Solve FLOW + BLOW + WOLF =  NOBLE?
115650
--5 2 0 1 -----  F L O W\n--4 2 0 1 -----  B L O W\n--1 0 2 5 -----  W O L F\n-------------------------\n1 0 4 2 7 ----- N O B L E\n\nBut still N and W value are coming same...does this problem have solution. If you have then can you please send the answer to my mail id hssiddhu@yahoo.com
904672
let u = x^2 + 1, then you have du = 2xdx, => dx = 1/2x du,\n\nsubstitute back you are on the way  to solve it
664257
Flow\nFloss\nFloral\nFloran\nFlorances
900600
listen closely!\nhttp://www.flowgo.com/index.cfm?action=view&id=5809&scid=0
188200
I had trouble with the same thing. The best way, and it's worked till now in college, is to take the most complicated element used (i.e. CH3OOH + C2H3O9 ---> etc), figure out O's first since O is used the most. Then work backward using the 2nd most used element. You might get a non-reduced form of balancing, but you can check whats divisible (i.e.  4CH + 2 CH2H3O2 <---> 4 OH + 2 CH3; all divisible by 2 so it'd be 2, 1, 2, 1 respectively). See if this works, it's worked when I've helped other people in the past as well.
883909
draw FBDs of the 2 blocks independently\nbe careful\nfind the accelrations of the  2 blocks using CONSTRAINT relations\nand comes ur answer\n\ndon't wait for the solution\nit lessens ur thinking power n creativity\n\nif it's still problematic, consult ur lecturer\n\nconcentrate on ur studies \nur still a college guy
706663
The idea is that all forces at the nodes (connecting points) are balanced. Draw the two free body diagrams.\n\nLeft point:\nFx\n0=-T1cos53+T2cos11\nFy\n0=T1sin53-T2sin11-57\nRight point:\nFx\n0=-T2cos11+T3cos42\nFy\n0=T2sin11+T3sin42-W2\n\nBy inspection we see only the first 2 eqns are required. Solve the system:\n0=-T1cos53+T2cos11\n0=T1sin53-T2sin11-57\n\nT1=83.62 N\n{=}8.5 kg
473829
The solution set is (-infinity, -7) U (-3/2, infinity)
1127531
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.
12859
The Shaum's Outline Series book will show you how to do the problems.\n\nhttp://books.mcgraw-hill.com/search.php?keyword=fluid+mechanics&template=&subjectarea=ALL&search=Go
244208
let\nl= 4h - 20,\nw= 2h + 10, and\nv= l*w*h,    then\n\nv= (4h - 20)*(2h + 10)*h\nv= (8h2-200)*h\n\nv= 8h3 - 200h
962694
Tehe answer is:\nAg + HNO3 -> AgNO3 + H\nAgNO3 + KBr -> AgBr + KNO3
306191
a) cube...final answer
771697
Try Google or Yahoo.  You'll find worked examples there.  I think it needs diagrams to understand; they and the accompanying explanations are too complex to attempt to reproduce here.
463421
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
974937
please complete the equation
713683
v1^2/2 +p1/rho =v2^2/2 +p2/rho\np=F/A\n(340m/s)^2/2 +F1/A/rho =(290m/s)^2/2 +F2/A/rho\n(340^2-290^2 )=2(deltaF)/(A*rho)\n(340^2-290^2) *70m^2 *1.29kg/m^3 / 2 = deltaF = 1 422 225 N,\nmg=1.4 MN=> m=142 tons (if 70m^2 is total wing area),\n(284 tons if 70m^2 is single wing area)