where does 9.8 m/s to gravity (squared) come from? from what equation? from who?
818751
Newton.  Freely falling bodies fall at a constant rate. Distance/Time.
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gravity is an inverse square law type graph , you can plot it at y=1/x^2\ntry this site http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html\n\nanother graph is available here\nhttp://csep10.phys.utk.edu/astr162/lect/light/intensity.html
864950
Gravitational Potential Energy = m*g*h joules\n        where   m - mass of the object in kg\n                     g - accelaration due to gravity in m/s^2\n                     h - height of the object from the reference in m\nHere the height from reference is 500m\ng is taken to be a constant and its value is 9.81m/s^2 though it may vary slightly with latitude, altitude and other such factors.\nMass is given as 0.400kg\nThus Potential energy = 0.400 * 9.81 * 500\n          Potential Energy = 1962 joules\n\nHope so u got wat u needed.
860576
Let the tension of string attached to 8.7 kg be T1 and that attached to 2.1 be T2. Let common acceleration be a. Cos 23 = 0.921\n\nequations are:\n\n8.7a = 8.7g - T1\n1.7a = T1 - T2 - 1.7gx0.14xCos 23 = T1 - T2 - 0.219g\n2.1 a= T2 - 2.1 g\n\nSolve thees equations first for a then for T1 and T2\n\nI f you have doubt about the validity of any one of the above equations  you may ask your specific doubts.
700115
Very simple, in your case you only need to convert from N(Newton) to Kg(kilogram) so just divide by 9.8 (gravity constant)\n\nN/m2---- /9.8 ---> Kg/m2\nN/m2<-- X9.8 ---- Kg/m2
825491
x=ut+(1/2)*a*t^2\n\nso, x=(0*16)+0.5*9.81*16^2=1255.68m\nwhere x is distance, u is initial velocity, t=time , v=velocity, a=acceleration(in this case due to gravity, 9.81m/s^2)\nthats a pretty deep well....... Are you sure its 16 seconds and not 1.6?\n\n:-)
364689
One common unit of acceleration is g, one g being the acceleration caused by the gravity of Earth at sea level at 45° latitude (Paris), or about 9.81 m/s².\nThe acceleration due to gravity denoted g (also gee, g-force or g-load) is a non-SI unit of acceleration defined as exactly 9.80665 m/s², which is approximately equal to the acceleration due to gravity on the Earth's surface at sea level.\n\nThe symbol g is properly written in lowercase and italic, to distinguish it from the symbol G, the gravitational constant, which is always written in uppercase; and from g, the abbreviation for gram, which is not italicized.\n\nThis conventional value was established by the 3rd CGPM (1901, CR 70). The total acceleration is found by vector addition of the opposite of the actual acceleration (in the sense of rate of change of velocity) and a vector of 1 g downward for the ordinary gravity (or in space, the gravity there). For example, being accelerated upward with an acceleration of 1 g doubles the experienced gravity. Conversely, weightlessness means an acceleration of 1 g downward in an inertial reference frame.\n\nThe value of g defined above is an arbitrary midrange value on the Earth, approximately equal to the sea level acceleration of free fall at a geodetic latitude of about 45.5°; it is larger in magnitude than the average sea level acceleration on Earth, which is about 9.797 645 m/s². The standard acceleration of free fall is properly written as gn (sometimes g0) to distinguish it from the local value of g that varies with position.\n\nThe units of acceleration due to gravity, meters per second squared, are interchangeable with newtons per kilogram. The quantity, 9.806 65, stays the same. These alternate units may be more helpful when considering problems involving pressure due to gravity, or weight.
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The acceleration due to gravity is independent of the mass of the object.\n\n The speed of any mass thrown up will decreases by an amount 9.8 m/s per second. \n\nIf the initial speed is u, then its kinetic energy is ½ m u u. Its potential energy is zero.\n\nThe total energy is½ m u u + zero = ½ m u u.\n\nAs it goes up its kinetic energy decreases whereas its potential energy increases keeping the total energy same.\n\nWhen it goes up through ‘h’ meter,  its total energy is [ ½ m u u + mgh].\n\n[ ½ m u u + mgh] = ½ m v v.\n\nm u u + 2mgh = m v v. \n\nCancelling m on both sides,\n\nu u + 2gh = v v\n\nThus we have the familiar equation vv = uu + 2gh.\n\nWhen the ball reaches 9 m its speed becomes zero.\n\nHence,  v v = 2 x 9.8 x 9\n\nSolving we get v = 13. 3 m/s . \n\nThe speed with which it is thrown up is 13. 3 m/s.\n\nWhen the height is 8 m, now having known the initial speed we use the same equation for solving for uu.\nvv = uu + 2gh\n\n13.3 x 13.3 = uu + 2 x 9.8 x 8 .\n\nSolving for u, we get  u u = 196 or u = 4.43 m/s.\n\nThe problem can be viewed in other way.\n\nWhen the ball is at its heighest point namely at 9m its potential energy is mgh. where h is 9m\n\nWhen it falls through one meter its kinetic enery increases and potetial energy decreases , keeping the total energy as the same.\n\nmg x 9 = ½  m vv + mg x 8.\n\nOr mg ( 9-8) = ½  mvv \n\ng x 1 = v  v/2\n\n2 g = v v. \n\nv v = 2 x 9.8 = 19.6 (m/s)^2. \n\nv = 4.43 m/s.\n\n Note that the mass gets cancelled and hence we need not know the mass of the ball.
390283
coefficient of friction and specific gravity
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1/2 * mass * velocity^2.
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Equatorial surface gravity \n1.35 m/s²\n0.14g
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Gravity slows down the ball at 9.8m/s/s. After 1s, its velocity is 12-9.8=2.2m/s
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
878213
f = ma = kN = kmg; so a = kg = .15*9.81; where a is the deceleration, k = .15 the friction coefficient and g = 9.81 m/sec^2 the acceleration due to gravity at Earth's surface.\n\nWork = fd = kNd = 1/2 m v^2 = kmgd; where friction force f = kN times distance (d) is work (energy in Joules), which in this case comes from kinetic energy 1/2 m v^2 where v = 5 m/sec and m is the mass of my damn physics book.  Upon rearranging factors to find d, we have:\n\nd = 1/2 v^2/(kg) = v^2/(2kg); where d is the stopping distance, v = 5 m/sec, k = .15 the friction coefficient, and g = 9.81 m/sec^2 from gravity.\n\nYou can do the arithmetic.  The important physics lesson learned here is that neither answer (a or d) depends on the mass of that textbook...and both depend in large measure on k in that larger k means greater deceleration and smaller k means longer stop distance.
866141
You have to calculate p=massXspeed before the "collision".  The quantity remains constant, so if the mass increases, the speed must decreases by the same ratio for the p to stay the same.  (First, you have to convert the 8N weight to a mass using the 9,8N/kg gravity constant)
289324
simple F=m*a\nm=0.8\na= the acceleration due to gravity (9.8 ms^2)\nWeight is 7.84 N\nthe conversion should be in your textbook from newtons to lbs and from lbs to oz
909370
Well, the equation for the time period of a pendulum is\n\nT = 2*pi*sqrt(l/g)\n\nWhere pi is the mathematical constant, 'l' is the length of the string, and 'g' is the acceleration of gravity.\n\nSo substitute your values, and you then have the equation\n\n0.80 = 2*pi*sqrt( l  /9.8 )\n\nYou could solve for the length and plug that back into the equation.\n\nT = 2*pi*sqrt( l / (.55*9.8) )\n\nSubstitute the length in for 'l' and solve for T