what is the formula used by a mud engineer to figure mud gradient on a mud report?
110505
What kind of mud?  Butt-Mud?
26736
look up stuff like static friction, and your coefficient of friction...
347282
The concrete formula has a very low "slump".  "Slump" is a measure of how much the concrete will move while its wet.
29063
http://www.eskimo.com/~jbm/calculations/traj/traj.html click this link it'll calc the answer for you
846447
http://www.bbc.co.uk/schools/gcsebitesize/chemistry/classifyingmaterials/formulae_and_equationsrev4.shtml
688944
View source of the that page.\nYou will see the formulas.
799401
Darcy's Formula is the general equation for flow of fluids. I was going to list it here, but the notation would be too difficult to express in this format.\n\nFind a copy of the book Flow of Fluids Through Valves, Fittings, and Pipe, Technical paper No. 410 by Crane Valves North America
948476
I won't show in details since it's hard to type out the work, but you'll get the rest, here it goes:\n\na)\nWork = mgd  *d=change in height\nW = (10kg) (9.8m/ss) (sin15) (-7.5m) = -190J\n\nb)\nW=(friction coef)(natural force)(distance traveled)\nnatural force = (cos15)(10kg)(9.8m/ss)\nW= (0.4) (cos15) (10kg) (9.8m/ss) (-7.5m) = -280J\n\nc) \nW=Force x distance\nW=(100N)(7.5m)=750J\n\nd)\n(initial KE)+ (puller work)=(gravity work) + (frictional work) + (final KE)\n(final KE) - (initial KE) = 750J - 280J - 190J = 280J\n\ne)\n(final KE) - (initial KE) = 280J\n*solve for final v\nv = squareroot (2 x [280j+ (1/2)(10kg)(1.5squared)] / 10kg) =7.63m/s
10661
Depends on weight of the block angle of inclination, friction. Get resultant velocity by getting velocity along X (w cos theta) and velocity along y (w sin theta).... blahblahblah... check other sources but basically it goes like that.
958491
i think we can convert the log with subsrcipt 3 to natural log\nlogx (with subscript 3) = lnx/ln3\nso derivative is\n1/(xln3)\nhope this helps
120944
dy/dx = -4x \n\nAt x=2, gradient is -8\n\ny - y(1) = dy/dx(x - x(1)) [x(1), y(1) means x,y coordinates]\n\ny - 1 = -8*(x - 2)\n\ny = -8x + 17
125054
WATER DISPLACEMENT  THEY GOT THE RIGHT FORMULA THE 40TH TRY
416567
Pressure = heat\n\nThe overpressurized zones compress the mud and force the atoms to move at a incresed pace causing the heat.
416901
A-a??\n\nA gradient is a vector field consisting of derivative vectors of a source of data (like a moving ocean)\nYou could use a gradient to show how fast the water is moving across the ocean surface. The derivitaive vectors that make up the gradient are instantaneous calculations of the rate of flow at several points on the surface of the ocean)
787859
To prevent the wearer from slipping over in the mud
882649
First solve for the force of friction between the block and the table using the formula,\n\nf=uR, where f is the force of friction, u the coefficient of static friction, and R is the force normal to the table. R therefore is the weight of the block equal to 31g.\n\nf=0.425*31*9.8\n =129.1N\n\nThe weight of the sand added to the bucket plus the weight of the bucket itself should be just slightly equal to 129.1N so that the system will begin to move. At that point, however, the block will begin to accelerate because the force of friction acting on it will no longer be static friction but kinetic friction, which is lower in magnitude. Thus there will be a net force equal to 129.1 minus the force of kinetic friction on the block. Thus,\n\nNet force, F=129.1-0.320*31*9.8\n                 =129.1-97.2\n                 =31.9N\n\nF=ma, where F is our net force of 31.9N, m the total mass of block plus mass of sand and bucket, and a is the acceleration of the system.\n\nMass of sand and bucket=129.1/9.8\n                                    =13.2kg\nSubstitute known values:\n\n31.9=(31+13.2)a\na=31.9/44.2\n  =0.72m/s^2
502267
Transpiration and concentration gradients