Determination of the acceleration due to gravity?
963091
I'll take you through the algebra step by step so we don't miss anything.\n\nT=2pi sqrt(L/g)\n                         divide both sides by 2pi\nT/2pi = sqrt(L/g)\n                          square both sides\n(T/2pi)^2 = L/g\n                          multiply both sides by g\ng(T/2pi)^2 = L\n                          divide both sides by (T/2pi)^2\ng = L/(T/2pi)^2\n                          Expand the square term\ng = 4pi^2 * L / T^2\n                          divide both sides by 4pi^2\ng / 4pi^2 = L / T^2\n                           invert both sides\n4pi^2 / g = T^2 / L\n\nRight...\nYou've figured out that T^2 / L = 4.025 yeh, so we can put that in on the right.\n\n4pi^2 / g = 4.025\n                          multiply both sides by g\n4pi^2 = 4.025 g\n                          divide both sides by 4.025\n4pi^2 / 4.025 = g\n\npi^2 = 9.87,    *4 = 39.44\n\n39.44 / 4.025 = g =      9.8\n\nSounds about right to me!\nGood luck
152239
AIR RESISTANCE CANCELS OUT THE FORCE DUE TO GRAVITY , THERE IS NO ACCELERATION BECAUSE THE OPPOISTE FORCE IN THE SAME.\nSO YOU CAN SAY 0
35590
I concur with previous answerer. The acceleration is 1.5 m/s^2. Rearrange Newton's Second Law, to solve for the acceleration. Then, substitute what you know.
945010
Equatorial gravity 3.69 m/s2\n(0.376g)
60123
The formula for acceleration is A = W^2*R, where W (omega) is in radians per second and R is the radius. This will be in the same length units as R, so if you give R in feet the acceleration is in ft/sec^2. For Gs you divide ft/sec^2 by gravitational acceleration (32.2 ft/sec^2).\nWith your example, given A (5000G) and R (6 inches) you need W = sqrt(A/R) = sqrt(5000*32.2/0.5) = 567.45 rad/sec = 567.45*60/(2*pi) = 5419 rpm. The Gs at 4089 rpm = (2*pi*4089/60)^2*0.5/32.2 = 2847G.
974496
Equatorial surface gravity \n1.35 m/s²\n0.14g
301536
Weight = Acceleration*Mass\n\nAcceleration of the free fall is the earth gravity that is 9.797 645 m/s²\n\nWeight = 9.797645*Mass\n\nfree-falling object is an object which is falling under the sole influence of gravity; such an object has an acceleration on Earth of 9.8 m/s/s, downward. This numerical value for the acceleration of a free-falling object is such an important value that it has been given a special name. It is known as the acceleration of gravity – the acceleration for any object moving under the sole influence of gravity. As a matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it – the symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s/s. \n\nThere are slight variations in this numerical value (to the second decimal place) which are dependent primarily upon altitude. The Physics Classroom will use the approximated value of 10 m/s/s in order to reduce the complexity of the many mathematical tasks performed with this number. By so doing, you will be able to better focus on the conceptual nature of physics without sacrificing too much in the way of numerical accuracy. When the moment arises that you need to be accurate (such as in lab work), use the more accurate value of 9.8 m/s/s. \n\ng = 10 m/s/s, downward\n\nthe acceleration is the rate at which an object changes its velocity. Between any two points in an object's path, acceleration is the ratio of velocity change to the time taken to make that change. To accelerate at 10 m/s/s means to change your velocity by 10 m/s each second.\na = dv/t = 10/1 = 10\nthen if a body falls in the first second his speed is 10m/s and in the 2nd second is 20m/s and so on
748368
first get the time from the horizontal motion  t = .51 / 1.2 s (uniform motion, no acceleration here)\n\nthen use h = 1/2gt^2 for the vertical motion (with initial velocity zero)
77860
hey, didnt i answer something very similar to this for you? you need to do F=ma so 43000000N/2100000kg then you get an acceleration, but then you have to subtract gravity from this because thats the acceleration produced by the rocket but there is still gravity pulling with 9.8 m/s^2 in the opposite direction. so just do (43000000/2100000) - 9.8, and that's your answer :)
162313
Let V = speed (the magnitude of the initial velocity)\na = angle, t = time, \nVx = horizontal (x) component of velocity\nVy = vertical (y) component of velocity\nAx = acceleration in x direction\nAy = acceleration in y direction\nDx = net distance traveled in x direction\nDy = net distance traveled in y direction\n\nVx = Vcos(a)\nVy = Vsin(a)\n\n  Dx = 1/2 Ax*t^2 + Vx*t\nbut Ax = 0, so Dx = Vx*t, t = Dx/Vx\n\n  Dy = 1/2 Ay*t^2 + Vy*t\n(Ay = -g, the acceleration of gravity, negative for downward)\n -1/2*g*Dx^2/(Vcos(a))^2 + Vsin(a)*Dx/(Vcos(a)) = Dy\nMultiply by V^2, and collect terms with V^2 on the left:\n(Dx*tan(a)-Dy) V^2 = 1/2*g*(Dx)^2/(cos(a))^2\nV = sqrt(1/2*g*(Dx)^2/(cos(a))^2/(Dx*tan(a)-Dy))\nsubstitute a = 30 degrees, Dx = 6.05 m, Dy = 3.05-2.2 = 0.85 m, g = 9.8 m/s^2\nV = 9.51 m/s\nFor a = 60 degrees,\nV = 8.63 m/s
125809
The struggle here is that you provide too little information to recognize this problem as coherent.\n\nWe may assume that the acceleration of gravity is 9.8 m/s/s or 32 ft/sec/sec.\n\nThe velocity(V) of an object is equal to its initial velocity(v) plus its acceleration(a) times time(t). \nV = v + at or V = v - 32t (in feet per second)\n\nYour object will reach maximum height when the velocity is zero or when the initial velocity is equal to 32 ft/sec/sec * time. With your list of times, you'd just multiply by 32 to solve for the initial velocity, if that is what the problem is actually looking for. There are many other permutations of this data, but hopefully that is a start for you.\n\nGood luck.
364646
Acceleration due to gravity on the Earth is 9.81 m/s^2 or 32.2 ft/s^2.\n\nUnless you mean the acceleration of the Earth?  If you assume a circular path (not true, but I'm guessing you're not doing rocket science, so I won't worry too much), acceleration on the earth by the sun would be the square of the velocity at which the eart travels divided by the distance from the earth to the sun. a = V^2/R.
22095
Let's try to anser the question without looking it up and without using a calculator.\n\nWe know Newton's gravitation law: F = G*m*M/r^2\nAnd we know the gravity on earth: F = m*g\n==> m*g = G*m*M/r^2\n==> M = r^2*g/G\nNow, the acceleration on the earth surface is approximately 10 m/s^2\nThe earth's radius is approximately 6*10^6 m\nAnd I remember G to be approximately 7*10^-11 Nm^2/kg^2\n\nM = 6^2*10/7 * 10^12/10^-11 kg\n\nThe earth weights about 5*10^24 kg\n\nIf you look it up in Wikipedia, it says 5,9736*10^24 kg. \n\nSo we're close enough if you take into account that we used rough estimates for the constants.
831091
w = mg\n(w is weight, m is mass, g is acceleration due to gravity (9.8m/s^2))\n :. m = w/g\n        = 300 / 9.8\n        = 30.6 kg
701530
Plug the numbers into the Newtons Equation :\n\nF =  (GMm) / r^2\n\nF = Gravitational Force / weight  \nG = Universal gravitational constant   6.673*10^-11 (note the - sign on the exponent)\nM = mass of earth (kg)                 5.98*10^24 kg\nm = mass of satellite (kg)             150kg\nR = radius of earth    (m)             6,378,137 m\nh = height above earth (m)             12,800,000m\nr   = h+R\n\nThe answer will come out as the weight being @ 162.67N\n\nNot weightless, although someone standing on the satellite would feel weightless as its in continuous free fall.\n\nEver been in an elevator and felt lighter as it falls or been weighless on the big dipper as it plummets?\n\nThe gravitational force (weight) of 162.67N is what keeps it in orbit and prevents it from flying off at a tangent. Basically it pulls the satellite towards the earth but the satellite is travelling so fast it misses and continues round in its orbit.\n\nAdded:\n\nWeight is measured in Newtons and is the attractive force between the two masses. (M & m)\n\nIn this case the earth and the satellite. A 1 Kg mass on the surface of the earth will be pulled towards the earth's centre with a force of 9.8N, therefore its weight at ground level is 9.8N. If you plug take m as 1kg, h as 0 and plug them into Newton's formula should get the answer 9.8N, which is where we get the constant g from. \n\nAs you get further away from the centre of the mass the gravity field decreases in inverse proportion to the distance squared. Hence the divide by r^2 on Newtons fornula. Therefore the weight of a 1Kg mass at 12,800km is 1.08N.\n\nBecause the satellite is in a stable orbit the opposing forces of the satellite trying to escape orbit and the gravitational pull towards the earth are equal so yes the weight is 162.7N and the centrefugal force is 162.7N. This is the correct answer to the problem. \n\nAs an aside; for simplicicities case in this scenario the mass of the satellite can be regarded as constant due to its relatively low velocity compared to the speed of light. \n\nHowever as you come closer to the speed of light mass increases according to Einstein's theory of relativity, eventually reaching infinity at light speed.
1082685
Simply equate gravity function with centripetal acceleration.\n\nGMe/(r+h)^2 = V^2/(r+h)....where r is earth radius and h is height\n\nsolve for v
537369
If your mass is m, multiply it by g to convert it to N, where, g is the acceleration due to gravity of the place where you currently are.\nFor most of the places on earth, we take the average value of g = 9.81 (m/s)/s.