how does one convert newton metres to foot pounds?
605852
multiply n-m by 0.737562
367853
F = m*a \n\nm = 2 kg (two masses; two strings)\na = gravitational constant which happens to be 9.81 m/s^2 on the surface of the earth (or 32.2 ft/s^2 in EE units)\n\nsum the forces in the y-direction:\n\nT = F = 2 * m * a = 2 N\n\n(assuming massless strings)
69858
Weight is a force.  The National Institute of Science and Technologiy defines it here as such.\nhttp://physics.nist.gov/Pubs/SP811/sec08.html\n\nA force is acceleration times mass.  Weight is the accleration due to gravity times mass.  On the Earth's surface, this is approximately:\n\nWeight = Mass x 9.8 meters/second^2\n\nFor a accurate conversion, you would need to use the acceleration due gravity minus the accelration due to centrifugal force at your exact location.  \n\nThis is why physicists more commonly use mass than weight to describe an object.
606521
Try this:\n\nhttp://education.yahoo.com/reference/weights_and_measures/
481398
sum the forces in the direction that they are applied and it will be equal for a static object. \n\nRule of thumb on this one is that you have 100N down ward, then you need 100 N's up ward.\n\nGivens, legs are equal lengths, same material, and no angles to calculate.\n\n4 legs - 4 forces - each at 25 Newtons.  \n\nNow if you want to get fancy, break out the strain gauges and let us see what we really find.  But it helps if you have preformed a few material tests and experimenters first.
997326
Use Newton's universal gravitation equation, then sum the forces.\n\nF = G*m1*m2/r^2\n\nF = force (newtons)\nG = gravitational constant (6.6742 x 10^-11)\nm1 = mass of object 1 (kg)\nm2 = mass of object 2 (kg)\nr = distance between objects (meters) ==> don't forget to convert the units\n\nIn this case, you do it twice:\nF1 = G * 390 * 390 / 0.11^2 = 0.84mN to the left\nF2 = G * 390 * 390 / 0.39^2 = 0.067mN to the right \n\nNet Force = 0.84 - 0.067 = 0.772mN to the left
79947
The unit of force is the Newton (N).\n\n1 newton is defined as the force needed to accelerate a 1kg object at 1m/s*2.(from Newton's 2nd Law of Motion: F = ma)   \n\nSometimes it is given in its base units as kgm/s*2
855447
if it is equations of motion then \nv=u+ft\ns=ut+1/2ft^2\nv^2=u^2+2fs\nhope this is what you wanted
970548
A stone is an English measure and is = 14 pounds.\n\na kg = approximately 2.2 pounds\nThus 1 stone = approximately 14/2.2= 6.3636 kg.\n\n1 meter = 39.37 inches, so\n1 cm =  .3937 inches\n12/.3937 cm =12inches = 1 foot\n30.48 cm = 1 foot
1009332
Convert everything to inches, then multiply.  LxHxW.  They need to be the same units, though.
105246
Psf (pounds per square foot) is a pressure unit, Newton is a force unit, so there is no equivalent.  But 0.224808943 pound = 1 Newton, and 1 foot = 0.3048 meter, so that's enough info to convert 8000 psf to Newtons/meter^2, which is a pressure unit.  I googled the conversion, 8000 psf = 383042.072 Newtons/meter^2.  How do you google conversions?  Just enter "8000 lb/ft^2 to newton/m^2".
398008
You multiply your measure in feet by 0.3048 and you get your measure in meters; EXACTLY. \n\nNote that US and British Feet are exactly the same.\n1 Inch = 2.54 cm EXACTLY by re-definition of US standards in 1959.\n\nTherefore 1 feet = 12 inches = 30.48 cm = 0.3048 meters.
903949
Yes to the first question.\n\nYou have to do like this:\n(ai+bj+ck) ___\n\nIn the ___ you put the vetorial unit. Like: meters, cubic meters, newtons, m/s, etc.\n\n₢
57076
m = 7.42 kg\na = 4.02 m/s^2\nF = ?\n\nF = ma = (7.42kg)(4.02m/s^2)\n\nF = 29.8284 kgm/s^2 = 29.83 N\n\nN stands for Newtons
471462
Multiply meters by 3.28084 to get feet.
222710
there is a formula for the direct conversion of meters to feet but i forgot, but you can use this instead;\n1 inch = 2.54 cm\n1 foot = 12 inches\n1 foot = 12 x 2.54 = 30. 48 cm\n1 meter = 100 cm\ntherefore\n1 foot = .3048 meter\n\nso \n2.36 meter = 2.36/.3048 feet\n2.36 meter = 7.74 feet
1045430
Momentum - P = mv (P=momentum, m=mass, v = velocity)\nThree Laws of Newton - F = ma (F=Force; a=acceleration)\nEquations of motion - 1) v = u + at\n                                    2) S = (1/2)ut + at^2\n                                    3) v^2 = u^2 + 2aS\n(v=final velocity; u = initial velocity; S=distance)\n\nIf this can help you please mail me back !\nBEST OF LUCK !!!!