what is the formula for drift, like in a lifeboat?
111743
There is a great Cal Tech Video on that.  The one dealing with vectors.  It should be at most larger libraries.
787054
Well, the simplest way would be to assume that Alex swims at a constant speed and takes no time at the turnaround.  The resulting graph would be two straight lines - one from the origin to (12, 25) and one from there back to the x-axis (the t-axis) at (24, 0).  You'd need two equations.\n\nFor the first, the slope will be 25/12 (25 yards in twelve seconds), so the equation would be\n\ny = (25/12)x where 0 <= x <= 12\n\n(You can replace the x with t, if you prefer.)\n\nThe other line would have a slope -25/12 (25 yards back in the remaining 12 seconds).  We'll have to use the point/slope form for that one:\n\ny - y1 = m(x - x1)  using m = -25/12 and the point (24, 0)\n\ny - 0 = (-25/12)(x - 24)\n\ny = (-25/12)x + 50\n\nAgain, you can replace x with t.\n\nHope that helps.
718408
All objects are pulled by earth with constant acceleration\n\nAcceleration is change in velocity per unit time = change in velocity / time\n\nIn one second the velocity changes by g unit.\nIn t seconds it velocity reduces by gt.\nIf u is the initial velocity and if in t seconds it is reduced to zero then u = gt.\n\nOr t = u/g.\n\nThe time to go up is u/g.\n\nThe height it has gone up is given by average speed x time. = u/2 x t = u/2 x u/g = uu/ 2g.\n\nWhen it falls down it starts with zero speed, but falls with the same acceleration g.\n\nWhen it reaches the ground let its speed is v and time of fall is t2.\n\nv= gt2.\n\nSince the distance is = average velocity x time = v/2 x t2 = gt2/2 x t2= gt2t2/2  \n\ngt1t2/2 = uu/ 2g. or t2t2 = uu/ gg  ot t2 = u/g .\n\nThus time of fall is equal to time of up.
1008231
Set the equation of motion.\nTaking upward direction as positive\ns=vt-1/2gt^2\nat maximum height s=10,v=0\ng=-g since upward direction is positive\nyou get t= sqrt(2s/g)\n=sqrt(2*10/9.8)\n=1.428 secs
535174
There is often a wide divergence between theory and application.  I remember one application where lengthy mechanical linkages and cables were used to transmit rudder position for a huge seagoing ship in the even that the steering gear room became flooded (and electronic indication was lost).  The system had too much lost motion to indicate rudder position even though it looked good in theory.  I suggested connecting a vertical rod to the rudders that pierced the overhead into the ("unflooded") compartment above which could be connected to a reliable electronic transmitter (similar to existing "flooded" transmitters) and that worked much better.  Even electronic control systems can exhibit unexpected hysteresis when built.
1111523
I think that "D" is a combination of the drag coefficient, the denisty of air, and the cross-sectional area of the skydiver.  I make that assumtion because they don't provide any of that information, and the units work out.  I don't know if it encapsulates the "2" in the terminal velocity formula or not.  The numbers look prettier if you assume it does.\n\nSo use the formula: Vt^2 = (m x g) / D\n\nVt = Terminal Velocity\nm = mass\ng = acceleration of gravity (9.8m/s^2)\nD = "the constant"\n\nsolve for D:  \nD = (85 x 9.8)/(42^2)\nD = .472\n\n\nfor part 2, solve for Vt:  \nVt = sqrt((40 x 9.8)/.0.25)  \nVt = 39.6\n\nBy the way, the first answer is really really wrong.
257784
SIne they travel at right angles to each other, the distance between them at any point of time will be the hypoteneuse of a right angle triangle. The path of each boat from the dock will form the other two sides.\n\nLet teh speed of the slower boat be x mph\nHence, the speed of the faster boat = (x+1) mph\n\nUsing Pythagoras' Theorem\nx^2 + (x+1)^2 = 5^2\n\nSolving the equation,\nx^2 + x^2 + 2x + 1 = 25\n\n2x^2 + 2x = 24\n\nx(x+1) = 12  (divided by 2 throughout)\n\nx^2 + x - 12 = 0\n\nx^2 + 4x - 3x - 12 = 0 (FInd two numbers whose sum is 1 and product is -12)\n\nHence, \nx(x+4) -3(x+4) = 0\n\n=> (x-3)(X+4) = 0\n\n=> x = 3 or x=-4\n\nSpeed cannot be negative. Hence, x = 3\n=> x+1 = 3+1 = 4\n\nHence,\nSpeed of slower boat = 3 mph\nSpeed of faster boat = 4 mph
102873
Archimedes Principle.  The bouyant force pushing up on an object in a fluid is equal to the weight of the fluid displaced by the object.  So as long as the ship displaces more than its weight in water, it will float.  This is why ships made of steel and other dense materials tend to be very big and hollow.
1184219
an extra floatation device for added safety
1102604
This is an oddly specific question just to be asked out of curiosity.\n\nThe mass of the cannon ball is 13.6 kg, thus it has a weight of about 133.4 Newtons (assume g = 9.81 m/s^2).\nThe force of weight is pointed down...however, since the cannon ball in not in a vacuum and/or not a point mass, it has a non-zero buoyant force acting upward on it.\n\nThe buoyancy force acting on the ball is dependent on the weight of the fluid (water in this case) it displaces.\nIron has a density of about 7874 kg/m^3, thus a 13.6 kg sphere would have a volume of about 1.73 E-3 m^3.\nIf we assume the salt water of the ocean has a uniform density of about 1028 kg/m^3, then the weight of the displaced water exerts a force of about 17.42 Newton’s upward.\n\nThus, the net force acting on (stationary) cannon ball is about 116 Newtons downward.\nAccording to Newton’s 2nd law,\nF = ma\nThis force will cause the 13.6 kg cannon ball to accelerate at a rate of 8.53 m/s^2.\n\nAssuming there is no "water" resistance / drag, then using the formula,\nd = 1/2 at^2,\nwhere the ball stars from rest and experiences a uniform acceleration for a time t as the ball covers d distance,\nThe cannon ball should take about 50.57 seconds to reach the bottom.\n\nHOWEVER,\nJust as there would be air friction if this ball was launched into the air, there is most defiantly "water" friction as this ball accelerated downward....eventually reaching a terminal velocity.\nThe water will exert a drag force upward on the ball, causing its acceleration to be non-uniform.\n\nThe drag force can be calculated as,\nF_d = C_d * ((p * v^2) / 2) * A\nwhere F_d is the drag force, C_d is the drag coefficient, p is the density of the medium which the ball is falling in, v is the instantaneous velocity of the ball, and A is the cross sectional surface area of the ball in the direction of travel.\nIn order to calculate the drag force, one must know the drag coefficient of the cannon ball in water, something which usually must be calculated experimentally.\nThe ball will accelerate non-uniformly until it reaches it terminal velocity in the water, at which time, acceleration will cease and the ball will continue falling at its terminal velocity.\n\nUnless more information is know (the drag coefficient), the best (guess)/estimate of the how long the ball will take is sometime longer than 50.57 seconds...probably on the order of well over a minute since the ball will not be accelerating nearly as long as it did in our mathematical model calculate earlier.
704408
wikipedia.org\n\n"On Floating Bodies (2 volumes) \nIn the first part of this scroll, Archimedes spells out the law of equilibrium of fluids, and proves that water will adopt a spherical form around a center of gravity. This was probably an attempt at explaining the observation made by Greek astronomers that the Earth is round. Note that his fluids are not self-gravitating: he assumes the existence of a point towards which all things fall and derives the spherical shape. One is led to wonder what he would have done had he struck upon the idea of universal gravitation. \nIn the second part, a veritable tour-de-force, he calculates the equilibrium positions of sections of paraboloids. This was probably an idealization of the shapes of ships' hulls. Some of his sections float with the base under water and the summit above water, which is reminiscent of the way icebergs float, although Archimedes probably was not thinking of this application. "\n\n\nOK - I thought that this might be easy to research, however, your teachers have come up with a good one as it does not seem that you can use a direct search to find much.  So, now we try some other ideas:\n\n* my gut feel is that you should turn up the edges of the foil to create a pan type struture (think of your mothers deep pans with 3" to 4" side - roll the edges for strength and bold the corners in).  This would be a version of a flat boat.  You should practice to ensure that it works.\n\n* also, why not try to contact some actual boat engineers!  - try searching under "aluminum boats" and see if you can sent them emails to try to talk to an actual boat engineer.\n\n* I would also suggest trying to contact makers of aluminum foils to see if any of their engineers could give you any ideas.\n\n* Since this type of contest is run at other locations, you might also try tracking down some of the past winners to see what they learned and what their secrets were.\n\n* then try looking for foil boat !!!\nhttp://juniorengineering.usu.edu/workshops/buoyancy/boats.html\nhttp://www.new-sng.com/foilboat.cfm\nhttp://www.sciencenetlinks.com/Lessons.cfm?DocID=143\nhttp://pbskids.org/zoom/activities/sci/boatsafloat.html\nhttp://www.newton.dep.anl.gov/askasci/phy00/phy00291.htm\nhttp://www.nps.gov/archive/miss/programs/brj/brjactivities/aluboat.html\nhttp://education.jlab.org/beamsactivity/6thgrade/designandengineering/designandengineering.pdf\nhttp://www.sciencenetlinks.com/lessons.cfm?BenchmarkID=3&DocID=143\nhttp://www.lessonplanspage.com/ScienceWillTheBoatSinkOrFloat4.htm\nhttp://www.monroe2boces.org/programs.cfm?subpage=955\nhttp://www.nps.gov/archive/miss/programs/brj/brjpdfs/aluboat.pdf\nhttp://users.marshall.edu/~bady/rsi/4th/4bouyancy.html\nhttp://www97.intel.com/en/ProjectDesign/UnitPlanIndex/FloatThatBoat/\n\n\n* Ah Ha - and the secret is (as I suspected):\nhttp://www.nps.gov/miss/forteachers/brjralumboat.htm\n\nNote to teacher:\n\nGive students the opportunity to be as creative as possible with the hull shape. Make no suggestions, unless to show how to fold up one side to make an edge. The best designs are flat barges with small sides. The placement of pennies also makes a difference in the number held. The record for one sheet of 8" x 12" foil is around 280 pennies. Recycle the foil along with your pop cans.\nScience Museum of Minnesota, 2001.\n\n\nOK, now that you know how to win the contest, I would still recommend contacting the engineers at the boat companies and the aluminum foil companies and be ready to talk about what you learn as you win the contest.  You will impress the judges so much with your knolwedge and skill of floating all those pennies that they will have to give you the prize.\n\n\nGood Luck.
689832
The object must be less dense than the fluid to float, and more dense to sink.  If it is exactly equal density, then it can be suspended.
253323
the weight is the same because of water displacement the relative weight of the barge is the same as the water it displaces so it stays the same .mole uk
607846
Apply conservation of momentum. MV + mv = 0. Initial momentum is 0, call M the mass of the astronaut, V his velocity, m the mass of the wrench, v the velocity of the wrench, all after the throw. Now we know the wrench is thrown in the exact opposite direction, so we can say v is negative making the formula: MV - mv = 0. Now rearrange to get V = mv / M and put in the numbers to get V = (0.5*20) / 85 = ~0.118m/s. So he is moving at a constant velocity of 0.118m/s and needs to travel 30m. To find the time just divide the distance by the speed:\n\n30/0.118 = 254 seconds.
64850
A correction to the first answer. The formula terminal V = SQRT(2W/(Cd*R*A)) is correct, as it expresses the fact that W is balanced by the drag force V^2*Cd*R*A/2. However, for two same-sized smooth balls of different materials, only W changes; Cd stays the same. The air density R depends only on altitude.\nSince air density increases with decreasing altitude, the terminal speed decreases during a fall. If dropped from high enough, the actual fall speed, which initially = 0, increases until the (decreasing) terminal speed is reached, and then the sign of the net force (gravitational + drag) changes so the falling ball slows. For the remainder of the fall, the terminal and fall speeds both decrease, but the fall speed is slightly faster. This speed difference provides the net upward force to slow the ball.
613715
First find the volume of the object, i.e metal, by placing it in water and then u know the amount of water displaced.\nIf it displaces half the water than it's weight, then u know that it's specific gravity is twice that of water.
255989
A stationary body submerged in a liquid is effected by two forces, gravity and buoyancy. Gravity acts at the center of mass and it is always directed straight down. On a rigid body, the center of gravity can be considered stationary in reference to the body. Buoyancy, on the other hand, acts at the center of buoyancy, and the center of buoyancy moves as the body tilts and rolls. The size of the gravity force is;\n\nFg = Mass * Gravity\nThe size of the buoyancy force is proportional to the volume of liquid that it displaces, the force of gravity and the density of the liquid.\n\nFb = (Volume * LiquidDensity) * gravity\nWhen an object is stable in the liquid, the gravity force and the buoyancy force cancel each other out. This means that the boat is displacing an amount of liquid that weight as much as the boat does itself.\n\nFg = Fb