i need somethin about a paragraph long. and something that explains what water pressure is. ALso i need another paragraph that explains atmospheric pressure.
129467
Water pressure is the pressure that water exerts on any object that is immersed in it. Picture holding an object over your head the object is exerting a force on you. In the same way water exerts a force on you if you in a pool, but a force on all parts of your body. Atmospheric pressure is similar to water pressure but it is force exerted by the air around us.
483997
http://www.anwr.org/topten.htm\nhttp://www.anwr.org/backgrnd/theissue.htm\nHope this helps!!!
202091
try these:\n\nhttp://www.maine.gov/portal/education/homeworkhelp.html\n\nhttp://www.meliving.com/weather/temperatures.shtml\n\nhttp://www.wunderground.com/US/ME/Portland.html
4081
Wow, we are sitting at the bottom of a sort of "ocean of air".\n\nHere is one simple source for the newspaper ruler air pressure demonstration.\n\nJust search yahoo for "air pressure rule newspaper" for others.
341030
There are two components to the pressure in the pipe. First is the pressure due to gravity. Lets assume for a moment that the pipe at point "b" is open, that is that the water flows out of the pipe at that point. There will be no pressure at point "b" due to gravity since there is no water above point "b". On the other hand, there is pressure due to gravity at point "a", as a result of the water in the pipe above point "a". \n\nNow we consider the component of pressure due to the water flowing from point "a" to point "b". As long as there is no friction between the pipe and the water (which of course there is in reality), the pressure due to the flow will be the same at point "a" as it is at point "b".\n\nSo, altogether, the pressure at point "a" will be higher, solely due to the "weight" of the water in the pipe pushing down on it.\n\nHope this helps.
166268
The answer is Pressure\n\n10 pts plz! im only level 1!
465343
30.5 m of water.\n\nConvert units to desired units.\n\n1 atm is 14.?? psi is about 32 ft of water.\n\nLook in reference for more detailed unit conversions.\n\nOr use P= &#92;rho g H.
430685
I got this from my physics prof.  Take a full glass of water, preferably a wine glass, and and some food colouring for the water.  Take a clear piece of acetate and cut it in a circle for the top of the glass.  The piece of acetate needs to be larger in circumference than the glass though.  Place the acetate on top of the glass and turn the glass over.    Explain why the water doesn't escape.  The answer might surprise you.  A hint is because of pressure but how much is there, where does it come from, and what is its dimension?
424116
these links might help:\n\nhttp://zahry.usm.my/CAI_CBL/15-42426/_main/assign3/312fou.htm\n\n\nhttp://environment.uwe.ac.uk/video/cd_hdef/topic%20foundations/section2.htm
646544
B. Capillarity\n\n"Capillarity causes the part of the surface of a liquid in contact with a solid, to be either elevated above (e.g. water), or depressed below (e.g. mercury), the rest of the surface. This trait is named for the behaviour of liquids in capillary tubes placed perpendicular to the surface. The forces operating within capillarity are cohesion, adhesion, and surface tension.\n\nCapillary action helps us in our daily lives. For example, it makes water rise through the fibres of a kitchen towel. If capillary action did not exist, we would not be able to blot up spills on a table, or ink from a page."\n\n"Expert: Spaceman\nDate: 3/9/2006\nSubject: mechanics\n\nQuestion\nDear Spaceman, \n\nI have a son who is in high school who needs some help in mechanics,and I'm really not good in this area. Please help! \n\n1. In which of the following units is acceleration expressed? \n\nA. Newtons \nB. Foot-pounds \nC. Kilograms \nD. Meters per second squared \n\n\n2. A stone falls from a ledge and takes 8 seconds to hit the ground. The stone has an original velocity of 0 m/s. How tall is the ledge? \n\nA. 1,254.4 meters \nB. 313.6 meters \nC. 78.4 meters \nD. 39.2 meters \n\n\n3. A tank with a flat bottom is filled with water to a height of 4 meters. What is the pressure at any point at the bottom of the tank? (You can ignore atmospheric pressure when calculating your answer.) \n\nA. 4.0 kPa \nB. 9.8 kPa \nC. 18.9 kPa \nD. 39.2 kPa \n\n\n4. The ability of a material to transfer heat or electric current is called \n\nA. insulation. \nB. porosity. \nC. conductivity. \nD. convection. \n\n\n5. A steel block has a volume of 0.08 m³ and a density of 7,840 kg/m³. What is the force of gravity acting on the block (the weight) in water? \n\nA. 5,362.56 N \nB. 6,150.64 N \nC. 6,700.56 N \nD. 7,600.18 N \n\n\n6. When an automotive battery is fully charged, the sulfuric acid and water mixture will have a specific gravity of about \n\nA. 0.15. \nB. 1.0. \nC. 1.3. \nD. 2.5. \n\n\n7. If an object has a mass of 20 kg, what is the force of gravity acting on it on earth? \n\nA. 196 N \nB. 2.04 kg \nC. 32.67 N \nD. 1.96 kg \n\n\n8. If a small-diameter open glass tube is partially immersed in a vessel containing mercury, the surface of the mercury inside the tube will be \nA. lower than the mercury in the vessel, and shaped like the outside of an umbrella. \nB. higher than the mercury in the vessel, and shaped like the outside of an umbrella. \nC. lower than the mercury in the vessel, and shaped like the inside of a bowl. \nD. higher than the mercury in the vessel, and shaped like the inside of a bowl. \n\n\n9. How much time should be allowed for a 605-mile car trip if the car will be traveling at an average speed of 55 miles per hour? \n\nA. 8.75 hours \nB. 9 hours \nC. 10.5 hours \nD. 11 hours \n\n\n10. A machine is supplied energy at a rate of 4,000 W and does useful work at a rate of 3,760 W. What is the efficiency of the machine? \n\nA. 92 percent \nB. 94 percent \nC. 96 percent \nD. 97 percent \n\n\n11. What is the kinetic energy of an object that has a mass of 30 kilograms and moves with a velocity of 20 m/s? \n\nA. 2,940 J \nB. 5,880 J \nC. 6,000 J \nD. 12,000 J \n\n\n12. If a gas has a gage pressure of 156 kPa, its absolute pressure is approximately \n\nA. 56 kPa. \nB. 100 kPa \nC. 256 kPa. \nD. 300 kPa. \n\n13. According to Ptolemy's model of the movement of celestial bodies, \n\nA. the earth rotates around the sun. \nB. the sun is the center of the universe. \nC. the earth rotates around the moon. \nD. planets orbit in circular paths around the earth. \n\n\n14. A car with a mass of 1,200 kilograms is moving around a circular curve at a uniform velocity of 20 meters per second. The centripetal force on the car is 6,000 newtons. What is the radius of the curve? \n\nA. 160 meters \nB. 80 meters \nC. 32 meters \nD. 16 meters \n\n\n15. What force is required to accelerate a body with a mass of 15 kilograms at a rate of
420005
Water is so different than every other fluid, there are too many reasons to list.  Books are written on this topic, so I'll point you to an interesting link.
345090
Density x Height = pressure on base\n\nPressure at bottom of largest side is the same as the pressure on the base.\n\nPressure at the top of the water level ( 2 meters) is zero.\n\nThe pressure varies linearly from zero to the base pressure over a height of 2 meters.  Which gives you a right triangle 2 meters tall by base pressure wide.  The center of mass of  a right triangle is 1/3 the height or 2/3 meter.  \n\nForce = Pressure x area\n\nShould be able to work it from there.
37585
compressing 2 liters of water into a 1 liter volume\n\nLiquids are not compressible.
148007
hey man, dont worry ill help. first of all our given:\nh=280m\nsg=1.03\ngc=9.7m/s^2\n\nreq'd Pressure(absolute)\n----------------------------------------------\nsol'n:\nfor DENSITY of subs:\nformula\nSpecific Gravity=density of subs/density of water\ndensity of subs=1.03 x 1000(density for water)\ndensity of subs=1030 kg/m^3\n\nFOR PRESSURE GAGE:\nP(gage)=Vh\nP(gage)=density x gravitational constant x height\nP(gage)=1030 x 9.7 x 280\nP(gage)=2,797,480 Pa\n\nFOR PRESSURE ABSOLUTE:\nP(abs)=P(gage) - P(atm)\n\nwhy minus? bec. the submarine is pressurized, meaning its vacuumed\n\nP(abs)=2,797,480 Pa - 1atm(101.325x10^3 Pa)\nP(abs)=2,696,155 Pa\n\nconvert Pa to Kpa\n=2,696,480 /1000\n=2,696.155 Kpa\n\nconvert to bar\n=2696.155kPa x (1.01325 bar / 101.325 kPa)\n=26.96155 bar\n\ndont mind the others, they dont even know how to answer this simple question. They just want to earn points!
351432
The volume will be 549.5 mL.  Let me know if you want to see the work.   Use this equation PV/T=P*V*/T*.    P,V, and T are just the pressure, volume, and temperature for the initial system (they are usually represented as P1,V1, and T1 but I left out the numbers which should be subscripts for simpliofication because yahoo does not have many text options.)  P*,V*, AND T* are the pressure, volume, and temperature after the system has been changed.  You could use any units pretty much for the pressure and volume as long as you keep the units consistent.  However always use Kelvin for the temperature.  Kelvin = 273 + Celcius Temp.
549098
p = 136(1 - e^(-0.51t))\n\n(i) Find the initial value of the pressure\n\nInitial always means at t = 0\nWhen t = 0,\np = 136(1 - e^(-0.51*0))\n= 136(1 - e^(0))\n= 136(1 - 1)\n= 0 bars\n\n(ii) Find the value of the pressure after 2.39 minutes\n\nWhen t = 2.39\np = 136(1 - e^(-0.51*2.39))\n= 136(1 - e^(-0.51*2.39))\n= 136(1 - e^(-1.2189))\n≈ 136(1 - 0.295555)\n≈ 95.80 bars\n\n(iii) Find the value of the pressure once it has settled after a long period of time\n\nAs t gets large  e^(-0.51t) = 1/e^(0.51t) = 1/(large nimber) and so gets smaller and smaller.\nTherefore (1- e^(-0.51t) gets closer and closer to 1\nSo p=136(1- e^(-0.51t))\ngets loser and closer to 136 * 1\nie it will settle on p = 136 bar\n\neg when t = 30 minutes (½ hour)\n\np = 136(1 - e^(-0.51*30))\n= 136(1 - e^(-15.3))\n≈ 136(1 - 2.266 x 10^(-7))\n≈ 136
277715
Atmospheric Pressure \nforce exerted by the weight of the air \n\nAtmospheric pressure is defined as the force per unit area exerted against a surface by the weight of the air above that surface. In the diagram below, the pressure at point "X" increases as the weight of the air above it increases. The same can be said about decreasing pressure, where the pressure at point "X" decreases if the weight of the air above it also decreases