Object A has a mass 2.0kg and is traveling to the right at 3.5m/s when it collides with object B which has a m
95497
This a momentum question.\n\nRemember momentum is:\n\nm * V\n\nm = mass\nV = velocity\n\nSo the conservation of momentum is:\n\n(m1 * V1) + (m2 * V2) = (m1 * V3) + (m1 * V4)\n\nMomentum (before collision) = Momentum (after collision)\n\n(Notice mass does not change after the collision, but velocity does.)\n\nSo let's plug in the variables:\n\nm1 = 2.0 kg\nV1 = 3.5 m/s\n\nm2 = 4.5 kg\nV2 = -5.0 m/s \n(negative indicates object is traveling to the left)\n\nV3 = ?\nV4 = -1.8 m/s\n\n(2.0 * 3.5) - (4.5 * 5.0) = (2.0 * V3) - (4.5 * 1.8)\n\n(7) - (22.5) = 2(V3) - (8.1)\n\n-15.5 = 2(V3) - (8.1)\n\n-7.4 = 2(V3)\n\nV3 = -3.7 m/s \n\n(the negative sign just indicates direction with relation to the axis; in this problem, negative indicates the object is traveling to the left)\n\nSo Object A will be traveling to the left at 3.7 m/s after the collision.
573522
kinetic energy is 1/2 mv² so it depends on mass and velocity
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)
700987
The energy taken to slow the object from v1 to v2 is\n\nE=FS=(1/2)m(sqr(v2)-sqr(v1))\n\nand given v2=(1/2)(v1)\n\nWe get  v1=sqrt((8/3)(FS/m)) = 12m/s\n\nAlso from v2=v1+at     and          F=ma\n\nt=v1/(2a)=mv1/(2F) = 1 second
1120781
do your own homework\n\nits probably 0.26 kg M/s ^2 since its a velocity\n\nyou also probably need to convert it to rad/s whatever that is
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
997326
Use Newton's universal gravitation equation, then sum the forces.\n\nF = G*m1*m2/r^2\n\nF = force (newtons)\nG = gravitational constant (6.6742 x 10^-11)\nm1 = mass of object 1 (kg)\nm2 = mass of object 2 (kg)\nr = distance between objects (meters) ==> don't forget to convert the units\n\nIn this case, you do it twice:\nF1 = G * 390 * 390 / 0.11^2 = 0.84mN to the left\nF2 = G * 390 * 390 / 0.39^2 = 0.067mN to the right \n\nNet Force = 0.84 - 0.067 = 0.772mN to the left
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
792153
‘impulse’ or momentum\n=m*(v2-v1)=0.42*(-17-4.1)=-8.862 kg*m/s
125621
B) Increase\n\nThe difference in the two objects' velocities would be constant since that are both accelerating at the same rate, but the first object already had a higher initial velocity (from Object B' point of view).\n\nThe equation relating the distance traveled with respect to initial velocity, accelertion and time is,\nd = v_i * t + 1/2 a*t^2\nSo for both objects, after B is released the 1/2 a*t^2 part of the equation is the same, so we can ignore that part and focus on the v_i * t part.\nIf we start t=0 at the moment object B is released, object A's velocity is some non-zero number while B has no initial velocity.  So as time goes on, the distance between the 2 objects inceases.
217138
the particles move faster.
824196
conservation of linear moment:\nm1*v1 = m2*v2\nm1*v1/m2 = v2\n(100.0 kg)(3.0 m/s)/(175.0 kg) = v2 = 1.71 m/s
754995
Nicole I think that to understand what is to be done here I suggest to assume that a coordinate system be attached to one of the objects. I-don’t-remember-who says, but << any reference system at rest or moving with linear constant speed has no privilege with others the like>>.\n(x) m2=4kg is reference system #2, where v2=0, and m1=2kg is moving with speed v=8+12=20 ms, impact by m1 on m2 M1=m1*v=40kg*m/s\n(y) m1=2kg is reference system #1, where v1=0, and m2=4kg is moving with speed v=8+12=20 ms, impact by m2 on m1 M2=m2*v=80kg*m/s, so R=40/80=0.5
759333
all object fall 4.9m after 1 sec\n\nX=1/2(9.8m/s^2)(2s)^2\nx=19.6m\n\nx=(1/2)(9.8m/s^2)(3s)^2\nx=44.1m
744405
a) (speed is actually closer to 3.97...not a big deal)\n\nb) Conservation of energy:\nThe sum of potential and kinetic energies when the bullet first hits the block and when it is at the top of the swing is the same:\nU(bottom) + K(bottom) = U(top) + K(top)\nNow, the speed at the top is zero, so the ewuation becomes:\nU(bottom) + K(bottom) = U(top)\nSolve for the change in potential energy:\n[U(top) - U(bottom)] = K(bottom)\nThe change in potential energy due to gravity is: (m1+m2)gh\nThe kinetic energy at the bottom is: (1/2)(m1+m2)v²\nSo the equation becomes:\n(m1+m2)gh = (1/2)(m1+m2)v²\nThe (m1+m2) terms cancel:\ngh = (1/2)v²\nSolve for h:\nh = v² / (2g) = 0.8 m
788625
This is a momentum question.  If the end result is comming to a halt, then then each vehicles' momentum was equal and opposite.  Momentun (M) = mass (m) * velocity (v),  therefore you are left needing to solve the equation m1*v1 = m2*v2...you can do the rest I hope....
685470
the speed = distance/time