for estimating of wave hight in a pond with the water depth of less than 30 cm (in paddyfield),which formula or method can be used for determining of wave hight
204068
You could place a vertical rod in the pond (on a stand or stuck in the mud?) and start the experiment when the rod is dry.  After you create the wave and the surface returns to rest, measure the height of the wave on the rod.  This will be the maximum height of the waves at the given point and will depend on the location of the disturbance (radiates outwards like a bullseye?).  Also, there should be nothing near the rod to magnify the wave (including shore effect or irregular bottom?).  Applying chalk to the rod (especially blue chalk?) may make the wetted height more visible and resist drying too fast.  Including a scale on the rod may allow you to read the wetted height above calm surface easier, especially using binoculars.  The trough should approximately equal the wave height.  Good luck.
471971
it is the form of measurement that is used to measure the "loudness of sound"\nIt's the amplitude of the wave
28654
Wave length is often represented by the Greek lower case letter lambda (your stick figure).  Replace that in your equation with "wavelength" or W until it makes sense then substitute lambda back into the equation.  It is just a shorthand notation that scientists have agreed upon (and nothing very magic).  See the link for more.  Good luck.
477829
To find wavelenght \n             wavelenght = c/v\n                                = (2.9979x10^8) / 9.73x10^14\n\nTo find energy \n           Energy = (plancks constant) x Frequency\n                    = (6.626x10^34) x (9.73 x10^14)
162423
Measuring the height of a wave is typically based on lowest to highest (after all, the wave could start at any point on the wave!)  The technical term for the vertical component of a wave is the amplitude, which is the average of the lowest and highest points of the wave.
120837
The height is 30 yd.  Just plug in what the problem gives you into the formula, and solve for the variable or the missing information.  \n\nArea = 1/2(Base.1 + Base.2)h\n\n450 = 0.5(20+10)*h\n450/0.5 = (30)*h\n900 = 30*h\n900/30 = h\n30 = h\n\n100% possitive
114439
in the unconstrained case (no floor), water from the lowest hole will reach the furtherest given enough time (vertical space to drop). With a vertical constraint (a floor), the optimal coverage will be from a hole one third the way up from the floor, not the top hole near the surface of the water, and not a hole at floor level.\n\nAlthough I really don't see how this applies to wind sailing:-)
652778
The pond is not likely round enough or a pedometer accurate enough for more than a rough estimate.\n\nA better alternative would be to use Google Earth.  It has a measuring utility that lets you define your path and will tell you distances more accurate than the above.\n\nGood luck.
540526
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
465343
30.5 m of water.\n\nConvert units to desired units.\n\n1 atm is 14.?? psi is about 32 ft of water.\n\nLook in reference for more detailed unit conversions.\n\nOr use P= &#92;rho g H.
512367
If we consider the volume of water is known.(If not known then use graduated cylinder and find the volume)\nMass of water can be calculated using following formula:\n\nDensity(ρ)=mass(m)/volume(V) , where\nρ is the object's density (measured in kilograms per cubic meter) \nm is the object's total mass (measured in kilograms) \nV is the object's total volume (measured in cubic meters) \n(Temp (°C)/ Density (g/cm3) \n30/ 0.9957 \n20 /0.9982 \n10 /0.9997 \n0 /0.9998 \n−10/ 0.9982 \n−20 /0.9935 \n−30/ 0.9839 )
313536
We finished covering sound and resonance a few weeks ago. In order to get a standing wave inside the tube, a node must be at the surface of the water (since it doesn't vibrate very well) and an antinode must be lightly above the rim of the tube for us to hear. It's up to you to figure out the smallest waves that have those characteristics (in terms of the wavelength).\nSince wave speed is frequency times wavelength, it's not hard to solve for wavelength. Plug it into your previous results and you have your answers!
432189
Can someone help me with my precalculus problem? Can you please read the following problem and figure out the?\n\nCan someone help me with my precalculus problem? Can you please read the following problem and figure out the equation for me along with how you figure this out?\n\n\nA tsunami (commonly called a “tidal wave” because of its effect is like a rapid change in tide) is a fast moving ocean wave caused by an underwater earthquake. The water first goes down from its normal level then rises at an equal distance about its normal level, and finally returns to its normal level, the period is about 15 minutes.\n\nSuppose that a tsunami with an amplitude of 10 meters approaches pier Honolulu, where the normal depth of water is 9 meters.
67233
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]
254510
Well as far as a formula i can't help you there. i just happen to own a program that allows me to draw perfect shapes of any size and diminsion them accordingly (auto CADD). and at the top of your 300 ft house, or where ever it is that your measuring from the horizon at sea level is a tad over 21 miles away (21.20389 to be analy exact). first you must know the the earths circumfrance (i know i can't spell) which is 20,890,637 ft, next you have to know the distance you are away from the earth (which you do, 300 ft). now you have to know the formula for finding a line that intersects your hieght and crosses tangentally across the earths surface. know you have your triangle and apply the Pythagorean theorem. you have an answer.
228310
Multiply the number of cubic feet by 7.481 to get the number of gallons (U.S.). \n\nThis formula applies to "liquid gallons", which are the kind commonly used today.
373842
E = hc/wavelength\n\nE = 6.62 10^(-34) * 3 10^8/ 5.95 10^-7 = 3.34 10^(-19)J\n\ni am qite sure