I'm looking for an answer for a high school physics class, just need to know the standard measure.
215285
Standard atmospheric pressure is 101 325 Pa = 101.325 kPa = 1013.25 hPa = 1013.25 mbar = 760 Torr. This definition is used for pneumatic fluid power (ISO R554), and in the aerospace (ISO 2533) and petroleum (ISO 5024) industries.
359839
Question is: why on earth would you want to make that calculation?  m/s^2 is the standard form for acceleration.
391001
C=2(Pi)r or diameter times Pi\nI don't know how to solve the problem cause not enough info is given. Does it say what C equals?\n\nv = lwh . H stands for the height. Does it say what v equals?\n\na= 1/2h(b1+b2) b1 is base 1. Does it say waht the area is?\n\nProvide more info please. :)
226476
This is more of a physics question...\nd final = .5*a*t^2 + v*t + d\nv final = a*t + v\nd is displacement, v is velocity, a is acceleration.\nEnjoy.
649787
Write it up and send it to a physics journal. You'd get credit for your discovery if it is valid. If it isn't, they'll let you know.
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!
412888
I teach students who are 11 years old in an American school. By the time they get to me almost all have had experience with measuring with the metric system. \n\nIn the U.S. public school I teach at, we teach both the standard and metric systems.
20910
The last poster needs to have his dynamic physics retaught, or stop trying to sound like an engineer.\n\nThe first mistake, is identifying newtons = force x distance.  This is actually a measure of work or energy and measured in joules.  The second mistake identifying force as being measured in joules.  Force is measured in Newtons.  Since it was Isaac Newton in the second law of motion that defined Force.\n\nYour question cannot really be answered without more information.  What is the direction of motion, horizontally or vertically?  \n\nVerically?  Then 150 lbs of force.  Pounds were once used to measure force.  If you want to use metric, convert the 150 lbs to kilograms, multiply by 9.8 and then you must exert at least this much force in Newtons to overcome the force of gravity.  150 lbs is 68.04 kg.\n\nHorizontally, what type of surface?  Determine the coefficient of friction, and multiply it by the mass in kilograms.  Then this is the amount of force you need to exert to overcome the force of friction.  There are two types of friction, static and dynamic.  In other words it takes more force to move an object at rest, but once the object is in motion it takes less force to keep it move (opposing the dynamic force of friction).\n\nWatts by the way is a measurement of power, which is energy derived by time, Power = dE/dt.  Joules per second.
291422
Go to your local physics professor. If you are correct, (s)he will no doubt help you to get it published. If you are incorrect, (s)he should be willing to show you where your math is off.
246077
Easy H*D*W    H=Height D=Depth W=Width *= times\n\nUse the units cubed, if it is meters use m^3 as the unit of measurement
480799
Do your own damn physics homework.
17544
I decided to go ahead and post here, despite the numerous posts already.\n\nI'm not going to use my own value for radius, I'm going to choose one that's already here. Namely 3963.19 mi. (This will be "R")\n\nLet:\n\nUnaltered circumference (C) = 2(pi)R\n\nAltered Circumference (c) = 2(pi)r\n\nThe length you need is: c - C\n\n==> c - C = 2(pi)r - 2(pi)R\n==> 2(pi)[r - R]\n\nr - R is the difference in radii, which is what you stated.....1".\n\nSo, r - R = 1"\n\nTherefore: 2(pi)(1") = 2(pi) inches!\n\n-or-\n\n6.28 inches.\n\nThis is how much you'd need to get that 1" gap between the earth and the string.
240244
http://www.onlineconversion.com/length_common.htm\n\nthat's for height.\n\nhttp://www.onlineconversion.com/weight.htm\n\nThat's for weight.
659043
H0: mu = 10 inches\nHA: mu < 10 inches\n\nThe null always has the equal sign in it.  The alternative has what you wish to show is true.
180046
no idea...but I think you use the formula (Gravitational constant x mass 1 x mass 2) / distance squared
495448
the most accurate answer is 170centimeters and 2 millimeters.. i got it from my meter tape
153491
Newtons, or sometimes kilograms