Find the least amount of material needed to build an open cylindrical vessel with a capacity of 400Pi cm^3.
119673
I assume this is a question from a calculus class.  The only way I know of solving this is using derivatives.\n\nsurface area = 2Pi*r^2 + 2Pi*r*h\n\nV = Pi*(r^2)*h\nV = 400Pi\nThus, 400Pi = Pi*(r^2)*h  Solve this equation for h by dividing both sides by Pi*r^2\n400PI/Pi*r^2 = h  Simplifying will give you\nh = 400/r^2\nNow replace h with 400/r^2 in the equation \nsurface area = 2Pi*r^2 + 2Pi*r*h\nLet S stand for surface area\nS = 2Pi*r^2 + 2Pi*r*(400/r^2)  Simplify\nS = 2Pi*r^2 + 800Pi/r\nNow take the derivative.\nS' = 4Pi*r - 800Pi/r^2\nSet this equal to zero and solve for r.\nI get that r = the cube root of 200 which is 5.84803547643\nNow you should test to see whether this is a max or min.  It should be a min; although, I did not check it.\nAfter you check it plug this value of r = 5.84803547643 into the equation h = 400/r^2\nh = 400/(5.84803547643)^2 \n= 11.6960709529\nThus,\nr = 5.8480354764cm and h = 11.6960709529cm
89721
use the following formula .. \n\ntotatal surface area = \n[(H/D)*4(pie)(0.5D*0.5D)] + [ (H-t)/(D-2t)*4(pie)*(0.5D-t)(0.5D-t)] + [(pie)(0.5D*0.5D)-(pie)(0.5D-t)(0.5D-t)]\n\nthis is considering D = tank diametere is from outer surface and not the cavity diameter \n\n\nouter surface area of cover = (H/D)*4*(pie)(0.5D*0.5D)\ninner surface area of cover = [(H-t)/(D-2t)*4*(pie)*(0.5D-t)(0.5D-t)] \n\nthe base line are = [(pie)(0.5D*0.5D)-(pie)(0.5D-t)(0.5D-t)]
510144
use the formula\nA=2pir(h+r)\n=2pi*7*21\n=294pi
325028
Assuming that NO2 is an ideal gas is these conditions (which is a reasonable assumption at such a low P), the number of moles into that flask is: n = PV/(RT) where P = 432 mmHg = 0.568 atm, V = 0.5 L, T = 60 C = 333 K, R = 0.082 L atm/(mol K).\nThen: n = 0.01 mol and, taking into account the Avogadro Number (in number of particles/mol) the number of molecules is NM = 0.01*6.022x10^23 = 6.022x10^21.
458132
Use the gas law\nPV=nRT
364369
1. volume = h*2*pi*r^2 =1000 so r^2= 1000/2/pi/h\n\nsurface area = 2*pi*(h+2*r^2)\nsurface area / volume = (h+2*r^2)/h/r^2 \n=1/r^2+2/h  = 2*pi*h/1000+2/h  which is a minimum when h is about 17.84  [(1000/pi)^0.5]
264363
Okay, I'll try this one -\n\nRun this experiment twice (or more times), each time with a piece of metal having different density.  The metal piece should be of adequate size (a sewing needle isn't going to do it), it needs to be big enough to significantly affect the temperature of the water.  You will need to know the mass of the metal (i.e., weigh it) and have a thermometer capable of reading from room temperature up through boiling or so.  Please note this is not going to give you a highly accurate answer, but can give you a ballpark idea (I'm guessing you're probably not doing this in a 500 million dollar lab).\n\nThe procedure:\n\n1)  Measure and record room temperature as T1.\n\n2)  Weigh the metal, call it M1.\n\n3)  Bring a known quantity of water (weigh it, or measure it and convert to weight, call it M2) to a temperature just short of boiling (say 180F to 200F, use the same temperature for each run of this experiment).  Record the temperature as T2, and remove the heat source (take it off the burner, etc.).\n\n4)  Drop in a metal piece at room temperature.  Allow a couple of minutes for heat to exchange.  Make sure you use \nthe same amount of time for each run of this experiment.\n\n5)  Measure and record the new temperature of the water as T3.\n\n6)  Repeat with your other metal samples.\n\nYou can now estimate the specific heat capacity for each metal.  It will not be super-accurate (heat escapes your system, and temperatures won't have completely equalized), but should give you some idea.  Just assume all temperatures did equalize and that no heat escaped for calculational purposes.\n\nSince (you are assuming) the heat which left the water went into the metal, then the product of mass, specific heat capacity, and temperature change should be the same for both the metal and the water (but in opposite directions).\n\nThis means that:\n\n(M1)(c1)(T3-T1) = (M2)(cw)(T2-T3)\n\nwhere c1 is the specific heat capacity of the metal, and cw is the specific heat capacity of water (look it up on Google to get the units you need).\n\nSolving this, you get:\n\n......(M2)(cw)(T2-T3)\nc1 = -----------------\n........(M1)(T3-T1)\n\n(sorry about the dots, I'm trying to keep everything aligned)\n\nRepeat the calculations for each of the metals.\n\nYou will need to calculate (or look up) the densities of each of the metals.  Then, graph density vs. the specific heat capacities you calculated, and see if your hypothesis is true.\n\nHave fun!
43120
13 and a third hours is the number i came up with...\n\nJust plug some fake numbers in, say, for example, the tank that these two pipes are filling is 500 gallans.  then you can figure out how much each pipe put into it, take that five hundred, divide by 8, then multiply by 3, that's how many gallens the smaller pipe put into the tank in 5 hours...   now, take that number and divide by the 5 hours, that's how much that smaller pipe can pour out into the tank in one hour...  to take that 500 gallens and divide by the amount that can be poured in one hour and it'll give you how many hours it'd take to fill at that slower rate...\n\nThis is not a standard way of figuring out this problem, but it seemed like it worked to me, it should have taken more then double the 5 hours, but less then 3 times...   does this help at all??
457651
Look at like this. Whatever the amount of the first component\nis, consider it 100%, then for the second component, it would be 80% of the first.\nExample:\nYou have 500 gm of component 1,\nyou put 80%of that or 400 gm of component 2.\nThe total gms would be 900 gm, 100 parts component 1 to\n80 parts component 2.\nHope this helps.
363502
1. obtain formulas for surface area and volume of cylinder\n2. let volume equal 1000 and solve this volume equation for h\n3. substitute the expression from previous step into surface area    equation to get equation with only r's\n4. differentiate surface area formula with respect to r set this equal to 0 and solve for r i got   r=10/3rd root(pi)\n5. substitute value for r into volume equation solve for h  i got h=10/(pi)^(2/3)\n\nincidentally, i used 2nd derivative test to conclude r is minimum\n\n\n\ni'll get back to u with more
522027
The point to note is that the mass of the water in the cylinder is continuously varying.\nMass of can = M (constant)\nMass of water = M*(x/H)   -- because it is proportional to the height of the water column.\nCentre of mass of can alone (from the bottom)  = H/2\nCentre of mass of water = x/2\na) Centre of mass of the system = (Mass of can * cm of can+Mass of water * cm of water) / (Mass of can+Mass of water) \n = {(M*H/2) + (M*x/H)*x/2} / (M+M*x/H)\n = 1/2 (H^2+x^2)/(H+x)\n \nb) The value of x for the minimum height of the cm can be found out by putting d(cm)/dx = 0\n This gives, x = H*(sqrt2 - 1). \n At this value of x, the cm turns out to be:\n cm = H*(sqrt2 - 1).
449015
let x be the length of the square base, and y be the height.\nGiven x^2 * y = 32000\nAssuming same thickness of the material, \nArea of the box, A = x^2 + 4xy = x^2 + 4x * (32000/x^2)\n= x^2 + (4*32000)/x\n\nTo minimize the material used means to minimize A.\nSo, find what value is x when dA/dx = 0\n2x - (4*32000)/x^2 = 0\nx = cuberoot(4*32000/2)\nx = 64000^(1/3) = 40\nIs this a minimum turning point for A? check d(dA/dx)/dx when x = 40.\nd(dA/dx)/dx = 2 + (2*4*32000)/x^3, thus when x = 40, d(dA/dx)/dx > 0, thus at x=40, A is minimum.\nFind y, y = 32000/x^2 = 32000/1600 = 20\nTherefore, the box's dimension is 20 x 40 x 40.
445616
The required depth = 0.3536m, or about 35.4cm\n\nThis can be solved with or without reference to the angle between the trough sides. The volume V = 6 A where A is the area of a triangular end. So we just have to find when A is a maximum.\n\nA = b*h/2 (where h is the required depth)\n\nAlso, b = 2*sqrt(0.5^2 - h^2)\n\nso A = h*sqrt(0.5^2 -h^2)\n\nNow find h, such that: dA/dh =0\n\nor, more simply:\n\nThe angle betwen the 2 pieces which maximizes the volume is 90 degrees(see footnote). Once you know this you can determine h since h = 0.5*cos(45) = 0.3526\n\nFootnote: Express the the end area in terms of the angle between the 2 pieces of wood. Use the area formula: A = (1/2)a*b*sin(x) where x is the angle between the two pieces. In your case, a=b= 0.5m and so  A = 0.5*0.5*sin(x)/2 = 0.125sin(x).\nSo the volume V=  6*0.125*sinx(x) = 0.75sin(x) cubic m. This will be a maximum when x = 90 degrees since we know the sine function value is a maximum when x=90 degrees, but formally:\n\ndV/dx = 0.75 d/dx sin(x) = 0.75cos(x)=0 when cos(x) =0 when x = 90degrees\nMax volume = 0.75sin(90) = 0.75 m^3
481358
Volume = pi r^2 h \n\nMeasure a Coke can to get r and h and calculate the volume.  Plug it in for V in  the volume formula and solve it for h.\n\nThen take the surface area formula, SA = 2 pi r^2 + 2 pi r h,\nreplace the h with your answer from above, find the first derivative d(SA)/dr, and make it equal to zero to find the minimum.
339780
Solve for the smallest tank, witha volume of 10,000 m^3 \n\nSurface area of the tank would be\n(height of the tank x circumference of the tank) + piR^2 \n\nvolume of the tank = 10000m^3 \nh*piR^2\n\nAs you change the values of H and R to get you volume the surface area will change. Solve for a minimum value.\n\nThe dimensions of the small tank then determines the radius of the hemisphere that it would take to hold that tank.
447900
The work done by the engine in lifting the object is m*g*h = 400*3.5 = 1400Joules. This the work output of the engine.\n\nNow, efficiency = (work output)/(input) * 100%\n\nTherefore, 15 = 1400/input * 100\n\ninput is thus = 140000/15 = 9333.33Joules
536735
PART 1:\nIf the ratio of the diameters is 1 : 2 : 3, then that means the diameters are 18 : 36 : 54.  You can solve this with similar ratios:\n\n1/18 = 2/x\nx = 36\n\n1/18 = 3/y\ny = 54\n\nSo the respective diameters are 18", 36" and 54".  (The radii would be half this or 9", 18", 27").\n\nHowever, the answer is to be expressed in feet so divide by 12:\n54 inches / 12 inches/ft = 4.5 feet.\n\nPART 2:\nVolume of the *middle* sphere.  The radius is 18" (one half of 36"), so just plug this into the formula:\nV = (4/3)(pi)(18)^3\nV = 7,776 pi (cubic inches)\n\nPART 3:\nThe volume of the *small* sphere = (4/3)(pi)(9)^3\nThe volume of the *middle* sphere = (4/3)(pi)(18)^3\n\nTo get the ratio between these two, just divide smallest / biggest\n= (4/3)(pi)(9^3) / (4/3)(pi)(18^3)\nThe 4/3 and pi cancel out\n= 9^3 / 18^3\n= 729 / 5,832\n= 1 : 8\n\nThis makes sense because the ratio of lengths is linear (1 : 2 : 3), the ratio of areas is squared (1 : 4 : 9) and the ratio of volumes is cubic (1 : 8 : 27).\n\nPART 4:\nWe have the 3 diameters (18", 36", 54").  the total height is the sum:\n18 + 36 + 54 = 108"\n\nConvert this to feet (divide by 12):\n108 inches / 12 inches/ft = 9 feet.