Motion of a submerged sprinkler device?
465987
Actually, the sprinkler rotates in the same direction as before.\n\nThe motion of the water as it gets into the sprinkler exerts a force on the inner walls of the tubing that has a similar direction as in the first case. Considering the previous posted answer, the thing to consider here is the friction. In a frictionless system, it may not move at all, but since our world is not frictionless, the contact of water with the walls implies some momentum exchange.\n\nIt's hard to explain without images, but try the following. Having the sprinkler out of the water, try blowing some air inside one of the openings with your mouth or a syringe. Obviously the whole tubing will start spinning away from the direction of the blowing air. Well, the same happens with the sucked water, but instead of it being "pushed" inside, it's "pulled" inside. The process may be different in cause, but the effect is the same.
835572
Gotta approach this using basic principles.\n\nEnergy is the integral of force applied over a distance.\nTheres two sets of forces working:\nfluid and gravitational.\n\nSo you make your control volume, and look at it.\nYou want to raise the water in it, thats mgh or rho*v*g*h\n\nYou also have to exert a pressure to push against the water ahead of it.  The downward pressure acting on your control volume is f/a so the force you must overcome is p*a and you are traversing distance h so the energy is p*a*h\n\nTherefore to raise the fluid vertically against a down pressure and against gravity the minimum energy (real world is lossy) is \nE = h*(mg+pa)
695604
The only way I can see this being done is to have a piston or other way to separate the water and air; otherwise, the air will simply bubble up through the water.\n\nAssuming there is a piston or other separation device and the water will overflow at the top, the initial pressure needed to begin the displacement of the water is the height of the water in inches  times its density in pounds per cubic inch. Taking the density of water as 62.4 pounds per cubic foot and converting to pounds per cubic inch the density becomes .036111. The height is of course 100 ft time 12 inches per foot or 1200 inches. This yields an initial pressure of 1200 time .036111 or 43.33 psi. As the piston moves upward, the required pressure will decrease directly in proportion to its height or .4333 psi per 12 inches of rise.
599546
water always follow the less resistance route...
537660
The vertical rise must not exceed the head of water, or the riser will not work.
121975
The machine consists of a screw inside a hollow pipe. The lower end of the device is put in the water, and the screw is then turned (usually by a windmill or by animal or human labor). As the bottom end of the tube turns, it scoops up an amount of water. This puddle of water will slide up in the spiral tube as the shaft is turned, until finally it falls out from the top of the spiral tube and feeds the irrigation system.
1099280
What you are seing are crops irrigated by center pivot irrigators.\n\nCenter pivot irrigation is a form of overhead irrigation consisting of several segments of pipe (usually galvanized steel or aluminum) joined together and supported by trusses, mounted on wheeled towers with sprinklers positioned along its length. The system moves in a circular pattern and is fed with water from the pivot point at the center of the arc. These systems are common in parts of the United States where terrain is flat.
287325
try casting in the back yard also try putting your thumb lightly on the spool just before it touches the water to slow the spool down. then let it sink...
702956
Okay I think the other 2 answers you already had when I started this misunderstood your question so I'll get it right for you.\n\nTwo things happen in the 16 seconds.\n1. rock falls s meters\n2. sound travels back up s meters from water surface to well opening\n\nFirst let's deal with the #1 part, the rock's fall.  For this let's use a variable called t1 for time.\n\nWe need to use this formula: s = (v1)(t) + ½(a)(t)²\ns = displacement, v1 = initial velocity, t=time, a=acceleration\n\nSince the rock is dropped it has no initial velocity.  Let down be positive.\n\ns = (v1)(t1) + ½(a)(t1)²\ns = (0)(t1) + ½(a)(t1)²\ns = ½(a)(t1)²\n\nOkay so we figured out what we can in this step about t1.  Now let's go to step 2 and see what we can figure out about t2.\n\nThe sound travelled from the water up to the well opening (s meters) in some amount of time we don't know yet (t2).  We can describe the sound's movement with this formula:\n\ns = (v)(t) (we use this because the sound is not accelerating)\n\ns = displacement, v = velocity, t = time\n\nNow let's use our variables (t2, our s) and use v2 for the sound's velocity and see what we can put together with the sound.\n\ns = (v2)(t2)\n\nOkay so we know something about the sound and rock now, let's gather what we found and look at it.\n\n1. s = ½(a)(t1)²\n2. s = (v2)(t2)\n\nThere is something else we know too and can write down as an equation.  We know that the time it took for the rock to hit the water and the time it took for the sound of the splash to travel back up to the opening add up to a total of 16 seconds.  So we can write it as:\n\nt1 + t2 = 16s (let's isolate the t2 and call it equation 3)\n3. t2 = 16 - t1\n\nSo all together we have:\n\n1. s = ½(a)(t1)²\n2. s = (v2)(t2)\n3. t2 = 16 - t1\n\nWe have enough equations now to solve for the number of variables we have.  Start by substituting t2 from #3 into #2 to get equation #4.\n\n4. s = (v2)(16 - t1)\n\nNow set 1. and 4. equal to each other (because they are both s so they are equal so we can do that)\n\n(v2)(16 - t1) = ½(a)(t1)²       (expand it out now)\n\n16(v2)-(t1)(v2) = ½(a)(t1²)\n\nNow put in what we know (v2 = 345m/s, a = 9.8m/s)\n\n16(345)-(t1)(345) = 4.9(t1²)   (move it all to one side and it should look familiar)\n\n4.9t1²+345t1-5520=0  (it's a quadratic equation and we can solve those, stick it in your calculator)\n\nt1 solves to\nt1 = 13.436\nor\nt1 = -83.844 (forget this one, the rock didn't hit before you dropped it)\n\nSo t1 = 13.436m (we only know this to 2 significant digits)\n\nNow we can finally solve for s by using equation #1\n1. s = ½(a)(t1)²\ns = ½(9.8)(13.436)²\ns = 884.57m (only to two sig. digits)\n\ns = 880 meters
65968
The water in the toilet is inside a "knee", basically an s-shaped combination of curved tubes. If the far side of the tube develops underpressure, some of the water will be pulled over into the far side, and water level will sink.\n\nPressure on the far side can oscillate. Usually the waste water system is connected to outside pressure via a separate chimney. On windy days, this chimney may act like a jet pump, reducing the inner pressure of the canalisation, so that water will be sucked into the falling side when the oscillation goes that way, lowering the water level.\n\nAnother possible source of oscillation are huge amounts of water flushed at once, moving down the tube pushing all the air below and leaving lower pressure above.\n\nBarometric pressure shouldn't really come into this, since canalisation and the open air are connected. However high or low the pressure, it ought to be the same inside the canalisation unless strong wind makes the chimneys act like jet pumps.\n\nNow, at which air pressures do you observe strong wind?
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.
104409
This is a classic problem from hydrodynamics textbooks. The short answer is, it is approximately simple harmonic motion. You can compute the restoring force caused by the displacement of the liquid if you depress a floating boat (Archimedes principle) and that then, balanced with the weight of the boat, creates the conditions for simple harmonic motion. You can approximate the period and amplitude of motion etc. \n\nHowever it gets more complicated than this of course:\n\na. There are damping forces caused by the motion of the boat through the water and the air\n\nb. The water is not completely incompressible, and the density of the water varies some with depth (a tiny effect)\n\nc. There are surface tension effects and entrainment effects\n\nd. The motion of the boat will be more complicated than simple linear motion in one dimension\n\ne. There will probably be deformation effects on the boat, changing its shape\n\nf. gravity is not constant as the boat is moving up and down a small distance in a gravity gradient\n\nand so on. \n\nHowever, the REASON that sines and cosines are so prevalent everywhere in physics is because simple harmonic motion is such a good approximation for so many things. \n\nWhy is this?\n\nWell consider Newton's Second law: F=ma, where F is force, m is mass and a is acceleration. Newton's second law is not always true, in the quantum world and in relativistic situations, but it is true under many conditions, so it is a good place to start. \n\nIn general, the force F will be some sort of nonlinear function of position x:\n\nF=F(x)\n\nHowever, if x is small, then we can approximate F(x) by a McLaurin series:\n\nF ~ F(0) + x F'(0) + h.o.t. \n\n(h.o.t. is a common acronym for higher order terms).\n\nThis then produces the equation for simple harmonic motion, when substituted into Newton's 2nd law:\n\nF(0) + xF'(0) = m d/dt(d/dt(x))\n\nMany other equations in physics like Schrodinger's equation, or wave equations with forcing, or the heat equation, or the inhomogeneous Laplace's equation include 2nd order derivatives, and can similarly be put in some form where simple harmonic motion is the result. So as strange as it seems when you are just learning about SHM in physics, it really is everywhere in all kinds of unexpected places! \n\n\nOne of my favorites is the waves of heat and cold that are travelling down into the earth from the seasonal forcing produced at the surface. Drilling down, one can observe these waves of hot and cold (with some diffusion) going back many years.
810879
The problem is not the weight of the water, it's the drag of the fluid. Unless you can find some magic way to eliminate the drag in a fluid like water which is so much more than that of a gas, you will be stuck with lower speeds. Even if you had some "magic" bubble around the sub, you would still need to displace the liquid to move the bubble.
963866
Cover one end of your garden hose with clear plastic (like Saran Wrap) and wrap rubber bands around it to hold the plastic in place.  Then remove your garden hose from the faucet and cover the faucet end of the garden hose with plastic wrap and use rubber bands to hold that in place as well.  Lay down the garden hose over a relatively flat area of the ground where you wish to determine level elevation.  Then look at each end for air bubbles.  (Make sure the garden hose isn't draped over a hill, or this procedure won't help.)\n\nIf you find air bubbles at one end of the hose, it means that end is at a higher elevation than the other.  Move the bubble end of the garden hose, without moving the other end until the bubbles leave the plastic.  Then check the other end of the garden hose.  If you find bubbles, move that end only until the bubbles leave the plastic.  \n\nRepeat this procedure until you don't see bubbles at either end.  That is when the ground is level along the length of the garden hose.
1184219
an extra floatation device for added safety
1018827
It will depend on the water pressure and the size of the orifice of the sprinkler head.
1181391
if you statically charge some sort of object, a glass rod works very well, by rubbing it on material in one direction only then hold it next to a stream of water the water should be repelled and bend away from it.