A tank is 6m deep. how much its bottom will appear to be raised if n of water is 4/3?
610353
4/3 = real depth/apparent depth\n4/3 = 6/x\n4x = 18\nx = 4.5\n\nApparent depth of the bottom = 4.5m\nBottom will appear raised by : 6-4.5 m = 1.5m
941831
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).
169646
The water pressure t a given depth is give as the weight of the water above that depth divided by the area it is being applied.\n\nSo imagine a column of water extending from the top of the lake down the depth.  Calculate the volume of this column of water then using the density of water (1 grams / mL) find the mass.  The take the mass of the water and find its weight (W = mg).  The weight it is the force exerted by the water, but pressure is the force of the water divided by the area it is applied.  So divide the force by the area of the base.\n\nEDIT:\nI see you added the density of water (even though I think my density [1 g/cc] is better).  I will leave you to do your own homework.\n[I also refuse to work with such an ugly unit, I prefer metric]
361907
If the tank is a cylinder, the Volume is πr²h.  If you get the volume in cubic feet, multiply by 7.5 to get gallons.
883842
the pressure will be higher but the density won't change, it can't possibly change if it's a sealed, rigid container, density is mass/volume and you aren't changing the mass or the volume, therefore the density will be unchanged
946895
Let the height be h (inches), radius r.\n\nNow by proportion, h = 5r at every level.\n\nHence volume is given by\n\nV = (1/3)pi*r^2*h which is\n       (5/3)*pi*r^3\n\n"Rate at which water is drained" means find dV/dt\n\ndV/dt = 5*pi*r^2*(dr/dt),\n\nand dr/dt = (1/5)*(dh/dt)\n               = -2/5\n\nYou can finish from there.
748998
Assume smaller pipe fills the tank in x hours and the larger one in (x-2.5) hours. Taking the given inf that both together can fill the tank in 8.4 hours, develop a quadratic equation in x, and solve it. \n\nLarger pipe fills the tank in 15.65 hours whereas the smaller one takes 18.15 hours to fill it.
856026
Use the gas law\nPV=nRT
1162775
if u evr seen a water tank they r usually round\nhttp://www2.warnerbros.com/main/homepage/homepage.html\n\nvolume = sqr ft x ht\nsqr ft  (of a circle) = pi x half the diameter(radius) squared\n\nvol = pi x (1/2x38) ^2x 100\nvol=3.1415926535897932384626433832795 x (1/2 x 38)^2 x 100\nvol=113,411.49479459153590850142613639 cubic feet
874895
The static pressure (P) of the water in the tank is\n\nP = p*g*h\n\nwhere p is the density of water, g is the gravitational acceleration, and h is the height of water. When the water exits the hole at the bottom of the tank it has kinetic energy per unit volume given by\n\nK = 0.5*p*v^2\n\nwhere v is the exit speed.\n\nThe jet travels a horizontal distance L hitting the ground. \n\nL = v*t\n\nwhere t is the travel time, which can be solved for by knowing how high the tank is above the ground. The water falls through a distance Y under the influence of gravity since there is no y-component to the initial velocity.\n\nY = 0.5*g*t^2.\n\nSolving for t yields\n\nt = sqrt(2*Y/g)\n\nwhich gives a speed of \n\nv = L/t = L*sqrt(0.5*g/Y).\n\nEquating the fluid potential energy per unit volume (pressure) to the kinetic energy we have\n\np*g*h = 0.5*p*v^2\n\ncanceling out p and substituting v gives\n\ng*h = 0.5*L^2*0.5*g/Y\n\nSolving for h, the height of the fluid, yields\n\nh = L^2/(4*Y)\n\nY = 1 m according to the picture provided. Of course, this is a dynamic problem as fluid exits the hole the height of the fluid will change as will the exit speed and impact distance. The height here is for the initial jet of water.\n\nThe links below provide more information regarding pressure, Torricelli, and Torricelli's law.
601468
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).
220777
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)]
723851
Dis you mean 2.4 gal?\nif so\n2.4gal*3.78l/gal=9.072l\n9.072l*.692kg/l=6.28kg
1003334
Assuming D is diameter, calculate the volume of the tank, subtract the volume of the cylinder and convert the remaing volume to gallons.  There are 128 cubic inches (0.0741 cubic feet) per gallon.\n\nyou do the math, that is obviously the intent of your assignment
346689
assuming you are going to fill it to about 4" from the rim, you will use approx. 5455 gallons of water
416654
This answer will depend on the density of the water that is used.  According to Wikipedia the density of Water is \n1000 kg/m^3.  The equation for density is as follows:\n\nDensity=Mass/Volume\n\nMultiply both sides by volume gives us the following equation\nMass=Density*Volume.\n\nGiven: \nVolume=1 cubic foot\n\nDensity of Water=1000 kg/m^3 \nYou most likely want how many lbs of water you need; thus we need to convert the Density of water from kg/m^3 to \nlbs/ft^3\n\nThus using an online conversion:\n1 kilogram/cubic meter = 0.062 427 962 pound/cubic foot\n1000kg/m^3= 62.43 lb/ft^3\n\nThus finally\nMass=Density*Volume\nMass=62.43 * 1\nMass=62.43 lbs approximate!\n\nNow since your designing it you want to design over 25%\nso therefore design for 78.04 lbs.  Worry about the pressure on the sides more when designing this system.
415302
You need to use Pythagorean theorem.\nMeasure the lenght of the lotus plant which is above the water.(X)\nBend the plant to any one side until it is just submerged under the surface of water.\nMeasure the Dist. between the initial and final positions of the plant(Y)(i.e. the initial position and final position where it is submerged)\n\nLet the actual lenght be L.\nThen (L x L) = (Y x Y) + (L - X)(L - X)