find k, the spring constant?
955845
The object has kinetic energy (1/2)*m*v^2 when first at x=0. The work friction does consumes all that energy as the spring compresses and pushes the object back to x=0. Let the distance the spring is compressed be x1. Then \n(1/2)*m*v^2 = Ff*2*x1 = mu*m*g*2*x1\n\nWhen the spring is compressed to x1, the friction has decreased the energy in the system by half and the remaining half is all stored in the spring compression because the object has come to a stop - about to start the rebound. So\nFspring*x1 = k*x1^2 = (1/4)m*v^2\nIf you solve the 1st equation for x1 and substitute the result into the 2nd, you should get the required result.
966154
The magnitude of a force is measured in newtons(N)
929836
v = b - u/k\nv - b = u/k\nk(v-b) = u
454576
Because the weight applies a force to the spring.  F=-kx, so x = -F/k.  The great the force, the greater the stretch.  This is because many metals have memory and try to restore themselves to previous states.  If you bend it, the metal will try to bend back to where it was.
307097
Ksp = [Ag+]^2[CrO4 2-]\nsince [Ag+] = 2*[CrO4 2-]\nKsp = 2*[Ag+]^3\nsolve for [Ag+] then plug it in this equation\n[Ag+] = mass /0.010L H2O
188896
K = mv^2 = (10kg)(4.5 m/s)^2 = 202.5 J\nI would suggest you try doing your homework, asking for all the answers will only hurt you in the future.
321421
0.215 kL = 215L\n(1kL = 1000L)
710178
λ = h/m*v\n\nwhere h the Planck's constant, m the mass of the proton and v its velocity
930657
So they told you the relation is:\nd=kh\n(Since they said "varies directly", there is no offset: d=kh+d0)\nThen: 8.45=6.5k\ntherefore k=1.3 (unitless, ft/ft)\n\nThen you want to find:\nd=5.2 = kh=1.3h\nTherefore:\nh = d/k = 5.2/1.3 = 4 ft
879746
Ek = 1/2.m.v2   (1/2 mass times velocity squared) - \n\nsorry can't do superscript on this site
1185900
H2SO4 + 2KOH --> K2SO4 + 2H2O
1145060
Using the equation U = 0.5*k*x^2\n\nU = potential energy (joules)\nx = compression distance (meters)\nk = force constant (N/m)\n\nx = sqrt(2*U/k) = sqrt(2*3.1/1600) = 6.22cm\n\n\nFor part 2, assume the book transfers all of its potential energy into the spring.  First figure out the book's potential energy.\nUsing the equation: PE = m*g*h\nPE = Energy (joules)\nm = mass (kg)\ng = gravitational acceleration (m.s^2)\nh = height (meters)\n\nPE = 1.1 * 0.9 * 9.8 = 9.702 J\n\nThen calculate compression same as in part 1:\n\nx = sqrt(2*U/k) = sqrt(2*9.702/1600) = 11.0 cm
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.
182644
° K = ° C + 273.\n\nIn this case:\n\nK=5C+273\nK=278
812261
You are very close. If K = -0.032 is accurate enough, then use it.\n\npA = Ae^(19 * -0.032)\np = e^(19 * -0.032) ~ 0.544\n\nIf you leave K = ln(.5)/22 then\np ~ 0.5496
700115
Very simple, in your case you only need to convert from N(Newton) to Kg(kilogram) so just divide by 9.8 (gravity constant)\n\nN/m2---- /9.8 ---> Kg/m2\nN/m2<-- X9.8 ---- Kg/m2
147821
- -----------O\n------------ . .-----------  _\n  K+ ( O: Mn  :O )\n-------------. .\n------------ O\n\nomit the long dotted lines