Roughly 4.3 psig or 29 kPag on Earth near sea level ("g" meaning gauge pressure, additional to atmospheric pressure).
10 feet of water depth is about 0.3 atmospheres.
4410 psi
The pressure of water increases about 0.445 psi per foot of depth. If we "zero" our meter so we have "no" pressure at the surface (ignoring the normal 14.7 psi of air pressure at sea level), at 18 feet we will have 0.445 psi/ft times 18 feet, which is 8.01 psi, or right at about 8 psi.
About 415psi at 900ft
The perfect answer requires understanding of the density of the water at different temperatures and pressures at all levels of depth. NOTE: The distinction between PSIG and PSIA is that PSIG is pressure as compared to local (atmospheric) pressure which becomes negative for a vaccum (pressure less than the local pressure). O PSIA is a perfect vacuum and normal atmospheric pressure is ~14.7 PSIA (0 PSIG when referenced to atmospheric local pressure) Without getting too "deep" into the perfect answer, using an approximation thumbrule of 1/2 psi increase per 1 foot of water depth, the answer is approximately 328 PSIG or ~ 343 PSIA.
"If you are 15 ft. under water, the pressure will be the same no matter how large the body of water is" is a true statements about fluid pressure.
Per the ICAO (International Civil Aeronautics Organization) standard atmosphere, the pressure at 1000 ft is 0.9644 times sea level, which is 1013.25 mb. So at 1000 ft, the "standard" pressure is 977.18 mb.
'Hydrostatic Pressure' is the Term used for 'the force exerted by a body of fluid at rest. The pressure increases with increase in depth.There are two ways to Calculate water (clean water) pressure at any depth (both yields almost same results):1. The Hydrostatic pressure of water is 0.433 Psi/ft (Pounds per square inch Per feet). So at 5000 feet, the pressure is: 0.433 Psi/ft. * 5000 ft = 2165 Psianother way to go about it is:2. Water pressure increases at 14.7 psi every 34 feet depth. Thus Pressure at 5000 ft will be: (5000 ft / 34 ft) * 14.7 psi = 2162 Psi.
Approx. 62.42lbs and it all depends on the temperature, the salinity, at what depth and thus the pressure
30 ft
the fluid pressure 10 ft under water is
its 10000 ft² its 10000 ft²
5 to 10 ft
The pressure of the water varies as a function of depth. To calculate the pressure at a given depth take a column of water terminating in some area at the depth you want. For instance, take a 1 in^2 area that is 30 ft deep. The volume of water in this column is 360 cubic inches. Multiply this by the density of water to get the weight of the water in this column. That weight (the force due to gravity) divided by the area (taken to be 1 square inch) is equal to the pressure. Now that we can calculate the pressure as a function of depth, we can then find the pressure for a small horizontal band on the wall with an area equal to the a small increment of height times the width of the wall. This multiplied by the pressure gives the force on that small band. Sum up all the bands (or, really, integrate over the vertical axis) with the pressure calculated at each depth as outlined above.
2.31 feet of water depth exerts a presssure of one pound per square inch. A column of water 10 ft high exerts 4.33 PSI (static pressure) To prove take 4.33 times it by 2.31 = 10.00 ft
133.64psi
Pressure changes very easily. The deeper that you go there is more pressure. For example, when you dive into the ocean, there is not only a whole sky-worth of air pushing down on you, but you are also being pressured by the tons of water above you. On the flipside, when you travel to a greater altitude, pressure decreases. For exapmle, when you climb a mountian, there is less air above you pushing down on you then there is on sealevel.
Given that this stands out a mile as almost certainly a school homework question, to answer directly would be to make me complicit in cheating. So I will tell you how to calculate it, which would appear to be the point of the question: 1) The relationship between depth and pressure of water is linear. 2) If water X ft deep exerts a pressure of P lb/in2, then water of Y ft deep will obviously exert a pressure of P(Y/X) lbs/in2 Given thats information you can now solve the original question.
wind that start currents, different pressure in the water, depth from one point to the other