An orderly exerts a horizontal force of 50 N on a gurney with a patient on it. If the gurney?
102595
50 N - friction = m*a= 90*0.35 N\nfriction = 50N - 31.5N = 18.5N
50723
its his weight * gravity. so the tension is: 75*9.8 which is about 735 N
354034
Forget the 6000N horizontal for for a second...\n\nYou have 5000N acting downward (negative) and 7000N acting upward (positive). The total net force acting vertically to the A/C is 7000-5000= +2000N so upward. We know that the horizontal force of 6000N is acting at a 90 degree angle to the upward force so the resulting forces make a right triangle. Use Pathagorean theorem.\n\n2000^2 + 6000^2 = x^2\n\nx is your MAGNITUDE.\n\nNow find the angle:\n\nUse trigonometry: \n\nTan(Angle) = opposite/adjacent\nwe know the opposite is 2000 and the adjacent is 6000 so,\ninverse Tan(2000/6000) = 18.4 degrees, upward and forward.
942588
zero, \n\nno acceraltion = no net force\n\nmathematically  Net F = ma\n\nif the right side is 0, left side is too
200510
The answer given by "Amanda" is actually quite funny, I like it.\n\nWork = Force * Distance applied\nIf a constant force of 45 Newtons is applies over a distance of 4 meters, the work done would be 180 Joules.\nI hope we can assume this 45 Newton force is being applied parallel to the ramp due to the weight (force of gravity) of the object.
903675
The frictional force acting against the motion is F\n\n= (mu)x mg x cos a. \n\n= 0.2 x 22 x 9.8 x cos 25 = 39 N. \n\nThe net force acting along the inclined plane\n\n= (mg sin a +39) Since both the forces are acting opposite to the motion.\n\n= 22 x 9.8 x sin 25 + 39\n\n= 130. 12 N. \n\nIf x is the distance moved along the plane, the work done by this net force is Fx.\n\nThis work done is used to remove the kinetic energy of the body = 0.5 m v v.\n\nFx = 0.5 m v v \n\nx = 0.5 x 22 x 3.9 x 3.9 / 130.12 = 1.28m.
40169
The force on the system is 3.0 kg x g. (the only force is due to gravity acting on the hanging mass) The mass of the system is 3.0 kg + 5.0 kg = 8.0 kg.  The acceleration of the system is a = force/mass = 3.0 kg x g / 8.0 kg   =  3 / 8 g or 0.375 x 9.8 = 3.675 m/sec^2
724756
For a body in circular motion there will be a centripetal force\n= M V V /R.\n\nIn this case the tension of the string provides this force.\n\nMaxinmum tension = 50N.\n\nThat is the maximum centripetal force is 50 N\n \nM V V /R = 50\n\nV V = 50 x R / M = 50 x 1/ 2 =25 (m/s)^2.\n\n The maximum speed is V = 5 m/s.
928343
Torque.\n\nThe weight of each particle is mg. The torque caused by particle 1 is mg * 20, counterclockwise, and the torque from particle 2 is mg * 80, clockwise. Therefore the net torque is mg * 60, clockwise.\n\nSo the force on the left particle is (mg * 60) / 20 = 3 mg, and the force on the right particle is (mg * 60) / 80 = 3/4 mg. Both of these are clockwise, so the left particle goes up and the right particle goes down. To get acceleration, just cancel m.
112424
the person to the left has the right a way
985678
Incoming momentum = p1 = mv1\nOutgoing momentum = p2  = mv2\nChange in momentum = p2 - p1 = m(v2-v1)\nThis change occurs in time, t.\nRate of change of momentum = m(v2-v1)/t = impressed force, f.\nm=0.149 kg, v1 = 43.4 m/s, v2 = -48.7 m/s, t = 1.1x10^-3 sec\n\nf = m(v2-v1)/t\nYou do the arithmetic.
590410
Emergency on the way home. An adult trying to take a child. The child may fall and need help up. The child may see another child injured. Are you the parent of the child?
813186
a counterbalancing force or system of forces.
58327
Pressure goes up as temperature goes up.
109486
This is an Static problem:\nThe forces in the wire are:\nIn the middle of the rope the mass of Jonathan: 50kg.\nRigth side: T1\nLeft side: T2\n\nThis is a Vectorial equation: T1+T2=mg\n\nSince Johnatan is in the middle of the wire T1=T2... \n2*T1=mg\nT1 has two components T1x and T1y (horizontal and vertical)\nThe vertical component of T1 is the one that act against weight.\nT1y=T1*Sin(10)\nThen:\n2*T1*Sin(10)=mg\n[using mks system g=9.8m/s^2]\nT1=mg/[2*Sin(10)]\nT1=50*9.8/[2*Sin(10)]\nT1= 1410 Newtons.\nThe tension at "this point" is 2T1 =2820 N.\nHope it helps
879813
you need to identify that there're two parts of the force, one acting horizontally, the other acting vertically.\nSo:\n1) the vertical part is:\n70sin30\n2) tje horizontal part is:\n70cos30
880761
People above neglected the vertical component of 460 N.  This must be added to the weight of the box.  Also, not sure why they took 350 N and multiplied it by acceleration.\n\nso F_vert=460sin35.6=267.78 N (downward)\n     F_hor=460cos35.6=374.0 N\n\ncalculating F_frict=mu x (F_normal)\n\nF_normal=weight of box + F_vert = 267.78+350=617.78\n\nso F_frict=0.57x617.78=352.13 N\n\nnow the net force on the box = 374.0 - 352.13 = 21.87 N\n\nand we need mass of the box, m=350/9.81=35.7 kg\n\nso a=21.87/35.7=0.6126 m/s2\n\nfinally,  d=V_initial x t + 0.5 x a x t^2, V_initial=0\n\nso t=((4.2)/(0.5x0.6126))^0.5=3.7 sec\n\n**********************************************************\n\nHowever, should you wish to ignore the vert component\n\nF_frict=mu x (F_normal)=0.57 x 350 = 199.5 N\n\nso net horizontal force is:\n\nF= 374.0 - 199.5 = 174.5 N, a=174.5/35.7= 4.89 m/s2\n\nplug a into:\n\nd=V_initial x t + 0.5 x a x t^2, and solve for t\n\nt=((4.2)/(0.5x4.89))^0.5 = 1.31 sec