(Select all that are correct.)\nKE = mv\n2y = (2 m)cos(kx), where k = 2 m-1\nvf = vi + ax \ndensity = volume/mass
209576
KE = mv\nWrong: energy <> momentum\n\n2y = (2 m)cos(kx)\nRight: a lenght =  lenghth x a number between -1 and +1\n\nvf = vi + ax\nWrong: x should be t\n\ndensity = volume/mass\nWrong: fraction should be inverse
487500
density = Mass/Volume\nDensity of water=1kg/m^3 or (g/cm^3)\ntherefore density of soda here =1\nvolume 2 L= 2 m^3\nso\nMass of the soda =density of soda X Volume of soda= 1X 2= 2 Kg
475672
Ek = 1/2.m.v2   (1/2 mass times velocity squared) - \n\nsorry can't do superscript on this site
473517
If you're using m to indicate the mass, the differential equation should be:\ndm/dt = km\nIf you're using y, then:\ndy/dt = ky\n\nThe error in your answer is that you are alternately using two variables, y and m, to indicate the mass. Pick one and stick with it.
431774
First write the balanced chemical equation:\n\nPb(NO3)2 + 2NaBr --->  PbBr2  +  2NaNO3 \n\nThen, find the total volume of the solution: 200mL + 200mL = 400mL \n\nFrom the density, the mass is:   density = mass/volume   mass = density * volume = 1.0 g/mL * 400 mL = 400 g\n\nQ = mC(Tf - Ti) ; where Q = heat released, m = mass of solution,, C = specific heat capacity, and (Tf - Ti) = change in temperature\n\nSo,  Q = 400g * 4.2 J/gK * (2.44 + 273)  (Since C is J/gKelvin, the temperature rise also has to be in Kelvin)\n\nQ = 462.5 KJoules\n\nHope this helps...
490170
A) I think I got it...Use energy, except for g, use   Fg cos(19 degrees), found from a free-body diagram.\n\nso 1/2 mv^2 = mgh\n\n1/2 (4.7)^2 =  9.8(cos (19))  h    solve for h\n\nB)  now that you have h...maybe you can use   \n\nd y = V(sub 0 sub y) t + 1/2 (a x cos (19)) t^2\n\nyea, tell me how it comes out.
488295
w=mg\nw=(.25kg)(-9.8m/s^2)\nw= -2.45N\nacceration is -1.67m/s^2\nF=ma\nF=(0.25kg)(-1.67m/s^2)\nF= -0.417N\nFnormal is 2.45N\nFfiction=Mu*Fnormal\n0.417N=Mu*2.45N\nMu=.17\nnot sure if did it right! but at least i try
261802
Start with the ideal gas law: PV=nRT.  Since no gas enters or leaves the system, n is constant, and R is the universal gas constant, the product nR has the same value at the beginning and as at the end.  So, collect it to one side, and move the other factors to the other side.\n\nP1 * V1 / T1 = nR = P2 * V2 / T2\n\nSolve for T2\n\nT2 = T1 * (P2 * V2) / (P1 * V1)\n= 300e4 K (.0003e15 atm * .046e-15 cc)/(50e3 atm * .0004e-7 cc)\n=2.07 * 10^7 K
388980
Volume = pi * r^2 * h\n\nwhere\nr is the radius\nand\nh is the height\n\nFor example:\nr = 0.9m\nh = 1.2m\nvolume = 3.14159 * 0.9 * 0.9 * 1.2 = 3.0536 cubic metres or 3053.6 litres\n\n_
384322
P=P (momentum before = momentum after)\nmv=mv\n(55kg)(4.6m/s) = (230kg)v\nv= ((55kg)(4.6m/s))230kg\nv=1.1m/s
339698
1)\nM = Mo e^(-kt)\ne^(-kt) = (M/Mo)\n-kt = ln(M/Mo)\nk = (-1/t)·ln(M/Mo)\n= (-1/18.8)·ln(57.9/93) = 0.025 yr^-1\n\n2)\nM = Mo/2 = Mo e^(-kt) ──► ½ = e^(-kt)\n-kt = ln(½)\nt = (-1/k)·ln(½)\n= (-1/0.025)·ln(½) = 27.5 yr
306791
n * MW / V = density\n\nTherefore, MW = density * V/n = density * RT/P\n\ndensity = 0.480 g/L\nR =0.08205746 L · atm · K-1 · mol-1\nT= 533.15 K \nP=0.140 atm \n\nMW = 0.480 * 0.08205746 * 533.15 / 0.140  = 149.996 g/mol\n\nIf your teacher is a stickler about significant figures, then the answer is 150. g/mol (don't forget the period)
572034
would you like to supply me with some formulas? im out of grade 12 physics and i forgot most of the formulas. trust me from experience, just plug numbers into a formula and things will work\n\nF=mg = ma\nFk=μkN\na= vf^2 - vi^2 / 2d
525780
Just remember M1*V1=M2*V2 where M is molarity and V is volume. A little Algebraic manipulation gives you:\n\nV1 = (M2*V2)/M1\n     = (0.195*15.5)/18.0\n     = 0.168 L = 168 mL
221425
KE=1/2mv^2\n=1/2(30)(20)^2\n=1/2(30)(400)\n=1/2(12000)\n=6000J\n\nC
640823
Surely you know the foumla for Re?\n\nRe=(density*velocity*length)/viscosity\n\nRe=(1000*1*25*10^-3)/1014*10^-3\n\nRe=24.655 (3dp)\n\nI think you've got yourself a laminar flow.
268117
Use the ideal gas law \n\nPV/T = constant for the final and initial states.