how do you measure the diameter of a water droplet?
88021
Using a set of callipers if the droplet is large enough. If not then using an eyepeice micrometer and a microscope.
649182
The size of the egg can be measured by two ways.\n\n1. By its breadth and width : Use any measuring scale with milli meter units to measure\n\n2. By its volume: Drop it in a filled water container and measure the displaced water in milliliter.
936045
Boy, you probably need to consider the liquid containing bottle's dimensions too. \nThe total average height of liquid will be 0.8 m + half the height in the bottle (say.\n\nAssuming the average height of liquid in the bottle = 100 mm = 0.1 m,\ntotal average height = 0.8 + 0.1 = 0.9 m\n\npressure exerted = potential energy= densityx acceleration due to gravityx height\n\nLength of needle can be neglected because they are usually installed horizontally and hardly add to the height. Also, it can be presumed that the flexible tubing and the needle were fiiled with the liquid at the comencement of the I.V drip. and the pressure difference if any due to this may be neglected.\n\nKinetic energy imparted = potential energy\n\n1/2 m v^2 = mgh\nv^2 = 2 gh\nv^2 = 2 x 9.8 x 0,9\nv= (2 x 9.8 x 0,9)^1/2\nv = 4.2 m/s= 420 cm/sec\n\nvolume to be injected/time in seconds\n= 490/ (35x60) = 1.4 cm3/ sec\n\nvolume flow (V) = speed (S)x cross section area\narea = V/S= 1.4/420 = 1/300 cm^2 \narea = 1/2 pi r^2\n1/300 = 1/2 x 22/7 x r^2\n\nr^2 = 1/300 x 2 x 7/22 cm2\n      = 7/ 3300 cm2\n\n  r    = .046cm = 0.46 mm\ndiameter (d) = 2r= 2 x 0.46= 0.92 mm................. Answer
33367
where the oak barrel is a cylinder:\n\nto get the weight, multiply 12*(the area-area of the tube)*density of water\n\np1v1t1=p2v2t2 to get the pressure of the water, where p1 is atmospheric, however you need to know how tall the barrel is to do this\n\np2*the area of the lid to get the pressure on the top
41200
If it is a high-volume, low pressure pump, one of the easiest ways is to get a long piece of clear tubing and make a U-tube manometer.  the pressure will cause the water to climb in the tube vented to the atmosphere, while the other end is connected to the pump outlet.  You measure the height of the water column, and can get the head directly in feet of water, a popular unit of measurement in the pump industry.  If it is a high pressure pump, a simple pressure gage from the local plumbing supply or hardware store will read in PSI.  In rough terms, 14.7 psi is approximately equal to 32.2 Feet of water column pressure.
734519
(1) Take a container and measure the length of the container. Let us say the length of the container is L. \n\n(2) Take a beam of light (preferably a monochromatic source like laser beam) and shine it through the empty container. \n\n(3) Measure the intensity of the light at the end where the beam has gone through the empty container. Let this be Io. \n\n(4) Now fill the container with the liquid whose molar extinction coefficient is unknown. Now measure the light intensity as the beam travels the length L of the container. The measured intensity is I.\n\n(5) Beer-Lamberts law is I=Io exp(-a L) where all the quantities have been experimentally determined except a which is the extinction coefficient. \nRearrange the above formula and we get, a=(-1/L)*log(I/Io) where log is to the base e.
409033
a vernier caliper should do the trick.  it can measure the inner and outer diameter
860910
Pour the liquid in the graduated cylinder and there will usually be mL markings along the side.  This measurement is volume
736682
The method that I would use would be:\n\n1. to completely fill  your five ounce measuring cup with water\n\n2. next, from the five ounce cup measure out 3 ounces of water   using the 3 ounce measuring cup and discard this water... leaving you with 2 ounces of water in the larger cup \n\n3. place this (2 ounces) water into the 3 ounce container\n\n4. next, pour water into the 5 ounce container again\n\n5. then, empty the bucket... leaving no water in it\n\n6. next, empty your 3 ounce container (which has the 2 ounces of water in it) into the bucket\n\n7. then, from the five ounce cup measure out 3 ounces of water   using your 3 ounce measuring cup and discard this water\n\n8. last, pour the water from your bucket back into the 5 ounce container, and you will now have 4 measured ounces of water in that container
724886
You can get measuremnts at 1000th of an inch from following page\n\nhttp://www.onlineconversion.com/ring_size.htm
905607
make a little "boat" that you can add weights to and float it in the different liquids.  Gradually add weight to the boat in each liquid to see how much it can hold before it sinks.  The more weight the boat can hold, the greater the density of the liquid!\n\nThis is basically a poor-man's hydrometer.
720092
While the answers above are correct, I will go a bit further.\n\nEstes rockets sells a nice tool for just this very job that will make your measurement fairly easy and accurate. It is called AltiTrak (TM).\n\nhttp://www.estesrockets.com/cgi-bin/products.cgi?view,296
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
554775
Put it in a graduated cylinder, record the water level, drop the rock in, and note the new water level. Subtract.\n\nYay science.
447183
Do you mean the experiment where you measure the size of an oil drop, then place it in a petri dish that has talcum powder or some such on it. The you measure the size of the pool of oil in the petri dish?\n\nIf so, this experiment was to calculate the approximate size of a molecule. \n\nEssentially, by measuring the diameter of the droplet of oil you can work out its volume (πr3). Then the nature of oil on water is to settle until it has reached a molecule's height and then to disperse. So, at its peak you measure the radius of the circle formed in the petri dish and you can work out how big the area is (πr2). \n\nThe volume of the new pool is then \n\ninitial_voume/secondary_area =  molecule height\n\n\nDoes this answer your question?
533825
Calculate the cross-sectional area of the water and multiply by the length of the cylinder.  The cross sectional area can be calculated with the depth of the wafer and the radius of the cylinder.  It is hard to explain without drawing a diagram but basically you draw 2 straight line connecting the center of the cycle (cross-section) to where the water surface contact the wall of the cylinder.  That form a sector.  Calculate the angle of the sector.  Multiply the total area of the circle by the angle of the sector divided by 360 degree to obtain the area of the sector.  Next, substract the area in the sector where there is no water and you will get the cross-sectional area of the water.
518403
You make a container 16 cm by 3 cm (and 1 cm thick) and pour water inside. The weight of the water in gm is your answer.\n\nAnother method is to use Napier's Bones. Just sliding and matching the numbers.