I would like to know the Physics involved when sailing?
140821
There is a lot of physics in sailing. The most important principle is Bernoulli's Principle, and it is the main reason that a sailboat moves. When the wind blows on a sail, it produces turbulent motion in the lee area of the sail, and lower pressures, which pull the sailboat forward. \n\n\nThere is also drag of the water and the air to consider, and buoyancy forces and their balance, wave propagation, and many other details. The following web site will get you started.
1038155
Bernoulli principle if you planning create a sailboat.
276362
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:-)
393880
It depends on its speed, weight, size, motor size, wind power, wave height and direction of the ship compared to the direction of the wind.
343204
It cannot go directly into the wind, rather it has to go back and forth at an angle to the oncoming wind, called "tacking".\n\nIf you want to go North, but the wind is blowing out of the north, you have to go northeast, then turn and go northwest, and keep going back and forth like that until you eventually get where you are going. \n\nThe keel in a sailboat makes this possible.  It allows the boat to take wind from the side and push it forward, not tip it over.  \n\nIt's easier to see using pictures.  See the ref.
516448
Sails can go against the wind but not ireactly into it. There is about 90degrees, 45degrees either side of the direction that boats with sails struggle into.\n\nThe sail is shaped much like an aeroplane wing and it is this shape that forces the air to move fater over one side of the sail than the other, decreasing air pressure on one side of the sail and pulling the boat along. This will pull the boat in a slighlty side on motion to the desired travel but a central board (dagger board) creates sideways resistance and forces the forward  motion to be dominant.
966112
u can consult ur physics book
1075806
I assume that you mean that the object is perpendicular to the velocity of the wind.\n\nConsider a flat plate, area A, against which a wind of speed v is blowing.  The air has a density of p.  All measurements are made in a consistent unit system of course.\n\nIn one second a mass of air will be brought to rest, this mass is given by A x p x v\n\nThe momentum change of this air is therefore A x p x v²\n\nwhich gives the force on the flat plate.  Your object will also have a drag coefficient (Cd) the more streamlined the object is the lower it's drag.  So the total force becomes:\n\nA x Cd x p x v²\n\nYou will have to look up the drag coefficient for your object in a text.
1026652
Here's a good explanation I found on Fuidzone.com:\n\n"A good turn involves shifting and using your weight. Slip back on your board a bit (assuming you are already up on the board for more speed) and slightly closer to the side of the board which is closest to the wave face. From there you need to apply weight to the inside rail. Pull up on your outside hand and also slightly up with the hand on the nose. Try to push your inside hip into the water more but keep your body over the board. You don't want to scoop water. The more you lean the sharper you will turn but you will also lose more speed."\n\n\nThat sums it up well, I think. But if you are talking about turning to do a 360 or cut back, it's a little different than adoing a bottom turn. For thos you need to get your legs out of the water and swing them around to help you turn the direction you want to go.
824337
Assuming that the boat is not moving under constant power, and neglecting the force of water resistence, the boat's new velocity would be 9.33 m/s.  \n\nWe know this because of conservation of momentum.  The momentum after the collision must be the same as the momentum before the collision.  Since momentum is equal to mass * velocity, the boat's momentum before the collision is:\n\n510kg * 11m/s = 5610kg*m/s. \n\nSince the momentum is the same after the collision, we just divide the total momentum by the new combined weight to get the new velocity:\n\nV = 5610kg*m/s / (510kg + 91kg) = 9.33m/s  \n\nThe speed with which Batman hit the boat does not factor into the equation, unless we're interested in the boat's angular velocity.  I'm assuming that you only want to know the boat's horizontal velocity over the water.  Since Batman had no horizontal momentum (he was falling straight down), his movement would not affect the horizontal motion of the boat.
783086
Gravity is pulling down on the object, air is resisting it (pushing up, if you will) and causing friction, and if there's any wind it's pushing sideways on the object. I doubt you'd have wind in your problem, though. Wind is complicated :)
459186
This sounds like a nautical question indeed!\nHmmmmm...
468214
Like wallablack says... Whilst going in the same direction as the wind you go in a straight line....but when you go into the wind you move in a large zig zag motion side to side edging forward in small increments using the tension between the board and the wind to propel you.
34167
Ur... I thought the question was HOW, not "WHY":\n\nYou draw arrows for each force.  They start from a point, an go in the direction of the force.  You draw them to scale so they are proptionately long for the different forces.\n\nSay you have a boat doing 20 mph due west, against a current 10 mph due north.  Where will the boat end up in 1 hour?\n\nScale:  1" = 10 mph\n\nDraw a line to the left (West) 2" for 20 mph\nDraw a line up from the same starting point, 1".\n\nTo find the end point, draw paralleles so that it makes  four-sided figure.  A line draw from you origin to the point across the diagram will tell you  the resultant force (from length of the line) and the bearing or direction it will go.\n\n(Its easier when you draw it, believe me)
452179
Latitude, Ocean currents, Winds, Elevation, Relief and Nearness to water.
216411
Sailing - There is nothing like the sensation when you head the boat up into the wind, hoist the main sail, fall off the wind.  Suddenly the sail snaps full of wind,  the boat responds by gently heeling over and lunging forward.  Now you are on your way with the only noise being the wind in the rigging and the wave at the bow..
890224
Velocities are vectors, so you´ll have to proceed according to vector rules for addition/subtraction.  Draw a horizontal line, with an arrowhead pointing to the left. This represents the (unknown) velocity due West. Now draw another line, vertical, this time.  Put an arrowhead in the lower end of this line.  This stands for the 50 kph wind speed to the South.  These two lines should make a right (90°) angle.  The lines haven' t to be drawn to scale.\n\nFinally, draw an oblique line connecting both remaining ends.  This is the aircraft speed; to indicate direction, put an arrowhead  at its leftmost end.\n\nNow, you have a triangle of which two sides are known.  This, and the fact it is a right triangle, suffices to solve the entire triangle. The angle you seek is that formed between the horizontal and oblique lines.  Let θ denote this angle.  From elementary trigonomety, sin θ = opposite side/hypotenuse = 50/205 = 0.2439... All you have to do now is to find the angle whose sine is 0.2439...  This is 14.1169...°.  From the figure you have drawn, this angle lies N of W.\n\nThere are several ways to figure out the speed to ground.  You may use Pythagoras' Theorem; or, having found the opposite angle, you may compute it as 205 × cos 14.117° = 198.809...\n\nJust follow the same procedure to solve the other problem.  You'll have to decide which direction is "river across" and which is "downstream"; I´ll suggest "right" and "down", respectively.  Just plug in the proper figures.