compare the accelerations of the two pins while they are in the air.?
913200
Both their accelerations will be the same and equal to the acceleration due to gravity, g=9.81 m/s/s...This is coz the only force acting on them is the force due to gravity
545558
Force = mass * acceleration\nForce = 1300 kg * 2.6 m/s2\nForce = 3380 Newtons
1040756
A. 3.3\nThe two parellel resistors receive the same voltage.  Just use ohms law to find the current.  40/12 = 3.3
118764
Newtons 2nd Law,\nForce = mass * acceleration\n\nIf you have two masses (m1 & m2) that accelerate at the same rate,\nF1 = m1*a --> a = F1 / m1\nF2 = m2*a --> a = F2 / m2\n\nset both acclerations equal to eachother, solve for the ratio,\nF1/m1 = F2/m2\nF1 = F2 * m1 / m2 --> F1/F2 = m1/m2\n\nSo the ratio of Force 1 (F1) to Force 2 (F2) must equal the ratio of mass 1 (m1) to mass 2 (m2) in order for both to accelerate equally.
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.
729735
both are the same if it is drop at the same height because the gravity speed is 10 ms-2 (9.81ms-2 more accurate)
660156
http://www.physlink.com/education/askexperts/ae25.cfm
77860
hey, didnt i answer something very similar to this for you? you need to do F=ma so 43000000N/2100000kg then you get an acceleration, but then you have to subtract gravity from this because thats the acceleration produced by the rocket but there is still gravity pulling with 9.8 m/s^2 in the opposite direction. so just do (43000000/2100000) - 9.8, and that's your answer :)
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)
634770
Yep! Pinkruth is right!! Pins = legs. "Do it yourself!"
124298
Answer to #2\nt=1.565 seg\nV= 2.556 m/seg\nt=1.565 seg, the same.
791525
questions are very easy. You must try them by yourself. I will only give hints-\n1> Instantaneous accl  = dv/dt\nAvg accl = (v2-v1)/(t2-t1)\n\n2> draw the vectors, take the components separately along x and y axes.\nremember accl component along x axis increases velocity along x axis, and accl component along y axis increases velocity along y axis.\n3> use the basic kinematics equations like v^2=u^2+2as etc
322347
it is the collision of two preasure sysems and the difference of the tempeture
524636
If the forces are both acting in the same direction, add them together. = 65 Newtons
45821
are you sure that you want acceleration instead of velocity??\n\nwell, the acceleration is going to be 0.  since the speed is going to be the same the whole time, there won't be any acceleration.
969523
if they are both going at the same speed, then the mass,size of the wheel and other considerations have already been take in account, so they both should arrive at the same time.\nGod bless,\ngabe
913014
F(net) = m x a\n\nF (net) = F (forward) - F(air friction)\n\nso  F(air friction)  = F (net) - F(forward)\n\nMass is given 20 kg\n \nnet acceleration given 2.0  m/s^2\n\n\nSolve for the Net Force, subtract the forward force, the difference is the force due to air friction.