the radis of the cylindrical tank is 2m. how high does the tank fill?
136650
Set the volume of rain from the roof (12m x 3m x .1m) equal to the volume of the tank in the form of pi x (2m)^2 x H, and solve for H.\n\nI got 0.2865m.\n\n*Edit: the person below is correct, if the dimensions given are only half the area draining into the tank, twice as much water would drain to the tank, making h = 0.573m
100966
You need the exact geometry of the hemispheric ends if you want the exact volume. Or fill the tank with water and measure how much liters you put inside.\nLPG tanks are filled up to an 80% only, they have to leave space for evaporation so I don't think there would be an important error if you take the volume of a cylinder plus the volumes of two spherical caps (the hemispheric ends)\n\nSee link for formula http://mathforum.org/dr.math/faq/formulas/faq.sphere.html
504226
A. Amount of Z in tank  = 4 lb\n\nB. After 10 minutes amount of Z = 4 lb\n\nC  Amount of water in tank at time t = (10 + 2t) gallons\n\nSo at time t, concentration of Z in tank = 4/(10 + 2t)\n= 2/(5 + t) lb/gallon\nSo 0.1 = 2/(5 + t) ie 5 + t = 20, t = 15 minutes
436756
V = πr²h\nA (top & bottom) = πr²\nEnd Area = 2πr²\nLateral Area = 2πrh\nCost of tube = 2πrhC, C being cost per unit area\nCost of ends = 2πr²C*1.2\nTotal cost, P = 2πC(1.2r² + rh)\nh = V/πr², so\nP = 2πC(1.2r² + rV/πr²)\nP = 2πC(1.2r² + V/πr)\ndP/dr = 2πC(2.4r - V/πr²) = 0 for max or min, so\n2.4r = V/πr²\nr³ = V/2.4π\nr³ = 5/2.4π\nr = 0.872"\nh = 5/π0.872²\nh = 2.093"\n\nCheck:\nC(2.4*π0.872² + 10/0.872) = 17.201050C\nC(2.4*π0.87² + 10/0.87) = 17.201444C (+0.000394)\nC(2.4*π0.874² + 10/0.874) = 17.201137C (+0.000087)
454129
The fall time T = sqrt(2*h/g) = 2.22 s. So only 4.22-2.22 = 2 s of water poured is in the pail. That's 2*.111 kg added to 1.07 kg = 1.292 kg. That mass * g = 12.66 N.
250579
take the 5 gallon container, and pour it into the 3 gallon container, leaving 2 gallons. dump the 3 gallon container out, and put the remaining two gallons into it. then, take the 5 gallon container, and use it to fill the remaining 1 gallon of the 3 gallon container. this leaves 4 gallons in the 5 gallon container.
359741
Dis you mean 2.4 gal?\nif so\n2.4gal*3.78l/gal=9.072l\n9.072l*.692kg/l=6.28kg
472041
V cone = ⅓πr²h\nh/r = 50/30 so h = 5r/3 and r = 3h/5 so when h = x, r = 3x/5\nThus V = ⅓π(3x/5)²x\n= 3/25 * πx³\nWhen h = x, dx/dt = (50 - x)\nSo dV/dt = dV/dx * dx/dt\n= 9/25 * πx² * (50 - x)\n\nFor the rectangular tank\n\nV = Ah\n= 400h\ndV/dt = dV/dh * dh/dt\n9/25 * πx² * (50 - x) = 400 * dh/dt\n\nSo dh/dt = 9/10000 * πx² * (50 - x) \nwhen x = 10\ndh/dt = 9/10000 * π * 10² * (50 - 10) \n= 9/10000 * π *100 * 40\n= 18π/5 in/min
584735
Obviously the question is "What is the volume of a fish tank whose dimensions are 116 mm x 149 mm x 87 mm?" \n\n1,503,708 cubic mms or approx 1.5 cubic metres
526471
Take a differential height dh at height h from the vertex.  The differential volume is dV = π*r^2*dh, where r is the radius of the tank at level h.  r@27ft is 10ft, r at 0ft is 0ft, so r = (10/27)*h.  Thus\n\ndV = π*2.7^2*h^2*dh.  Divide both sides by dt to get\n\ndV/dt = π*2.7^2*h^2*dh/dt.\n\nYou are given dV/dt, that is the volumetric flow rate into the tank = 10cfm   dh/dt is the rate of depth increase:\n\ndh/dt = dV/dt / (π*2.7^2*h^2)
341371
I'm going to assume the dynamic pressure of the outflowing water equals the static pressure just inside the tank:\n\n0.5ρV²  =  ρgh → V = √(2gh) = √(2*9.8*10) = 14 m/sec\n\nFlowrate = A*V → A = F/V = 2.9E-3/14 = 2.071E-4 m² and\n\nd = √(4A/π) = 1.624E-2 m = 16.24 mm
187995
Given that the volume of the cylindrical can is 100 cm^2...\n\n100 = (pi)r^2h\n\ndivide both sides by (pi)r^2 to isolate h\n\nh = 100/(pi)r^2\n\nSo, if the radius is 2 cm...\n\nh = 100/((pi)*4) - r^2 = 2^2 = 4\n\nh = 7.96 cm, rounded to first two decimal places, using 3.14 as the value for pi.
286245
The answer is (d) if this is the question:\n\nInitially the tank is full.  The outlet pipe is opened.  When the water in the tank is 3/4th full, both inlet pipes are opened.  Calculate the fractional level of the water in the tank after 2:05 hr.\n\nSince inlet A and outlet C cancel each other out at 50L/hr, only need to consider inlet B at 25L/hr.  Since 2:05 hr is the same as 2.083 hr (5 min = 0.083 hr), the amount of water added by inlet B is 25L/hr * 2.083 hr = 52.08 L.  Will use 52 L in the calculation.\nThe total amount of water in the tank after 2:05 hr is the initial 750 L + 52 L = 802 L.\nThe fraction level of the water is 802 L / 1000 L = 0.802\nSince 4/5 = 0.8 the answer is (d).
215270
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:)
413904
Yes, the answer is h = 2r\n\nThis is a classic optimization problem. You want the minimum surface area(SA) for a cylindrical shaped tin for a fixed Volume(V)\n\n(1) SA = 2pi r^2 + 2pi r h\n\nV = pi r^2 h (for simplicity set V = 1)\n\nso (2)    h = 1/(pi r^2)\n\nSubstituting this value for h into (1) gives:\n\nSA = 2pi r^2 + 2/r\n\nd/dr SA = 4pi r - 2/r^2\n\nd/dr SA = 0 when 2/r^2 = 4pi r\n\nthat is: (3)   r = 1/(2 pi r^2)\n\nTo get the value of the ratio h/r divide equation (2) by equation (3), which yields a value of 2 to 1.\n\nh = 2r
365830
I have made 2 liter bottle rockets using water, a tubeless tire valve, a bicycle pump and a 2 liter bottle.  It goes about 100 feet up if you do it right.
510076
This is pretty straightforward.\nFirst you have to write the formula fo rthe cost of the tank as a function of, say, its radius, r:\n\nThe volume of the cylinder is 100 cubic feet.\nIt is also pi*r^2 times the height, h\n\nso let's solve for h\n100=pi*r^2*h\nh = 100/(pi*r^2)\n\nnext, the cost:\nthe rectangle cost is 2.40*(h*2*pi*r),\nor 2.4*100/(pi*r^2)*2*pi*r\nor, simplifying, 480/r\nThe cost of the base and top are:\n1.50*2 pieces * pi*r^2\n\nSo the total cost is\n480/r+3*pi*r^2\n\nNow you need to differentiate this and find the roots, and make sure they are minima, not maxima!