Physics Homework. Help!!?
969410
Given: m=1.6*10^7kg\n           F=8.5*10^5N\n\nI only guessed that the above is what you meant.\n\nSolve for a, using the formula:\n\nF=Ma\n\n8.5*10^5=1.6*10^7a\na=8.5*10^5/1.6*10^7\n  =5.31*10^-2m/s^2\n\nv=70km/h =70*1000/3600m/s\n  =19.44m/s\nu=0\n\nTo solve for time, t, use the formula:\n\nv=u+at\n\n19.44=0+5.31*10^-2t\nt=19.44/5.31*10^-2\n =3.66*10^2\n =366s
396330
Another homework question.
539746
F = ma\n\nF - Force\nm - mass\na - acceleration
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 :)
1119730
ahh, good theory problem...\nlet's see, there's energy, friction, force (normal, tension, as well as the other ones mentioned), gravitation, mass, pendulum motion, gotta be more that I've missed, but no doubt someone'll be here to correct me.
800064
For these problems, it would be helpful to draw a free-body diagram to represent all the forces on this problem.  Here you have 4:\n\n1) Weight in the vertical direction going down, let's denote it W\n2) Force, moving in the x-direction, denote it F_1\n3) frictional force, moving opposite of Force, denote it F_2\n4) Normal force, perpendicular place of contact (opposite direction of W), denote it N\n\nNow use Newton's 2nd law to write equations in each direction:\n\ny-direction: \nN - W =0 (because the block does not move in the y-direction).  Therefore:\n\nN = W, where W = m*g\n\nx-direction:\nF_1 - F_2 = 0 [because the block is just about to move when F_1 is applied].  Therefore:\n\nF_1 = F_2, where F_2 = u*N\n\nMaking the necessary substitutions for F_1 and N, you get:\n\n30 = u*(4.4)(9.8).  Solving for u, you get --------->\n\nu = 0.7\n\n--------\n\nHope this helps
565790
g=m*a\nGravitational Force=Mass * Acceleration
75525
a) 72.072 cm/sec^2\n\nb) 82.63 cm/sec^2
923764
a) F = ma\n    F = 0.0412 kg * 9.8m/s^2\n    F = 0.404 N\n\nb) F = 0.0370 kg * 9.8 m/s^2\n    F = 0.363 N\n\nc) The difference in "weight" is equivalent the weight of the displaced volume of water...  So...\n   0.404N - 0.363N = V * 1000 kg/m^3 * 9.8 m/s^2\n   0.041N = V * 9800N/m^3\n   V = 4.18 * 10^-6 m^3   or  4.18 cm^3\n\nd) Oh Juliet you are my density.... (bad joke!)\n     Density = m/V\n     D = 0.0412 kg / 4.18  * 10^-6 m^3\n     D = 9848 kg/m^3
127000
This problem is solved with the application of work and energy principle which states that total work done on a moving particle is equal to the change of its kinematics energy:\nW=T2-T1\nHere the only work done on the particle is dot to gravitation force and is equal to -m*g*h:\n\n-m*10*0.45 = (1/2)*m*(1)^2-(1/2)*m*(v)^2 --->\nm*v^2=(4.5+0.5)*m*2=10*m --->\nv=sqrt(10) m/s\n\nv is equal to sqare root of 10.(approximately 3.162 m/s)
558215
F = KM1M2/r^2\n\nwhere K is the constant, M1 and M2 are the magnetic moments, and r is the distance between them.
398633
Part 1: Mike\nWork=Force x Distance(moved) x cos (theta) (angle between work and distance)\nF=215\nD=64.7\nTheta=35 degrees\n\nW=215*64.7*cos35\nW=11,394.815 Joules\n\nPart 2: Chris\nF=8N\nD=3.9\nTheta=90 degrees\nW=8*3.9*cos90\nW=0\n\nThats all you get...good luck
954796
51kg x 25 m = energy used\n\ndivide the energy used by 0.30 to get the second answer
748800
1)c because the forces are equal (a also for the same reason 0 is constant)\n2)b because the box has an unbalanced force in the horizontal direction and to the right\n3)d because the unbalanced force is to the right\n4)d because there is more force to the right\n5)b there is an unblanced force to the right
772638
try the formulas:\n\nforce = mass * aceleration\n\nwork = force * distance\n\n\n\n\nWell i dont speak english very well, but i hope i help u.\n\nKisses
984208
a)\n\nfirst you need is how much time passes at impact \n\nd=1/2at^2,  so  t=( 2xd/a)^0.5 = (2x10/9.81)^0.5 = 1.43 sec\n\n(v2-v1)/t=a,  v2=axt-v1=9.81x1.43 - 0  ( v1=0 )\n                                    v2 = 14 m/s\n\nb)  The force exerted by the water = Weight of swimmer\n\nW=mxg=65 kg x 9.81= 637.65 N\n\n\nCar Question:\n\nformula to use is:\n\ncoefficient of friction = (Force of friction )/ ( Normal Force acting down)\n\nwe need to find acceleration \n\nuse this formula:\n\n(v2)^2= (v1)^2 + 2ad,  d=25 m, v2=0 and v1=14 m/s\n0 = 14^2 + 2x25xa\nsolving a=  -3.92 m/s2\n\nso Friction Force = (mass of car)xa = 2500 x 3.92 = 9800 N\n\ncoefficient of friction =( frict force / Weight of car )\n                                  =9800 / ( 2500 x 9.81 )\n                                  =0.4\n\nDon't let that prick bother you !!!!!!!!!!!
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