TRue or False. Mgh=0.5MV^2?
476031
Well, yes, neglecting any energy loss to friction (e.g. air resistance)
107382
v^2-u^2 = 2gh\nHere, v=1, g=-10m/s/s, h=0.6m\nHence, u^2 = 1+12=13\n=> u = 3.606 m/s
963771
conservation of momentum (mv): mv_final=mv_initial:\n655kg*2.25m/s =1055kg*v_van_f +655kg*v_car_f (sign of velocity to be determined)\nconservation of energy: Ek_car_initial =Ek_van_final +Ek_car_final\n655kg*(2.25m/s)^2 =1055kg*v_van_f^2 +655kg*v_car_f^2. sub v_van or v_car from momentum eq-on into energy eq-on:\n\n655kg*(2.25m/s)^2 =655kg*v_car_f^2 + 1055kg * (655kg*2.25m/s/1055kg -655kg/1055kg*v_car_f)^2, expand and solve quadratic equation to get\nVcar=2.25m/s or -0.526m/s. the first one is nonrealistic, since it implies no collision ,ie velocity didn't change.\nso Vcar is -0.526m/s or opposite to where it was travelling before\nVvan is found from momentum eq-on: v_van = 655kg* 2.25m/s/1055kg - 655kg/1055kg*v_car_f=1.723m/s\nI checked this by plugging into the momentum formula.\nso \na) -0.526m/s (ie in opposite direction)\nb) 1.723 m/s\n\n[To Michael: the point of Yahoo Answers is that I answer the question and explain as good as I can/have time for. It's not up to me or you to judge educational value of my answer or to assume that this question is for homework.]
942436
F(t-t0) = m(v - v0)\nF = 0.25W = 0.25mg\n0.25*9.80662*t = 8 m/s - 5 m/s\nt = 3/(0.25*9.80662) s\nt = 1.2237 s
728641
acceleration can be found using\na=(m1-m2)g/(m1+m2) = 3*9.8/5 = 5.88 m/s^2
865501
This is letting spring energy = gravitational potential energy.\n\nkx²/2 = mgh→ k = 2mgh/x² = 2*1*9.8*.6/.02² = 29400 N/m
816057
again ... conservation of (mechanical only here) energy\n\npot energy = kin energy\n\nmgh = 1/2 m V^2\n\nso V = sqrt(2gh) -- see mass irrelevant\n\nV = sqrt(2*9.8*8.53) = 12.9 m/s\n\ng=9.8m/s^2 is the acceleration due to gravity\n\ng=10m/s^2 is not a bad approximation
819860
m(v)(v)/r= 7.18 mg  or m(v)(v)/11.5 = 7.18 mx9.8\nv = sqrt(7.18x9.8x11.5) = 28.446 m/s
954443
Hi Yo-Yo!\nEnergy conservation law: m*g*h=E1+E2, m=0.12kg, h=1.2m, g=9.8m/s^2, E1 is translation kinetic energy, E2 is rotational kinetic energy.\nE1=m*v^2/2, v is translation speed at full length h, E2=I*w^2/2, w=angular speed at full length h.\nNow v=w*r, r=3.2*10^(-3) m, thus m*g*h=m*(w*r)^2/2+I*w^2/2, hence w=sqrt(2*m*g*h/(m*r^2+I)). Plug numbers in and find w & v (be a good boy!).\nMeanwhile v=a*t & h=a*t^2/2, ‘a’ is linear acceleration.\nThus t=v/a & h=a*(v/a)^2/2, so a=v^2/(2*h). Plug numbers in and find a & t. Angular acceleration being w’=a/r. My regards!
378804
Vf=Vi+At\n5=12+A*2.3\nA= -3.04348 m/s^2
788598
just taking a shot at it the mechanical energy of the spring will turn to potential energy at the highest point so:\n0.5*k*x^2 = mgh\nx---->compressed distance\nso rearranging h = 0.5k*x^2/(mg) = 5.44m\nhope this is correct and it helps.
833966
The question is easy enough if you think about it properly.  What you want is the speed at which the centrifugal force (given by mv^2/r) equals the gravitational force given by the bank, and that will be mg sin(2.5 degrees).  Note that m cancels out in this, and you are given r, so v is the only unknown and you can solve for it with a scientific calculator.
1035224
v^2 = 2gh\n\nh=1/2 gt^2\n\nsubs this.\n v^2 = 2g (1/2 gt^2 )\nv^2 = 2 g^2 6^2\nv^2 =  100 X 36\n\nv= 60m/s
759307
here we go  kinetic energy at point of shot equals potencial energy at top of the squirt\n\n1/2 m v^2 = mgh\n\nm's cancel  (Rained in seattle)\n\nv = Square root (2gh)\n\ncheck my algebra... Note, you must change cm to meters before you can answer
976160
the number of moles in the 1 L  bulb is using pV=nRT\n\nn1 = pV /nR  = 649 *1 /RT = 649/RT\n\nfor the 0.5 L  we obtain n2 = 445 * 0.5/RT= 222.5/RT \n\nn1+n2 = 871/RT  which is put in a total volume of 1.5 l\n\nSo, the pression is  871/1.5 = 581.2 mm Hg\n\nNote that with my answer, you need not to know neiter the temperature nor to convert the pression in mmHg in pascals or atmospheres
688699
No it's not right...\n\n1 atm = 101.325 kPa = 760 mm Hg = 760 Torr = 14.7 psi\n\nall those are equal to eachother
1129109
0.004vi = (1.110 + 0.004)vf\n\n(1/2)(1.114)vf^2 = (0.220)(1.114g)(0.300)\n\nvf^2 = 2g(0.220)(0.300)\n\nI think you should be able to do the rest...