organisms, based on seawater density, is around 9.94 kilometers, although this figure is debated. Some participants in the discussion believe the initial calculations may have rounding errors or approximations affecting the results. Overall, the challenge lies in accurately determining the maximum depth at which these organisms can thrive under extreme pressure conditions.
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The maximum pressure most organisms can survive is about 1000 times atmospheric pressure. Only small, simple organisms such as tadpoles and bacteria can survive such high pressures. What then is the maximum depth at which these organisms can live under the sea (assuming that the density of seawater is 1026 kg/m3)?




Ptotal = 10^5 (1000)= 100000000 Pa
100000000 Pa = (1026)(9.81)(d)
d= 9935m= 9.94km?

This is wrong but I am unsure why...
 
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any ideas??:-p thanks!
 
ahh, I'm still workin on this one, here is my latest try:

1000 atm=101325000 Pa

101325000 = 101325 + 1026*9.81*d
d=909.64m
=.909km

this is still wrong but I've no idea why? Thanks
 
Hmm... that equation you came up with looks like
10^8\approx 10^5+10^4d
so it should work out to roughly
d \approx 10^4
but you come up with
d \approx 10^3

I'd guess that your first answer was correct modulo some approximation or rounding errors.
 
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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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