A large rocket has a mass at takeoff of 2.0 X 10^6 kg. What is its initial acceleration if its?
153995
F=m*a\na=F/m = 3.5 x 10^7N/(2.0x10^6kg) = 17.5 m/s^2 or 1.8g
439869
Found at Wikipedia:\n\nHeight: 184.2 ft (56.14 m) \nGross Liftoff Weight: 4.5 million lb (2.04 million kg) \nTotal Liftoff Thrust: 6.781 million lbf (3.076 million kgf) \n\nconvert kilograms to newtons by multiplying by 9.8
787981
let al' mass be m\n\nconserving momentum\nm*0.91=(158-m).829\nsolving for m we get\n         m=75.320kg
666681
1/2 * mass * velocity^2.
1163903
Set m1 = 0.145 kg, v1i = 0.720 m/s\nand m2 = 0.295 kg, v2i = 2.22 m/s\n\nConservation of linear momentum:\nm1v1i + m2v2i = m1v1f  + m2v2f\nSimplified\nm1(v1i-v1f) = -m2(v2i-v2f)\n\nConservation of energy (kinetic only):\n(1/2)m1v1i^2 + (1/2)m2v2i^2 = (1/2)m1v1f^2 + (1/2)m2v2f^2\nSimplified:\nm1(v1i - v1f)(v1i + v1f) = -m2(v2i - v2f)(v2i + v2f) \n\nSolving the two simplified equations:\nv1f =((m1-m2)/(m1+m2))v1i + ((2m2)/(m1+m2))v2i\nand\nv2f =((m2-m1)/(m1+m2))v2i + ((2m1)/(m1+m2))v1i\n\nTherefore,\nv1f = 3.315 m/s\nv2f = 1.23 m/s
939750
W=mg ... where g is the gravitational acceleration
83305
I suppose the problem was already due so I'll finish it for you now.\n\nF=m*a\na=F/m= 2*10^6N/1.2*10^6kg = 1.67m/s^2 * g/9.81m/s^2 = 0.17g
767660
A rocket is subjected to four forces in flight; weight, thrust, and the aerodynamic forces, lift and drag.\n\nat rest I will denote the launch pressure by the symbol ps. The pressure exerts an equal force on all sides of the pressure vessel, the cone of the engine. Because the rocket is free to move along the launch mechisims, the pressure exerts a vertical force given by: \n\nFp = (ps - po) * A \n\nwhere Fp is the pressure force, po is the atmospheric pressure, and A is the cross-sectional area of the rocket tube. The net force acting on the rocket is the difference of the vertical pressure force and the weight W of the rocket: \n\nFnet = (ps - po) * A - W \n\nTo simplify the analysis, we are going to assume that the length of the launch tube is small relative to the length of the rocket. As the rocket moves, the pressure remains fairly constant inside the tube. In reality, there is a slight decrease in pressure as the volume increase, but we are going to neglect that effect to simplify this analysis. With a constant pressure, and constant pressure force, we can use a simple algebraic form of Newton's second law to determine the acceleration a, velocity v and the distance y which the rocket moves along the launch tube: \n\nFnet = m * a \n\nm * a = (ps - po) * A - W \n\nwhere m is the mass of the rocket This equation can also be written as: \n\nW * a = g * ((ps - po) * A - W) \n\nwhere g is the gravitational acceleration, which is equal to 32.2 ft/sec^2 or 9.8 m/sec^2 on the surface of the Earth. The value of g is different on the Moon, and on Mars. The weight W is equal to the mass times the gravitational acceleration. Now divide both sides by the weight to determine the rocket acceleration along the launch tube: \n\na = g * ((ps - po) * A / W - 1) \n\nFor a constant acceleration, the velocity v and distance traveled y are given by: \n\nv = a * t \n\ny = .5 * a * t^2 \n\nwhere t is the time. Using the length of the launch tube L, we can now solve for the lift off time - TLO by using the second equation: \n\nTLO = sqrt (2 * L / a) \n\nand then using this time, we can use the first equation to solve for the velocity at the end of the tube: \n\nv = TLO * a \n\nv = sqrt ( 2 * L * a) \n\nAs the rocket clears the end of the launch tube (t = TLO+), the air inside the rocket, at pressure ps, will rush out the exit and the pressure will equalize with free stream inside the tube. In reality, this will provide an additional amount of thrust T given by: \n\nT = m dot * Ve \n\nwhere m dot is the air mass flow rate out of the tube, Ve is the exit velocity of the air. For an air rocket, with no nozzle and small volume, this additional thrust is small relative to the weight of the rocket, and we will ignore this effect. \n\nThe velocity v at the end of the launch tube is used as the initial velocity in the flight equations to determine the flight trajectory of a rocket.
716722
For launch velocity\nv0 = 4.47 m/s,\nlaunch angle\nθ = 66 degrees,\nThe horizontal range is\nR = 1.515173718712374 m.\nThe total time of flight is\nt = 0.83337718278825 s.\nThe peak height is\nh = 0.8507839727702984 m.
1001506
Dude! all you have to do is increase the velocity!\nOr Dude! I got it! E= MC2
545558
Force = mass * acceleration\nForce = 1300 kg * 2.6 m/s2\nForce = 3380 Newtons
835037
Let the two case be identified as having  moment-of-inertia I1 and I2. \nTorque T on to the each of the two bodies produces angular accelerations (alfa1 and alfa2) given  by the relation T= I x alpha \n\nalfa1 = T/ I1 and alfa2 = T/I2\n\nStarting form rest(wi =0), let the time taken to reach final angular velocity wf be t1\nusing:  wf =- wi + alfa x t  \n \nwf/alfa1 = t1 and wf/alfa2 = t2 \nleading to \n\nt2  = alfa1x t1/alfa2 \n\n     = I2 xt1/I1\n     = (I2/I1) x t1\n     =. ......Ans \n(Hoping you will put the values and get the answer you want)
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
725904
this is the amount of force.  Force equals mass times acceleration.  The mass is ur bodyweight in kilos and the acceleration is gravity 9.8m/s/s
35590
I concur with previous answerer. The acceleration is 1.5 m/s^2. Rearrange Newton's Second Law, to solve for the acceleration. Then, substitute what you know.
233618
1 mass\n2 magnitude of separation
564218
The acceleration along the plane is g cos30.\n \nInitial velocity is zer0. Distance moved is 1m.\n\nUsing v^2 = u^2 + 2as; v^2 = 16.974; v = 4.12m/s.\n\na) =4.12m/s.\n\nInitial momentum before impact = mass x velocity.\n\nMass of the first body = (24.3/9.8) kg; velocity = 4.12m/s.\n\nInitial momentum before impact = (24.3/9.8)x 4.12 =10.09 kgm/s.\n\nMomentum after impact = Total mass x velocity = {(24.3/9.8 + 35.7/9.8) x v\n\n These two are equal. Therefore, {(24.3/9.8 + 35.7/9.8) x v = (24.3/9.8)x 4.12\n\nOr (24.3 + 35.7) x v = (24.3) x 4.12; v = 1.67m/s.\n\nb) =1.67m/s.\n\nFriction is 5 N; therefore acceleration is Force / mass =- 5/ (24.3 + 35.7) = - (5/60).\n\nMinus is because the force is opposite.\n\nInitial velocity is 1.67m/s. Final velocity is zero. Acceleration is - (5/60) = 1/12\n\nUsing v = u +at; t = v - u /a; t = 0 - {- 1.67 x12} = 20.04 s.\n\nc) =20.04 s.\n\nUsing “v^2 = u^2 + 2as”; s = u^2’2a =1.67 x1.67 x 6 = 16.73 m.\n\nOr using s = average velocity x time s = 1.67 x 20.04 /2 = 16.73 m\n\nd) =16.73 s.