This is a physics question with pressures and fluids ?
969973
Atmospheric pressure at sea level is caused by the weight of the atmosphere pushing downward. (weight /area).\n\nIf you approximate the Earth as a sphere, calculate the surface area of the sphere, and then multiply the pressure times the area you will get an approximation for the weight of the atmosphere.\n\nStandard pressure is 100 kPa. \nThe Earth's equatorial radius is roughly 6400 km\n\nSurface Area of a sphere is 4*pi*r^2 = pi*d^2\n\nMass = A*P
914444
I will be honest I only know part of the answer but here's what i know...\n\nfluid pressure increases with depth because there is more fluid on top of it thus more weight and it casues compression (pressure)\nand thats about all I can expalin
1080952
mass, volume and temperature.
97560
Are you referring to a fluid mechanics problem, or?\n\nCan you restate the question?
1127531
Surely you know the foumla for Re?\n\nRe=(density*velocity*length)/viscosity\n\nRe=(1000*1*25*10^-3)/1014*10^-3\n\nRe=24.655 (3dp)\n\nI think you've got yourself a laminar flow.
939941
Yes you're on the right track, you know the momentum of the car along with the energy, both are conserved.  \n\nForce=Mass*Acceleration=Coefficient of Friction*Mass*Gravity\n\nYou can then solve for Acceleration=Coefficient of Friction*Gravity\nVelocity=Acceleration*t\nThen its simply a matter of filling t into this equation:\nDistance=1/2*Acceleration*t^2+Velocity*t
980208
ok, im going to assume that you know (rho)(g)(h) = pressure\n\nlets label 4 points:\nA is at the bottom of the 1.28cm mark\nB is at the top of the 1.28cm mark\nC is at the bottom of the hl mark\nD is at the top of the hl mark\n\nNotice 2 things:\n1. the pressures at A on the right and the left must be the same\n2.if you removed the same amount of blue from both sides, hl would still be the same measurement (so lets remove it for calulations)\n\nnow the blue on the left side is of a height of 1.28+hl and the green on the right is 1.28\n\nat the top they are equal, so:\n2.25*g*(1.28+hl)=8.41*g*(1.28)\n2.25(1.28+hl)=8.41*1.28         g cancels from both sides\n2.25*1.28+2.25*hl=8.41*1.28\n2.88 + 2.25hl = 10.765\n2.25hl = 10.765 - 2.88=7.885\nhl = 7.885/2.25 = 3.50435.....cm
380084
The answer is Pressure\n\n10 pts plz! im only level 1!
187607
The density of water is about 100 times the density of any gas .\nSo the gas will be lifted. (Archimedes law).\nBasicly the gravity energy decreases.
443139
BOUYANCY & SURFACE TENSION
652894
the amount of pressure exerted on an object
627120
a)\ncross-sectional of the small piston: A1 = 25 cm²\ncross-sectional of the large piston: A2 = 400 cm²\nMass on the small piston: m1 = ?\nMass on the large piston: m2 = 1200 kg\n\nby Pascal's Principle:\nF1/A1 = F2/A2\nm1*g/A1 = m2*g/A2\nm1 = m2*(A1/A2)\nm1 = 1200 kg (25 cm²/400 cm²) = 75 kg\n\n\nb)\nwhen a person enter into the car, the person on the small piston will be elevated a heigh h1 over initial level and the car is going down a lenght h2, and height difference in the fluid levels in the pistons is H = h1 + h2.\nMass on the small piston: m1 = 75 kg\nMass on the large piston: m2 = 1290 kg\n\npressure by 75 kg person + hydrostatic pressure by column of fluid = pressure by (90 kg person and 1200 kg car)\nm1*g*A1 + d*g*H = m2*g*A2\nH = (m2*A2 - m1*A1)/d\nH = {(1290 kg * 0.04 m²) - (75 kg * 0.0025 m²)}/(720 kg/m³)\nH = 0.0714 m = 7.14 cm\n\n\nc)\nvolume displaces is the same in both pistons:\nA1*h1 = A2*h2\nA1*(H - h2) = A2*h2\nA1*H = h2*(A1 + A2)\nh2 = H*{A1/(A1+A2)}\nh2 = 7.14 cm (25 cm²/425 cm²)\nh2 = 0.42 cm = 4.2 mm
887467
Im a little rusty on the thermodynamics, but lets see if this helps.\nV/T (initial)=V/T (final) since no gas is lost or added  Raising the temperature on a container that is free to expand does not change the pressure, so you can find the change in volume with temperature.  Since V=A*h, you can find h.\n\nNow, the problem is that the vessel is not entirely free to expand, and acts on a spring, with force equal to F=k*h.  This force is related to the pressure created by the expanding vessel as simply F/A.\n\nAll values are given, so I think you are set.
1105154
pressure is: \n p(density)gh / gc =y(specific weight)h\nflow of liquid flowing inside the pipeline is:\n use the bernoulli equation:\n [p1/p(density)g  +  v1^2/2g  +  z1]  =  [p2/p(density)g  +  v2^2/2g  +  z2  +  hf]\n\nnote: p(density) is rho and y(specific weight) ir gamma.
554866
air and water vapor\n\ntheir volume depends on temperature and pressure
725588
1. concentration and pressure\n2. see sources
107612
This is potentially a massive and complex task.\nSearch the web for "computational fluid dynamics".  Here's what I found in a couple of minutes:\nThe 1st ref. below has some interesting simulation ideas (but unfortunately for a liquid, not air).\nThe next discusses things you should know before even deciding whether to do this.\nThe last gives links to software downloads.