g = 9.81m/sec/sec what is the slope?
155071
9.81 m/s^2 is the acceleration due to gravity on the Earth's surface, it is a constant.\nIf you were to plot the acceleration of an object as it undergoes a constant acceleration, the slope of the line would be zero.
825498
Kinetic Energy = 1/2mv^2; Potential Energy = mgh\n\n1/2mv^2 = mgh or 1/2v^2 = gh\n\nv^2 = 2gh = 2 * 9.81m/s^2 * 0.49 m\n\nv = 3.1 m/s
889368
The first step is to set up two simultaneous equations for the acceleration upward and the acceleration downward\n\nFor upward:\nmg+ma=771.7\nfor downward\nmg-ma=318.8\n\nadd the two equations together:\n2mg=771.7+318.8\n\nm=(771.7+318.8)/(2*9.8)\n=55.64\n\nThen solve for a in either equation and check with the other\n\na=771.7/55.64 - 9.8\n=4.1\n\ncheck\na=9.8-318.8/55.64 \na=4.1\n\nj
943006
The balance of forces acting on you: m*a=f1-f2; m is your mass, f1 is the force pulling you down the slope, f2 is force of friction, both being parallel to the slope incline. Thus f1=m*g*sin(b), f2=m*g*cos(b)*k, while g=9.8m/s^2, b=31deg, a=1.16m/s^2, m*g being your weight. Mind! While f1 is a vector parallel to the slope, the force m*g*cos(b) is normal to the slope and should not be taken into account unless friction be present. \nNow m*a=m*g*(sin(b)-k*cos(b)), hence k=(sinb-a/g)/cosb=0.46
831310
It seems the 16 m/s is not relative to this question. That is velocity in the horizontal direction. Gravity only effects in the verticle in this case. \n\ntime = sqrt(2 x height/ g) = sqrt(2 x 45 x 9.8) = 3.03 seconds\n\nvelocity = time x acceleration of gravity = 3.03 x 9.8 = 29.694m/s
371794
slope = (-1-2)/(-1-0) = -3/-1 = 3
858740
I assume you mean on Earth where the gravitation constant (g) is 9.8 m/s²\nF = mg\n\n300 = 9.8 * m\nm = 300 / 9.8\nm = 30.6 kg\n\nIf you are using a rounded value for g (10 m/s²), you will come up with a mass of 30 kg.  This might be the intended answer if they want you to do it in your head.
743106
a) it is always same speed.\nb) Vf=at+Vi\nVf= -8m/s (because it is going downward)\nvi= 8m8s\na= -9.8m/s^2\n\n-8m/s=(-9.8m/s^2)t+8m/s\n-16m/s=(-9.8m/s^2)t\nt=1.63 s
508539
A) F = mg = 50 kg x 9,81 m/s2 = 0,49 kN.\nB) F = GMm/R2 = 6.67E-11 x 5.98E24 x 50 / (6.370E6)^2 N\n   = 0,49 N
1100415
1)Acceleration is the change in velocity (speed) per unit time\n\nVelocity = (meters/second)\nAcceleration = \n(meters/second) / second = meters/second-squared\n\nAcceleration = meters per seconds squared or D\n\n2)B\n\nFinal position = \n(initial position) + (initial velocity * time) + (-9.8/2 * time squared) / 2\n\nX is the height of the ledge\nInitial velocity is zero and so is final position:\n\n0[m] =  X[m]  + 0[m/s] - (4.9[m/sec sqrd] * 64[sec sqrd)\n\nX = 313.6 [m] or meters
332219
An egg is pretty aerodynamic, so we can neglect friction. Thus, a falling object satisfies the equation\n\ns(t) = -0.5*g*t^2+v0*t+s0\n\nwhere g is 9.82 m/s^2, v0 is the start velocity (in this case, 0) and s0 is the start height (in this case, 3 metres).\n\nSo, letting s(t)=0 and solving for t, we get\n\n0 = -0.5*g*t^2 + 3\n\nt = sqrt(6/g) = 0.78 seconds.
913283
T^2=4pi^2l/g\ng=4 pi^2l/T^2\n=4*(3.14)^2(1.5)/(180/27.3)^2\n=1.36 m/s^2
864791
A.) 3.02891 seconds\nB.) 290.1 m/s\n\nTrust me... =)
446591
1)X=V1t+1/2gt^2\nat the interval between getting to the highest point and being dropped back to the first place is half of the full time = 2 seconds\nV1 at the highest point = 0\nX=displacement\nX=0+1/2*9.8*4=19.6 m \n2)V2=V1+gt\nV2 = velocity of landing\nV1 = velocity at highest point\nV2 = 0 + 9.8 * 2 = 19.6 m/sec
75525
a) 72.072 cm/sec^2\n\nb) 82.63 cm/sec^2
418402
Using one of the equations of linear motion:\n\nv² = u² + 2as\n\nInitial speed u = 0\nAcceleration a = 32 ft/s\nDistance s = 9 ft\n\nSolve for final velocity v = sqrt( 2 * 32 * 9) = 6.2 ft/s\n\nThis assumes negligible air friction.
569320
v=vo + at\nv=0 + (9.8)(8)\nv=78.4\n\ny=1/2(vo + v)t\ny=1/2(78.4)(8)\ny=1/2(627.2)\ny=313.6m\n\nB