fractional water level?
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).
1165265
its all to do with weight displacement and the volume of water
470447
You need to know the specific gravity of the weight.  (Specific gravity is weight per unit of volume, typically expressed in grams per milliliter).  The specific gravity of water is generally accepted as 1 (meaning that it weighs 1 gram per milliliter).  The equation to determine volume of water displaced would be:\n\n[10g/(S.G. of the weight)] X 1 = Vol. of water displaced\n\nNotes:\nS.G. is short for Specific Gravity\n\nThe 1 in the equation is the S.G. of water.\n\nIn actuality, the S.G. of water varies with temperature, so you may need to adjust the S.G. of water to the correct value if the problem gives you that information.\n\nThe answer will be volume of water in grams.
904431
Just the take the [H+] ion concentration divided by the initial [HF]:\nfrom your info i gather it should be (10^-1.59)/1.0 \n\nConvert it into a percentage.
765689
Yes you are right to subtract the water pressure from the total pressure.\n\nThen use p1v1/t1 = p2v2/t2\n\nSTP is p2v2/t2\n\nSTP = 1 atm and 0C or 273K\n\nSolve for V2.\n\nKeep units the same on both sides!
580754
i haven't done this in a while.\n\nThe volume of the reservoir is equal to (1/2)*(height)*("base")*length.  the reservoir is a rectangular prism with the "base" being the distance across the surface of the water.  We are given the shape of the full volume so we can scale down to the level of the water to find what the volume in the reservoir at its current level.\n\nif you think about the cross-section of the volume, and use half of that forming a right triangle, we can determine the ratio of the surface (the "base") to the height (water depth).  In this case, it is (40/2)/(20*sqrt(3)) or 1/sqrt(3).  so ("base"/2)/(height) = 1/sqrt(3)  and then "base" = (2*height)/sqrt(3).\n\nsubstitute that into the volume equation so that\nvolume = [(height^2)*length]/sqrt(3) or literally,\nV = [(h^2)*L]/sqrt(3).  differentiate with respect to time,\n(dV/dt) = [2*h*(dh/dt)*L]/sqrt(3) and substitute the numbers in:\ndV/dt = 200 - 50 = 150\nh = 10*sqrt(3)\nL = 450\nand solve for dh/dt which is the rate of change of the water level in yds/hour.  these are odd units, you might want to change to inches/hour.\n\ni did it this way and got 1/60 yds/hr or 0.6 inches/hr.\n\nbut again it has been a long time....
52661
after looking around at the usgs fl page for groundwater levels and recent precipitation records, i would guess that your water table has risen somewhat. from what i gather, this is normal fluctuation.  my answer is nothing more than a guess, because the data i looked at doesn't really support any other cause.  (all that just to say "I don't know.)
354566
The osmolarity property.
1022856
A water gauge would have been used. A hosepipe fitted both ends with glass rod, filled with coloured water, and of course it would be level
784896
fractional distillation
373532
you have your water temperature at between 37degrees and 39degrees buy yourself a thermometre to place in the water to check before you get in ok
552177
A bucket, a measuring cup, and a stopwatch is all you need.\n\nOpen the tap until it has achieved a steady flow, then put the bucket under it while turning on the stopwatch.\n\nFill the bucket to a desired amount or until you have a fairly long time.\n\nMeasure the water in the bucket and divide it by time. Thats the volume flow rate.\n\nTo get the velocity of the water when it exits the tap, divide the volume flow rate you have computed above with the area of the (circulare) hole.
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.
708158
well aside from that you said the vapor pressure is 18.6 and put 18.9 in your calculations, it looks great! Besides, even with that difference, it doesn't change the answer at all, its still .877 atm with both values.
812261
You are very close. If K = -0.032 is accurate enough, then use it.\n\npA = Ae^(19 * -0.032)\np = e^(19 * -0.032) ~ 0.544\n\nIf you leave K = ln(.5)/22 then\np ~ 0.5496
411587
You are right!!! The 1% solid part remains the same weight (1kg). If the water on the second day is 98%, then the solid must now be 2%. If it is now 2%, it still weighs 1kg, so the water weight is now 49kg. 49 + 1 = 50. \n\nAnother way to look at it is, the solid part is now double the amount it was, but since it didnt grow (did it?), the water must be half what it was.\n\nLets assume that on day X the water is reduced to 50%. The solid part is now 50%, but still weighs 1kg, so the water must also weigh 1kg, 1kg + 1kg = 2kg.
966711
The volume of the water in the cup at any moment is:\n\nV = (1/3) pi [ (Rw)^2 + (1)Rw + (1)^2 ] \n   = (1/3) pi [ (Rw)^2 + Rw + 1 ] h\n\nRw = Radius of upper part of water\nh = height of water\n\nRw / h = 2 / 6\n\n6 Rw = 2 h\n\nRw = 1/3 h\n\nSubstitute in to the volume equation\n\nV = (1/3) pi [ (1/9) h^2 + (1/3) h + 1] h\n = (1/3) pi [ (1/9) h^3 + (1/3) h^2 + h ]\n\ndV / dt  = (1/3) pi [ (1/3) h^2 + (2/3) h + 1] (dh / dt)\n\nSolve for dh /dt  and plug in the values h=4 and dV/dt = 3