how do you prove that in fluids pressure against the wall say in a dam varies with level (height)?
605393
The pressure against the wall is equal to the weight of the water above it. If you are measuring pressure in pounds per square inch, the pressure at a given depth is the weight of a column of water 1 inch by 1 inch by the depth.\n\nYour explanation of what the solution should be makes no sense whatsoever. You cannot provide a proof of a pressure principle with electronics.
60733
simply use a hydrometer.
732533
D = M/V\n\nDensity= Mass/Volume
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).
759307
here we go  kinetic energy at point of shot equals potencial energy at top of the squirt\n\n1/2 m v^2 = mgh\n\nm's cancel  (Rained in seattle)\n\nv = Square root (2gh)\n\ncheck my algebra... Note, you must change cm to meters before you can answer
803552
it *is* Bernoullis principle.  Think of pressure as a form of potential energy and velocity as kinetic energy.  The total energy must remain constant so, if one increases, the other must decrease.\n\n\nDoug
1036468
At the top of the Dam, the water has 100% of its energy <or if you prefer energy per second - that is power >\n\n87% of power <=> 1900 MW\n100% of power <=> 2184 MW\n\nAt the top of the Dam, the potential energy the water possesses is what can be ideally converted into 100% power.\n\nDensity of water = 1000 kgm^-3\n\nLet the mass of water be y kg\nPotential energy of water = y * 9.80 ms^-2 * 170 m \n                                         = 1666 y Joules\n\nEach second it can deliver 1666y Watt of energy\n\nThat is 1666y W <=> 2, 184, 000, 000 W\n                           y = 1310924 Kg of water each second\n\nUsing density of water, above, each second, 1311 m^3 of water must flow from the top of the dam\n\nYour answer is ultimately 1311m^3/s\n\nI just love those physics exercises, feel free to contact me. I am a student in Science subjects
914444
I will be honest I only know part of the answer but here's what i know...\n\nfluid pressure increases with depth because there is more fluid on top of it thus more weight and it casues compression (pressure)\nand thats about all I can expalin
60322
Archimedes principle on bouyancy states that a body submerged in a fluis is subjected by an unbalance force called bouyant force equal to the weight of the displaced fluid. By formula, Fb = Vd\n;Fb = bouyant force\nV = volume of te dispaced fluid\nd = density of the fluid\n\nDimensional Analysis,\nFb must had a unit of force; Kg, N, lb\nV is volume; cu.m, cu.ft\nd is density; kg/cu.m. and so on.\n\nFb(Kg) = V(cu.m.)*d(Kg/cu.m.)\n\nu'll get Kg.
49359
At any given point in a fluid, yes, pressure is equal in all directions.  However, a given point is likely along a pressure gradient, meaning that the pressure changes in places directly adjacent to the first.\n\npressure is a scalar quantity, not a vector quantity.  Vectors have direction; scalars do not.
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
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]
372621
Since the speed is constant, the pressure difference is only due to the pressure exerted by the 4m column of liquid.\n\nThe pressure exerted by “h” column of liquid of density‘d’ is hdg.  (g= 9.81 m/s^2). \n\nhdg = 4 x 925 x 9.81 = 36297 pascal.
96913
Pascal's principle : Pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid, as well as to the walls of the container.  \nAn example: hydraulic lift.
936945
The best way to find the volume of a liquid is just to put it in a liquid measuring device, graduated cylinder or measuring cup. \n\nFor a solid, a displacement test is good when the shape is irregular. \nex. fill a graduated cylinder with enough water to ensure that when you place the object in the water it will be completly submerged. Note the volume. Place the object in the water, the water level will increase. The amount of the increase will be the volume of the object.
961939
The weights of the liquids inside 2 branches must be equal, therefore:\n\na) m_water + m_oil = m_mercury (the amount of mercury that raised in the other branch to keep the balance)\n\nm_water = D_water * V_water = D_water * h_water * Area\nm_oil = D_oil * V_oil = D_oil * h_oil * Area\nm_mercury = D_mercury * V_mercury = D_mercury * h_mercury * Area\n\nSubstitute this in the balance of weights above, we have\nD_water * h_water * Area + D_oil * h_oil * Area  = D_mercury * h_mercury * Area\n<-> h_mercury = (D_water * h_water + D_oil * h_oil) / D_mercury\n\nb) To bring mercury's surface two the same level again, the weight of the to-be-added water must be the same with the weight of both oil and water in the other branch.\n\nYou can do the same as above to find:\nh_water2 = (D_water * h_water + D_oil * h_oil) / D_water
983162
The force required is the pressure of the water at the leak times the area of the hole.  You are given the area as 0.012 m^2.  \n\nThe pressure of the water is dependent on the height of water above the leak, the density of the water and the acceleration of gravity (9.8 m/sec^2).  \n\nP = density x height x gravity \n= 1025 kg/m^3 x 3.1 m x 9.8 m/sec^2 = \n31,140 kg m/ sec^2 m^2.  \n\nA kg m/sec^2 = 1 N, so this is 31,140 N/m^2.  Multiply this by the area of the hole and you get 374 N.