Surface acceleration =acceleration per meter squared?
206777
Pressure is force over area and would be per m^2.  They use the center of the earth because the math works out such that all the mass of the earth could be treated as being at that point.  It doesn't have as much physical meaning as you might think.
279221
C, gravity is an acceleration, it is by definition not uniform then.  \n\nit's 9.8 meters per second per second, just in case you wanted to know, that's also written 9.8 m/s^2\n\nhere's a little more: it's not A, (see above), it's not B, since again this would mean that it's uniform, since 10 seconds is twice as great as 5 seconds, and it's not D because it would fall faster in a vacuum than in air since there would be no resistance.  if you drop a pen and a piece  of paper right now, the pen should fall faster since the air will resist the paper more, but in a vacuum both would hit the ground at the same time.
367003
W = F * d\nW = (400 newtons) * 2.15 meters = 860 joules
219798
Kinetic energy = 1/2 x mass x velocity^2\n\nYour velocity would be in meters/sec probably, but you don't mention a mass.\n\nmass is measured in kilograms\n\nEnergy is measured in joules.
14268
Well if I am going 10 miles per hour, that is an instantaneous speed (it is a scalar).\n\nIf I am going 10 miles per hour north, that is an instanteanous velocity (it is a vector).\n\nIf I am changing my speed by 5 miles per hour per second, that is an acceleration. Without a direction, it is a scalar. Of course, I could specify a direction, and then it would be a vector, so 5 miles per hour per second north is a vector expression of acceleration. \n\nOne can have an instantaneous vector acceleration, or an instanteous scalar acceleration, or an average vector acceleration or an average scalar acceleration. \n\nOne can also have a change of acceleration, called a boost, which can also be a vector or a scalar, depending on whether it has a direction associated with it or not. \n\nSo 2 miles per second cubed is a boost (scalar boost that is). If it as at some instant of time, it is instantaneous. If it is some integral over some time period, it is an average. And so on.
454773
final speed=initial speed+(acceleration*time)\n=10+(30*30/60)\n=10+15\n=25km/h\nhope that helps..\ngood luck
423841
1. Momentum=Massxvelocity=3600\n2.3600=.2KgxVelocity=18000\nThats all i know
36151
tried this problem, but when converting 2cm disance to meters, the acceleration was way out, too large.I used the equation V2 = U2 + 2aS to arrive at the acceleration (deceleration). This might have been the wrong approach.
472819
Currently, the scientific community uses the MKS system for most engineering, and beginning physics:\n\n      M = Meters (Length)\n      K = Kilograms (Mass)\n      S = Seconds (Time)\n\n...the answer is KILOGRAMS.
523709
Frictionless flat surface, so no angles to deal with.\n\nBut cant you just use the final velocity, and KE= 1/2 * m *v*v ?\n\nKE = 0.5*(m)*v2 \n\nKE = (0.5) * (7kg) * (0.8 m/s)2\n\nKE =  2.240 Joules     (kg*m*m)/(s*s)
342065
Have you been given the standard formulas?\nx = a cos (nt + c),  frequency = n, period = 2*pi/n, phase = c (Not sure of that -- it may be -c.  Look it up.) This gives you the answers to (d), (e) and (f) immediately.  [No, I'm not doing it for you: You can substitute values I'm sure.]\n\nx is the displacement, so just sub t = 2 in this equation.  If you need touse your calculator (you shouldn't!), make sure it's set in radians.\n\nVelocity is dx/dt.  You can do that I hope, and get\n\nv = - 18*pi*sin(3*pi*t + pi/3)  Notice I omit units.  I prefer to use equations relating to numbers only (though I know units are sometimes useful especially in a Physics context) and then state the correct units in presenting the answer -- which you are going to do!\nSub t = 2 to get v in m/s\n\nAcceleration is dv/dt, so differentiate and then sub t = 2\n\nAll the best
167291
use d=V1*t + [(a*t*t)/2]\nd=distance=490m\nV1=initial velocity = 0m/s\nt=time = ?\na=acceleration due to gravity = 9.81m/s*s\n\nusing equation above, t will come up to be:\nt=(square root)(2*d/a)\nt=(square root)(2*490m / 9.81m/s*s)\nt=9.99 sec or 10 sec.
379630
acceleration = v^2/r or v*v/r (this is m/s^2 so divide by 9.81 to get g's)\n\nwhere v is the velocity\n\nr is the radius\n\nBe careful, think about the one g the pilot feels from the pull of the earth. ;)
365572
Delta M = m(deltaV) = .2*(30-(-20)) = .2*50 = 10 kg m/s
534770
mu = 2d/gt^2 = 2*1000/(9.8*12^2)  =  .708 assuming constant acceleration
66626
assuming your statement is true, then\na=k/t\nwhich gives t=k/a\nso time is a nonlinear function of acceleration.\nGraphically, the plot of a versus t is not a straight line (it is a hyperbola)
623895
You should probably skip what I wrote and go straight to the link http://whatis.techtarget.com/definition/0,,sid9_gci529957,00.html :-)\n\nYou measure the distance fallen in meters.  Let us say that this distance as a function of time t in seconds since the ball was dropped in a vacuum is s(t) meters.\n\nThe "instantaneous" downward velocity of the ball at time t is just the derivative of s(t) with respect to t ... for example in the h seconds between time t = k and time t = k + h, the ball drops an extra s(k + h) - s(k) meters ... so the average velocity between times t = k and t = k + h is (s(k + h) - s(k)) / h, and the limit as h -> 0 of (s(k + h) - s(k)) / h is the velocity at time t = k and is the derivative of s(t) at t = k, and you can see that it has units meters per second because s(k+h) - s(k) is in meters and h is in seconds.\n\nSo distance is meters and velocity is meters per second, with v(t) = s'(t) the derivative of s with respect to t.\n\nNow acceleration is defined as the rate of change with repsect to time of velocity.  You will have a(t) = v'(t) = s''(t), by the same reasoning as before.  The velocity at time t = k is v(k) = s'(k) and the velocity at time t = k + h seconds is v(k + h) = s'(k + h).  The average acceleration between times t = k and t = k + h is then (v(k + h) - v(k)) / h in units of (meters per second) per second, or meters per second squared.  The instantaneous acceleration at time t = k is the limit of that as h -> 0 and is v'(t) = s''(t) at t = k.\n\nFalling in a vacuum at constant acceleration g = 9.8 meters per second squared means you have\n\nacceleration = constant g = 9.8 meters per second squared\n\nincreased downward velocity after falling t seconds = g t = (9.8 meters per second squared) time (t seconds) = 9.8 t meters per second.\n\ndistance fallen (if dropped with initial velocity zero) after falling t seconds is (1/2) g t^2 = (1/2) (9.8 meters per second squared) (t seconds)^2 = (1/2) g t^2 meters.\n\nThe rate of change of acceleration would be meters per second cubed.