how do you figure the formula of trench silo capacity?
148864
If your trench silo looks like the ones that I am familiar with, it's rectangular when you look down on it and  has sloping sides and sloping ends. The way to compute its volume is to break it down into pieces that are regular prisms or pyramids. I would break it down like this:\n\nStarting with the flat rectangular bottom, extend the edges of the rectangle to the boundry of the silo. That will divide your silo into 9 pieces.\n\nThe center section with a flat bottom is a rectanguar box.\n\nThe 4 corner pieces that slope in two dirctions. They are pyramids.\n\nThe 4 pieces sloping sides, two that lie along the long side of the rectangular box and two at the ends,  are triagular prisms.\n\nOf course, some of those silos get mounded up in the middle, so that adds another dimension to solving the problem.\n\nYours for better silage,\nGuru
811969
The rate at which they will work:\nX/30 + X /45- X/90, where X is equal to the tank volume per minute.\n(3+2-1)X/90= 4 X /90. That is the combined capacity= X/22.5.\nTherefore it will take 22 and half minutes to fill up the tank.
674116
It will depend on the specific resin.  Look up the resin, the capacity is usually given in the technical information.
474349
I love this.\n\nFill the 7 liter pail.\n\nPour into 5 liter pail.\n\nLeft with 2 liters in 7 liter pail.\n\nEmpty 5 liter pail.\n\nPour the 2 liters from the 7 liter pail into the 5 liter pail.\n\nFill 7 liter pail.\n\nPour into the 5 liter pail til full.\n\nLeft with 4 liters in 7 liter pail.\n\n:)
1066290
First, you need the volume. The formula yields: V = pi rh/3 = 20000pi/3 cubic feet. Since it weighs 50 pounds per cubic foot, the ratio 50 lbs/ft.^3 = 1. So multiplying this to our volume equation, V = 20000pi/3 ft.^3 = 20000pi/3 ft.^3 * 50lbs/ft.^3 = 100000pi/3 lbs. The work function is (25-x)100000pi/3 = W(x). Integrate this result from x=0 to x=25, and you get the answer.
141968
Since you have a circular hole you have to make the next operation:\n\nFirst multiply the diameter (wide) by PI (3.1415926535)^2 and divide the result by 4 (or multiply by half the wide  by pi^2 and don't divide by 4 it's the same result).\n\nThat's 30 feet by 9.8696044 and divided by 4 the result equals 74.02203300 square feet.\n\nThen multiply by 0.5 feet (that's 6") to get the volume (that's the cubic feet) and equals 37.01101650408 cu. feet.\n\nJust in case you need to know
196386
These are the equations used in PSpice:\n\nIn subthreshold: \n\nCgs = Cgd = 0 \nCgb = Cox \nIn saturation: \n\nCgs = 2 / 3 * Cox \nCgd = Cgb = 0 \nIn linear: \n\nCgs = 2 / 3 * Cox * Source Factor \nCgd = 2 / 3 * Cox * Drain Factor \nCgb = 0
1097124
Aplly the following formula:\n\nG = w * A, where:\n\nG - flow rate [m^3/s]\nw - speed of fluid [m/s]\nA - transversal section area of the pipe [m^2]
751939
you can either do the math, look it up on a chart or go here:\n\nhttp://leebrewery.com/hydromet.htm\n\nand use the java-calculator.
483088
this is very easy, just use the pythagorean therom (i might spell it wrong) the formula is: a^2 + b^2 = c^2 (c is the length of the pipe)\n\n15^2 + 8^2 = c^2\n\n225 + 64 = c^2\n\n289 = c^2\n\n17 = c\n\nyou need 17 feet of pipe
912896
There are 12/5 litres which must be multiplied  by 7/4. That gives us 84/20. Divide both numbers by 4 and you get 21/5, or 4 and 1/5.\n\nAt least I hope so :)\n\nAnd for Sazi, I hope I've explained it clearly enough that you can talk your teacher through it.
576519
tensile strength = (2*P)/(pi*t*d)\nwhere; P =maximum load you get from the machine(kPa); pi = 3.14; t = thickness/height of specimen (m) and d = diameter of specimen(m)
333658
BOXES not BOXS...FULL not FILL\n1) multiply the dimensions of boxes\n2) take the percentage of remaining space as a decimal\n         Ex: 100-13=87 = .87\n3) multiply the number from 1 with the number from 2\n\n4259.52 Cubic Inches\nPlus\n4723.20 Cubic Inches\nPlus\n3801.60 Cubic Inches\n\nTotal: 12,784.32 Cubic Inches of remaining space
788352
1) Rearrange so\ny = -(5/3)x + (2/3)\nThis means the perpendicular to this line (which the point must lie on) is of the form\ny = (3/5)x + n \nlet (a,b) = the point we require, so\nb = -(5/3)a + (2/3)   equation 1\nand -5 = -4(3/5) + n so n = -(13/5)   equation 2\nand b = a(3/5) + n  equation 3\nsubstitution of equations 1 and 2 into 3 give:\n-(5/3)a + (2/3) = a(3/5) - 13/5\nso a = 49/34\nand consequently b = -59/34\nso the point is (49/34, -59/34)\n\n2) let h = height of box and l = width of box = length of box (square base)\nso volume, v = hl^2\nand material area, a =l^2 + 4lh = 1300 cm^2\nso h = (1300-l^2)/(4l)\nso now v = (1300l - l^3)/4\nfor max volume \ndv/dl = 0 \nso dv/dl = (1300 - 3l^2)/4\n(1300 - 3l^2)/4 = 0 when l = sqrt(1300/3)\nwe take only the positive solution as the negative is meaningless in this problem.\nto check this is indeed a maximum value (as oppose to a minimum) we derive d^v/dl^2:\nd^v/dl^2 = -3l/2 so for l = sqrt(1300/3) this is always negative confirming this is a maxima.\nso now, v  = (1300l - l^3)/4\nsubstitute l = 20.817 cm\ngives volume, v = 4510.28 cm^3
703001
To find the force P, its equal in size, but opposite in direction to the weight of the crate,resolved in the direction of the slope:\n\nP=Wcos60 or P=Wsin30\nP=(80*9.81)cos60=392.4N\n\nR is equal in size, but opposite in direction to the weight of the crate, resolved perpindicular to the slope\n\nR=Wcos30 or R=Wsin60\nR=(80*9.81)cos30=679.66N\n\nPlease note you need to use weight, not mass ( W=m*g)\n\nGo do the math!
358957
Volume = pi * r^2 * h\n\nr is 1/2 the radius of the drilled hole = (D'/2)\nh is equivalent to the diameter of the original cylinder = D
245193
Use your buffering equation and plug in the numbers you know to get your answer.  You might have to do reverse calulations to find your answer.