A 65 kg person stands on an elevator, as the elevator accelerated upward 2m/s squared, what is his new weight?
957725
kg is a measure of mass and not weight.  Though, it often gets confused.  The mass of a person does not change unless he consume something or eject something from his body.  His mass would not change wether the elevator goes up, down or sideways.  Depends on what the question is looking for.  The most accurate answer should be: \n\n65kg x (9.81+2)m/s^2 = 767.65 N (newton).\n\nThe old weight of the person is:\n\n65kg x 9.81 m/s^2 =637.65 N\n\nIt would be ridiculous if the question asked for the weight in pounds, but here is a conversion:\n\n767.65 N x 0.2248 N/lbf = 172.56 lbf (as in pound-force, which is different from pound-mas (lbm), the common concept of measure in weight most of us are familiar with, 1 lbf = 1 lbm x 32.2 ft/s^2 )\n\nNow that I have confused everyone.  I just wonder if a physics teacher can explain it any better? lol.\n\n\nXR
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
942786
The tension is the weight + the force Fb required to accelerate the bundle along with the rest of the elevator. \nNeed to calculate the elevator's acceleration a\ny = (1/2)*a*t^2 where y = 1m and t = 1.6s. Solve for a.\n\nNow calculate Fb\nFb = m*a. Solve for Fb.\n\nT = W + Fb
537369
If your mass is m, multiply it by g to convert it to N, where, g is the acceleration due to gravity of the place where you currently are.\nFor most of the places on earth, we take the average value of g = 9.81 (m/s)/s.
727100
Remember Newton's second law:  F = ma.  You can rearrange that to read a = F/m.  So the original acceleration was equal to the original force divided by the original mass.  After the adjustments, you have a2 = 3.3F/2.5m.  Therefore, a2 = (3.3/2.5)a, or 1.32a.  The original acceleration was 6 m/s^2, so the new acceleration is (6*1.32) or 7.92 m/s^2.
321671
Ah, this brings me back!\nThis is conservation of momentum problem. That means that the initial momentum must equal the final. You must break up the values to x and y directions.\n\nThe 900kg mass is traveling 18mps * cos(30) = 15.6mps in +x\n                            18mps * sin(30) = 8.98mps in -y\nThe 750kg mass is traveling 0mps in x\n                            23mps in +y\n\nWith these values take the initial momentum as\n\ninit mom in x = 900kg(15.6mps) + 0 = final mom = 1650kg*(v)\nwhere v is final speed in x = 8.5mps\n\ninit mom in y = 900kg(-8.98mps) + 750kg(23mps) \n\nthat equals the final mom in y = 1650kg*(v)\n\nso v=+5.6\n\nIf you know the speed in the x and y directions, convert to vector notation by using s^2=vx^6+vy^2 = 10.2mps\nand the angle is atn(vy/vx)\n\nHope this helps
697796
I assume you mean Kilograms\n\n1kg = 2.2 lbs
456850
Do your home work. \n\nStill I tempt to answer this question.\n\nW= weight of the box\nmu= coefficient of friction.\n\nWe assume there is no acceleration. So force to upward movement is Fup = Wcos(20)*mu + Wsin(20),\n\nsimilarly Fdown = Wcos(20)*mu - wsin(20),\n\nso\nWcos(20)*mu + wsin(20) = 4Wcos(20)*mu - 4wsin(20)\n5Wsin(20) = 3Wcos(20)*mu\nmu = 5sin(20)/3cos(20)\nmu= 5/3 tan (20)\nmu = 0.606
1171207
Let us write coeff. of friction = k     \n                    acceleration due to gravity = g     m/ sq. s\n                    uniform velocity of motion = u      m/s\n                 \n                    Since k=F/R where R = weight of the box acting downward and F = Frictional force acting against the direction of motion.\n                   Hence, F = kR. Again F= ma, where a is the retardation on the box and R=mg.\n                    Therefore, a=kg. (m cancels out on both sides).\n\n                     By using the formula:\n                            v^2=u^2 - 2aS  ( here a is retardation) we have\n                      S= u^2/(2a) = u^2/(2kg).\n\n            Substituting the values:\n                         u=8 m/s\n                         k= 0.16\n                         g=9.8 m/sq. s\n                        The result is S= (8^2)/(2*0.16*9.8) = 20.41 m (approx.).
50475
A)come on, the weight of the man is just his mass times gravity.\nB)this sounds like a torque problem, you need to know where on the beam the man is standing... since the beam is in equilibrium, you know that the sum of all torque is 0. 9im going to use T for torque because i cant get the greek letter i usually use) T=Fr (force times the distance from the pivit point) so all torques directed up minus all torques directed down equal 0. so that means all torques pointing up equal all torques pointing down. ok, so that means that:\n Tbeam + Tman = Tcable.\n and the cable is at an angle so it would actually be FrsinX instead of just Fr.  so now since the beam is uniform, we say that it's weight is in the center, so r is half of the length of the beam. so\n Tbeam=Fr=(288)(2.6) and now we convert the mass of the man to newtons, just multiplying by gravity(which im rounding to 10 because i dont feel like getting a calculator) and getting 750 Newtons. so that means\n Tman=Fr=750r. this has to stay as r because you didnt tell me where on the beam the man is standing. it probably indicate's where he's standing on your diagram. r would be the distance between the wall and the man because we're making the wall our pivot point, because if the end with the cable was our pivot point, everything there(including the cable) would be 0 because it would be 0m away from the pivot point, so we couldnt solve for Ft... so you'll have to plug in the "r" for the man yourself.  and Tcable=FrsinX=F(5.2)sin67 where F is the force of tension which is what you are solving for overall, r is 5.2 because the cable is 5.2m (the length of the beam) away from the pivot point. so now, we plug into what i told you before(Tbeam + Tman = Tcable.) which is true because the beam is in equilibrium. so \n\nTbeam + Tman = Tcable\nFr    + Fr   = FrsinX\n(288)(2.6) + (750)(r) = F(5.20)sin67\n\nplugging in the r which should be on your diagram, you can solve for F whichi is the tension in the cable.
86433
Assuming no external force, the obivous choice is 56.5g because the bouyant force is equal to mass*gravity
1012426
Using 9.8 m/s^2 for gravity, the answer is 3,234 J.
143729
Final veloc. when he reaches the land:\nV=sqr(2*g*h)= 8.743 m/s\nKinetic Energy= Ec= 0.5 * m* V2 = 1872,780 \nThis turns into potential Energy Ep= 1872,780  = mgh\nHe stops in h=0.8m , hence: \nForce = m g = 1872,78 / 0,8 = 2340,795 Newtons\n\nHe has two knees, so F= 2340, 795 / 2 = 1170,488 Newtons in each knee.
883415
according to newton's second law, F=ma\nso, initially, let the force be F1, mass be m1 and the acceleration be a1\nthe final force is F2, mass is m2 and the accleration is a2\nF1=m1a1 and F2=m2a2\nas the force is decreased by a factor of 3.7, F2=F1/3.7\nand since the mass is decreased by a factor of 2.6, m2=m1/2.6\nF1/F2=m1a1/m2a2\nso, 3.7=2.6*a1/a2\nbut a1=8m/s^2\nso, a2=2.6*8/3.7\nso, a2=5.62m/s^2\nthis is the new acceleration
792099
And the acceleration of the elevator is e.\n\nm*g = 477.1 N\nm*(g+e) = m*g + m*e = 652.4 N\n\nSolve for e.
428368
(m)\n\nOne kilo equals 2.2 pounds
921575
9.8 m/s^2 is the acceleration due to earths gravity. calculate the volume of the paint bucket. multiply it by 7.65 to get the mass.