Easy Density and Depth Question- Will pick best answer?

Question

The accleration of gravity is 9.8m/s^2. Calculate the depth in the ocean at which the pressure is three times atmospheric pressure. Answer in meters

The density of sea water is 1025 kg/m^3
Atmospheric pressure is about 101,325 Pa



When a 55N stone is submerged in water, the spring scale (which measures apparent weight) reads 45N when attached to the rock. What is the density of the stone. Answer in units of kg/m^3

Answer 1

the presure in water is called hydrostatic pressure
P = h * d (water) * g
where h = depth from free surface of ocean
total pressure = 3* P (atm)
h d g + atm pressure (P) = 3 P
h d g = 2P(atm)
h = 2* P (atm)/d g = 2*101,325 /1025*9.8 = 20.17 meter
>>>>> corrected
-------------------
spring balance reading = apparent weight = actual weight - bouyancy force
---------------------------------------------- ^ bouyancy force
......................................................... O V 55N
------------------------------------------------ V Tension on rock = 45 N
45 N = 55 N - Bouyancy force
Bouyancy force =10 N = volume * water density * g
volume of stone V = 10 /1000*9.8
density of stone = mass/V = (actual weight/g) / V
density of stone = (55/9.8) [1000*9.8/10]
= 5500 kg/m^3

Answer 2

The pressure at depth = P = 3p = 2p(w) + 1p(a); where p(w) is pressue due to the column of water over the ocean at depth h and p(a) = 1 atm pushing on the ocean's surface.

rho' = mass density of water; so Rho' = rho' g, which is the weight density. So if we let A = 1 m^2, the area of the column of water, and V = Ah, the volume of water in the column, the pressure at depth is P = 2 rho' g A h + 1 atm = 3 atm; so that 2 rho' g A h = 2 atm and h = 2 atm/(2 rho' g A) and you have all the numbers for the RHS. You can do the math.

W = mg = 55 N, the weight of the stone in air. w = W - B = 45 N, the effective weight in water and B is the buoyancy. Let rho = m/V the mass density of the stone of mass m = W/g and Rho = rho g. We need V, the displacement volume, to find rho; so buoyancy B = 10 N = rho' g V, where rho' is the given salt water density, and V = B/(rho' g).

Putting this all together we have rho = m/V = W/g//B/(rho' g). Again the RHS is given; so you can do the math.

[NB: Note not all the P on the ocean at depth h comes from the column of water above. Some of it, 1 atm in fact, comes from, ta da, the atmosphere which pushes down on the column of water and adds to P. A solution that presumes all of P comes from the water is in error.]

Answer 3

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