A canteloupe is rolled off the edge of a building roof with a velocity of ?
960791
First I am assuming that the roof is flat.  This means that the canteloupe had a vertical velocity of 0 m/s when it started.  (All the velocity is horizontal)\n\nh = 1/2 a t^2\nh = 0.5*9.8*3^2\nh = 44.1 m high\n\nv = x/t\n5.4 = x/3\nx = 16.2 m away from the building\n\nThe mellon splatters.
641866
wind blows from an area of high pressure to an area of low pressure.
858717
oblique, obilisque shape
616835
The shape of a roof approximates the shape of an airplane wing. Air flowing rapidly over the roof generates lift, lifting the roof off the building.
882625
during the throwing motion the ball is accelerated in the opposite dirrection as gravity. once the ball leaves the hand of the thrower however, the acceleartion is constant, and is equal to that of gravity, g = 9.80 m/s^2
969870
It's difficult to know your options without knowing how big the building is and what shape it is. If the walls are at right angles then the length of the curved corner will be a quarter or the circumference from which you can find the radius. The radius would be twice the length divided by pi.\nIf not then, aside from taking a mould and working with tangents, I can't think of an easy way.\nThe answer above will work very well if you have access to the face of the curve.
919241
rolling on floor ....
268859
it gets its shape from the wind the wind pushes the top
757387
(a) Assuming the base is a half-circle and that there are no other forces acting on the mass (friction, drag, etc), then the kinetic energy at the base is equal to the potential energy at the top. Note that the change in potential energy is not affected by the semi-circular path the mass follows, but rather the net change in height. This net change is simply the radius of the half-circle container:\n\nKE = mgR = (.71 g * 1 kg/1000 g)(9.8 m/s²)(25 cm * 1 m/100 cm)\n\nKE = .00174 J\n\n(b) The kinetic energy associated with rotation is given as:\n\nKER = (1/2) I w², where w is the rotational velocity. Note that since the sphere rolls without slipping, the translational velocity of its cener of mass is simply:\n\nv = w * r\n\nSubstituting this into the expression for the rotational kinetic energy:\n\nKER = (1/2) (I / r²) v²\n\nNote that the expression for the translational kinetic energy is:\n\nKET = (1/2) m v²\n\nThus, the total kinetic energy of the ball at the base of the hemisphere is;\n\nKE = KER + KET = (v²/2)( I/r² + m)\n\nThe fraction of the rotational kinetic energy over the total is:\n\nf = KER / KE = (v²/2)( I/r²) / (v²/2)( I/r² + m)\n\nf = ( I/r²)/( I/r² + m)\n\nFinally, the moment of inertia, I, of a sphere is:\n\nI = (2/5)mr²\n\nSubstituting this into the expression for f:\n\nf = (2/5)m/( (2/5)m + m)\n\nf = (2/5)/( (2/5) + 1)\n\nf = 2/7\n\n(c) At the base of the hemisphere, there are two forces acting on the ball: the force of gravity and the centripetal force. At the base, these forces are coincident (act in the same direction). Thus the normal force, which opposes these two, is simply:\n\nN = Fgrav + Fcent\n\nN = mg + mv²/R\n\nTo find v, recall that KET / KE is 2/7. Using this, we can manipulate the terms to get v:\n\n(1/2) (I / r) v² = (2/7)(mgR)\n\n(1/2) ((2/5)m) v² = (2/7)(mgR)\n\n(1/5) v² = (2/7)(gR)\n\nv = sqrt((10/7)(gR))\n\nv = sqrt((10/7)(9.8 * 25 cm * 1m/100 cm))\n\nv = sqrt((10/7)(9.8 * 25 cm * 1m/100 cm))\n\nv = 1.871 m/s\n\nFinally, the normal force is computed as:\n\nN = mg + mv²/R\n\nN = (.71 g)(9.81 m/s²) + (.71 g)(1.871 m/s)²/(25 cm)\n\nN = (.71 g)(9.81 m/s²) + (.71 g)(1.871 m/s)²/(25 cm)\n\nN = .0169 Newtons\n\n\nGood Luck! :)
670367
Unidentified flying object.
162564
Potential Energy = stationary boulder on top of a building\nKinetic Energy = once the boulder is dropped and has speed
1069669
gravity is stronger, leave it down
863271
mass doesn't matter here, just the acceleration\nv=u+at\na=(v-u)/t\na=(9-0)/0.15=60 m/s^2
177575
If the fluid is brought to a rest by the beam then the drag coeff Cd is one.\nhttp://en.wikipedia.org/wiki/Drag_coefficient\n\nThis paper seems relevant to your calculations.\nhttp://www.artificialreefs.org/ScientificReports/CancunHotelRBstability.htm
735384
Using the law of conservation of energy determin\n\na) the vertical component of vector V(initial)\n\nIt went up 20 meters, mgh = 1/2mV^2, solve for velocity\n\nsqrt[2(9.8m/sec^2*20meters)] = v = 19.8m/sec\n\nb) the work done by the gravitational force on the particle during its motion from P to the ground.\n\nMaybe this is a trick question, I will take it at face value, when at P it is 60meters above the ground, the work of the gravity bringing it down from P is just mgh = 0.5kg*9.8m/s^2*20m=98J\n\nc) the horizontal and vertical components of the velocity vector when the particle reaches the ground:\n\nHorizontal velocity is constant at 30m/sec, nothing to calculate.\n\nVerticle velocity, keep in mind that the object reached its maximum height at 80 meters high (ie from a verticle perspective is is at zero velocity at that point, and begins to fall, the terminal vertical velocity comes from the potential energy of \nmgh = 0.5kg*9.8m/s^2*80m = 1/2 mV^2, solve for v\n\nvertical velocity is 39.6 m/sec downward\n\nplease check for calculation errors.
468173
it depends on how much the ground "gives". when it falls from a known height, it has potential enrgy=mgh, which translates to kinetic energy 1/2mv^2, so impact velocity is v = sqrt(2gh). final velocity (assuming inelastic collision) is 0. if you know how long it takes to deaccelerte to 0, the average force is f=ma=m*v/t = m*sqrt(2gh)/t. if you know how far it travels before stopping, v^2 = 2ax, a=v^2/2x, f=ma=mgh/x\n\n\ni realize an easier way to do this is to say that the work done by the ground is the same as the force on the object times the distance traveled through the ground. since \nWork = change in Energy\nEnergy = mgh\ndistance into ground = x,\nW = Fx\nW= E = mgh\nmgh= Fx\nF = mgh/x
429577
a donut!!!! it's round...so no beginning or end, and there isn't a middle LOL!!