How do I find the accelaration due to gravity at a place using the formula 'h=1/2g(t square)w?
341507
You mean find "g" as if you didn't know the constant? Okay.\n\nPretty simply really. Have the height measured that you're going to drop it from to where it's going to land (this will be your "h"). Make it a high one (one meter is enough). Measure the mass of the object (that's the "w"). Then when you do the experiment, time it. Start it at the instant you let it drop and stop it once it touches where ever it lands. That's enough information to find "g." Just solve for "g" then that would be the answer.\n\nActually, for an experiment, you should solve for "g" as a formula before you start doing anything else. So the formula would be like this:\n\ng = (2h)/[(t^2)(w)]\n\n(t^2 is "t squared")\n\nSo just have that formula then do the experiment: measure height of the drop, the mass of the object, then time it when you drop it (you need good reflexes when you drop it, as it falls fast).\n\nPS Penguins!!!
247359
Mass interacting with gravity
757331
The motion is deccelerated by the action of gravitational and frictional force. Hence, \nm*a=Fg+Ff\nm*a=w*sinx+f*w*cosx\na=w*sinx/m+f*w*cosx/m\na=g*sinx+f*g*cosx\na=g(sinx+f*cosx)----(1)\n0.5*m*v^2=m*a*s\ns=v^2/(2*a)---(2)\nInsert (1) in (2) and solve for s\ns=v^2/(2*g(sinx+f*cosx))\ns=3.1^2/(2*9.81(sin26+0.5*cos26))\ns=0.55 meter
808209
Using one of the equations of motion\n\n(v*v) - (u*u) = 2*a*s\n\nwhere \nv is final velocity\nu is initial velocity\na is accleration\ns is displacement\n\n\nOR \nuse \n\nv * (dv / ds) = a
827544
1) use conservation of energy:\n\n1/2*m*vb^2=1/2*m*va^2+m*g*h\n\nm=2\nva=4\ng=9.8\nh=6\nv=sqrt(16+2*9.8*6)\n=11.56 m/s\n\n2.\nSame concept\nm*g*(bounce)=1/2*m*va^2+m*g*(drop)\nmass divides out\nbounce=va^2/(2*g)+drop\nva=6\ng=9.8\ndrop=4\nbounce=36/(2*9.8) + 4\n=5.84 m\n\n3)  since the mass isn't moving it's offset the 300N force\n\n4)\nAgain, use conservation of energy\nNote that the change in potential energy will subtract from the kinetic energy of the start \n\n1/2*m*vc^2=1/2*m*va^2-m*g*delta-h\nthe mass divides out\nva=8\ndelta-h=2\ng=9.8\nvc=sqrt(64-2*9.8*2)\n=4.98\n\nj
755943
If you are assuming "uniformly accelerated rotation" then the following apply;  (please excuse the crude variables)\n\nw = dθ / dt\na = dw / dt = d^2 θ / dt^2\n\nwhere "w" stands for omega <-- angular velocity\nand "a" stands for alpha <--  angular acceleration.\n\nIf you have had any calculus you can take the first derivitive of your θ expression to get the angular velocity, and then the second derivitive to get the angular acceleration.\n\nsince we have a a simple polynomial the following says\n\nd/dx [x^n] = nx^(n-1)  , however in our case we will be deriving t\n\nTo yeild       w = 8t + 6t^2   , for angular velocity\nand              a = 8 + 12t      , for angular acceleration\n\n\n(a) if you plug in 0 for t then you get θ = 2radians\n(b) its angular velocity is zero when t = 0\n(c) w = 8(4) + 6(4^2) = 128 radians/second\n(d) a = 8 + 12(2) = 32radians/second^2\n(e)  I don't believe its angular acceleratin is constant.  Because if you do some testing with plugging in times for acceleratin and velocity they don't quite add up...   \n\nBut that could be because I am totally wrong.  Anyway that is my stab at it.  Hope it helps
1008231
Set the equation of motion.\nTaking upward direction as positive\ns=vt-1/2gt^2\nat maximum height s=10,v=0\ng=-g since upward direction is positive\nyou get t= sqrt(2s/g)\n=sqrt(2*10/9.8)\n=1.428 secs
849468
gravity at the earth's surface = 9.80 m/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
1012426
Using 9.8 m/s^2 for gravity, the answer is 3,234 J.
654739
The horizontal position of the projectile is described by x(t) = (v_x)*t + h since the horizontal force is zero. v_x is the initial horizontal velocity and h is the initial horizontal position.\nThe vertical net force is -mg, so -mg=ma, thus -g = a so the projectile is accelerated by -g (the acceleration due to gravity at sea level). Integrating, we get y(t) = -(1/2)gt^2 + (v_y)t + k where v_y is the initial vertical velocity and k is the initial vertical position.\nSolving for t in the first equation, we get t(x) = (x - h)/v_x. Since the times in both equations are the same, we plug this value into y(t) to get y(t(x)) = y(x) = -(1/2)g((x - h)/v_x)^2 + (v_y/v_x)(x - h) + k.\nThus, the path of the projectile is a parabola. Setting y=0 will get you the x-intercepts of the parabola, or where the projectile touches the ground.  Note that the projectile starts above ground and time only runs in one direction, so the projectile only touches ground on one root of the equation.
1001657
V= V(i) + AT Initial velocity plus acceleratin x Time\n\n\n   = V(i) + (F/m)T     Acceleration = Force/ Mass\n\nNote  use same system of units
24383
here are some simple mechanics formulas dealing with velocity:\n\nv=d/t\na=v/t\nvf^2=vi^2+2ad\nd=vit + 1/2at^2\n\nwhere v is velocity, vi is initial velocity, vf is final velocity, a is acceleration, and t is time.  also the i and the f after the v are just subscripts, and ^2 just means sqared.
845902
Let’s consider from viewpoint of physics. F=G*M*m/r^2 is gravitation force holding the sat on its orbit. On the other hand F=m*a – centripetal force. Thus a=G*M/r^2, where M is mass of Earth (I don’t remember), r=R+h, R=6.4*10^6m is radius of Earth, h=0.615*10^6m is height of the sat, a is gravitational acceleration at h (like that ‘g’ on the ground), G is gravitational constant  (?? 6.67*10^-11 ??  I don’t remember!), m was mass of the satellite. Another formula for centripetal acceleration for circular track a=r*w^2, w is angular velocity or w=2pi/T, T being period. G*M/r^2=r*(2pi/T)^2. Now T=2pi*r*sqrt(r/(G*M)) seconds
406596
no idea...but I think you use the formula (Gravitational constant x mass 1 x mass 2) / distance squared
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)
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.