As aerodynamic resistance is the square of the speed of the craft, how is hydrodynamic resistance?
402307
Hydrodynamic drag on a submerged body is calculated using the same formula as for aerodynamic drag (ref. 1). However, for surface craft, the energy put into formation of surface waves must be taken into account. For a non-planing (displacement) hull, the maximum practical speed is proportional to the square root of the waterline length of the hull. Above this speed drag increases much more rapidly. The "hull speed" rule-of-thumb formula is Hull Speed = 1.34 * (LWL)^1/2, where hull speed is in knots and waterline length LWL is in feet (ref. 2).
517321
cfm. Cubic feet per minute, a measure of air flow
1131554
Yes. \n\n(All of the following bars air resistance.)\n\nA trajectory would be the path that it follows. If an object has no initial velocity, it would fall straight down.\n\nBut say it did have an initial velocity. Let's say you threw a football. Does the football ever stop in mid-flight and fall straight down? No; it has an initial velocity, a sideways trajectory. If something has an initial velocity sideways, it's trajectory will look like an upside down parabola (ie y= - x^2) once gravity is applied.\n\nHope that helps.
909831
if you are going 4.5m/s then you have traveled in two directions, straight across and downstream. If you make a drawing of this it will create a triangle with 2 sides of the river and a hypotenuse of the distance traveled, 27.\nyou know one side on the river is 14, the hypotenuse is 27\nthe pythagorean theorem says that A squared + B squared = C squared for a right sided triangle\nso 14 squared + distance across river squared = 27 squared\nD squared  = 729 - 196\nd = square root of 533\nd= 23.08 \nso the distance traveled at 4.5m/s was 23.08\nwe know that d = r * t\nso t = 23.08/4.5\n\nt = 5.15 s\n\nwe also know that in 5.15s the boat went 14 m downstream\nagaing d=rt\n14 = r * 5.15\nr (speed of current) = 14/5.15\nr= 2.71m/s
915571
Initial vert velocity is zero\n\nfor free fall\n\nd=Vixt + 0.5xgxt^2, d=60, Vi=0, g=9.81\nt=(60/(0.5x9.81))^0.5\nt=3.497487084 s\n  \n\nIf there is no air resistance, horizontal velocity doesn't change
165147
When a bag of nails reaches its terminal speed, the force of the air is exactly equal to the force exerted downward, so the bag no longer accelerates.  Therefore, the amount of air resistance is 100 N.
893701
v(m/s) = Q(m3/s) / Nozzle Area(m2)
924224
No.  The design of the vehicle does not matter, it is the old e=mc^2.\n\nAs you accelerate, your vehicle gains mass, until, as you get closer to the speed of light, the mass gain approaches infinity, and it then takes an infinitely powerful push to overcome this infinite resistance.\n\nIn the case of the nested doll ships, if we use the external (larger ships), to begin the journey, you would find yourself in the strange position of continually needing greater power from ever smaller engines.
338219
the resistive force that air will exert upon an object moving through it at a given velocity.\n\nthe equation is Drag= 1/2 cpav2 (squared)\n\nc is a constant.\np is the density of the fluid (air in this case)\na is the cross sectional area of the object\nv is the velocity of the object.
1088769
P = d * Q * g * H * efficiency\nWhere,\np= power in KW\nd= density of fluid used\nQ= discharge\nH = head\nefficiency= given for a turbine
251868
Depends on the level of elevation.  The higher elevations have less air pressure.  At sea level it's about fifteen pounds for every square inch.\n\nConversion to mph is about 31.07 mph or 546.83 inches per second.\n\nHere are some drag formulas\nhttp://carini.physics.indiana.edu/E105/drag-force.html
1020782
Principle of all boats is that of buoyancy. Objects will float (will have buoyancy) when the volume of the liquid that they displace is equal in mass to the mass of the object. In a way, you can imagine a 1-to-1 tradeoff. When you place a boat of X volume in water, it displaces (pushes aside) X volume of water. That X volume of water, if you can take it out and weigh it, will equal the weight of your boat. So you can do some calculations for "displacement" (the volume that your boat takes up in water) and then calculate the mass of the water it displaces. If the mass of the water you displace is greater than the mass of your ship, it will float.\n\nNo matter what, make sure the boat is waterproof. In olden days wooden boats were sealed with tar and wax. Any waterproof substance (usually an oil-based resin) should help seal the boat.\n\nYour boat moves faster if it displaces less water. But wait..there's clearly another trade-off here: if you have a lot of displacement, you float better, but that also means your boat takes upa lot of volume, and that will increase water resistance and make your boat slower. So your task here is to find a volume of displacement that allows your boat to float while it doesn't take up so much volume that it will slow the boat down. One solution is to make the hull (underside that is beneath water) more hydrodynamic (more flowy). You can experiment with  different designs (if the materials permit) such as a standard hull or a catamaran. The principles of hyrdodynamics are similar to aerodynamics: sharper contours are less resistant than square shapes. Think of an arrow going through air versus a square brick.
242689
There is a boiler in Nuclear Subs, and a diesel engine in (obviously) diesel subs. The steam pressure from the boiler pushes through a set of mechanical turbine fans, which have output shafts driving a set of small gears. These small gears turn a large one, called the bull gear, which reduces the speed of the screw, or the impeller part (the big fan that spins) of the "hydrodynamic propeller", in order to provide torque (strength). If the propeller were to turn at high speeds, it would "tread water", because the low pressure it creates would be enough to vaporize the water (this is called "cavitation". The propeller would be simply stirring up steam and have no push. So the propeller has alot of strength at low speeds and is very large. It can push on an extremely large area of water. The water does not move, due to laws of inertia and mass, and gives the submarine something to push against, and therefore allowing it to move forward.\n\nA vague representation of this ability is the little experiment where you put a ruler on the edge of a table, and put something small and heavy on the end that sits on the table. Slap down on the end hanging off, and watch the object go flying. Next, spread a sheet of newspaper over the ruler, all the way to the edge of the table. Now slap down on the ruler again, and if you it hard enough, you will either break or bend your ruler. This happens because, thought the newspaper is much lighter than the previous object, it has the entire area of air above it, which has mass. The surface area of the newspaper is large enough to distribute all of the force to the mass of air, and, because of the law of inertia, the air holds the newspaper down.\n\nIn a diesel sub, it is basically the same concept, except that there is a direct drive between the engine and the small gears.
767660
A rocket is subjected to four forces in flight; weight, thrust, and the aerodynamic forces, lift and drag.\n\nat rest I will denote the launch pressure by the symbol ps. The pressure exerts an equal force on all sides of the pressure vessel, the cone of the engine. Because the rocket is free to move along the launch mechisims, the pressure exerts a vertical force given by: \n\nFp = (ps - po) * A \n\nwhere Fp is the pressure force, po is the atmospheric pressure, and A is the cross-sectional area of the rocket tube. The net force acting on the rocket is the difference of the vertical pressure force and the weight W of the rocket: \n\nFnet = (ps - po) * A - W \n\nTo simplify the analysis, we are going to assume that the length of the launch tube is small relative to the length of the rocket. As the rocket moves, the pressure remains fairly constant inside the tube. In reality, there is a slight decrease in pressure as the volume increase, but we are going to neglect that effect to simplify this analysis. With a constant pressure, and constant pressure force, we can use a simple algebraic form of Newton's second law to determine the acceleration a, velocity v and the distance y which the rocket moves along the launch tube: \n\nFnet = m * a \n\nm * a = (ps - po) * A - W \n\nwhere m is the mass of the rocket This equation can also be written as: \n\nW * a = g * ((ps - po) * A - W) \n\nwhere g is the gravitational acceleration, which is equal to 32.2 ft/sec^2 or 9.8 m/sec^2 on the surface of the Earth. The value of g is different on the Moon, and on Mars. The weight W is equal to the mass times the gravitational acceleration. Now divide both sides by the weight to determine the rocket acceleration along the launch tube: \n\na = g * ((ps - po) * A / W - 1) \n\nFor a constant acceleration, the velocity v and distance traveled y are given by: \n\nv = a * t \n\ny = .5 * a * t^2 \n\nwhere t is the time. Using the length of the launch tube L, we can now solve for the lift off time - TLO by using the second equation: \n\nTLO = sqrt (2 * L / a) \n\nand then using this time, we can use the first equation to solve for the velocity at the end of the tube: \n\nv = TLO * a \n\nv = sqrt ( 2 * L * a) \n\nAs the rocket clears the end of the launch tube (t = TLO+), the air inside the rocket, at pressure ps, will rush out the exit and the pressure will equalize with free stream inside the tube. In reality, this will provide an additional amount of thrust T given by: \n\nT = m dot * Ve \n\nwhere m dot is the air mass flow rate out of the tube, Ve is the exit velocity of the air. For an air rocket, with no nozzle and small volume, this additional thrust is small relative to the weight of the rocket, and we will ignore this effect. \n\nThe velocity v at the end of the launch tube is used as the initial velocity in the flight equations to determine the flight trajectory of a rocket.
449788
Doesn't matter whether it's dropped from a helicopter, skyscraper, or the goodyear blimp.  Terminal velocity is the velocity at which the air resistance of a falling object is equal to the acceleration due to to gravity, and the object can no longer can gain speed.  It's not a constant, as you get closer to the ground air pressure will increase, so the air's resistance will increase, and the teminal velocity will drop. The object will actually slow down.
1099052
21 %\n\nbcoz range is proportional to the square of velocity \n\n*********************************************************************\n\nR prpotional to square (u)\n\nlet u0 initial velocity ,r0 inital range\n\nu1 velocity with 10 % increase ,r1 new range\n\nlet u0=10  >>> then r0=100 (proportional to 100)\n\nwith 10% increase in velocity\n\nu1=11 >>>> r1 =121\n\nthus %age increase =21 %\n\n\n*********************************************************************
705060
If I understand the question, you are giving a velocity to the cube and don't want it to touch the hemisphere at any time in its flight. I think you want the minimum, not the maximum hor. velocity to do this, right? Because any arbitrarily high velocity will do it. I'll use a planar analysis where the hemisphere is a semicircle.\nAssuming the semicircle is centered at x=y=0, the trajectory will be a parabola y=r-kx^2 and at y=0, |x| must be >r. Another constraint arises when we consider the region of the semicircle near the top. Here the semicircle approximates a parabola, since 1-cos(small angle) approaches small angle^2/2. Thus deltay = r(1-cos(small angle)) = r-y approaches r*small angle^2/2 = r(x/r)^2/2, or x approaches sqrt(2/r*deltay*r^2) = sqrt(2r*deltay).\nDownward acceleration is g and horizontal velocity is v. When y=0, 0.5gt^2=r, so t=sqrt(2r/g), and x=vt must be >r. Thus the 1st condition (for clearance at the bottom) is v > r/sqrt(2r/g), or\nv > sqrt(rg/2).\nThe 2nd condition requires that when r-y=deltay is small, then and x=vt must be >sqrt(2r*deltay) where t=sqrt(2deltay/g), or  \nv > sqrt(2r*deltay) / sqrt(2deltay/g) or \nv > sqrt(rg).\nThis latter is the more stringent of the two requirements and thus is the minimum clearance velocity.