I am trying to figure out the draft of a boat. How deep will a boat sit in the water weighing 750 pounds?
134278
Let B equal the weight of the boat, say in pounds.  Then B pounds of water will be displaced.\n\nSince there are 62.43 pounds of water in a cubic foot, then a boat of weight B pounds will displace B/62.43 cubic feet of water.\n\nIn your case, with a boat weighing 750 pounds, the boat will displace 12.01 cubic feet of water.  To determine how deep the boat is in water you need to know the geometry of the boat.
535379
You must show not tell. Take a bowl filled to the top with water and put it on a plate to catch overflow. The put objects in the water and compare the overflow for sinking and floating objects of the same size. If you have a scale weigh the objects and the water if not just hold them and judge weight. Then explain density.
263362
I don't really know.\n\nI came here because I'm curious about what you're basing your expectations about their height on.  Do you have the dimensions of the ships?  \n\nI hope you'll post an additional piece of information about this.
624229
1 foot = 12 inches \n55 divided by 12 is roughly 4 and a half feet. Replace the remainder with inches so the correct answer is 4 feet and 7 inches
341997
The boats path as viewed from an observer on the bank will be a the hypotenuse of a right triange.  It moves along that hypotenuse at a speed of 4.5 m/s and has traveled 27 m on that path.  therefore the time it took to get across was 27/4.5 seconds.  In that time, it travelled downstream 11m.  The river current is what was carrying it downstream and moved the boat 11m in 27/4.5 second.  Velocity is distance/time so the river current is 11 / (27/4.5) m/sec
164983
I think that 13k is over priced but that's me I like to build things. But for a bought boat sounds about right. If you feel your paying to much shop around see if you can find cheaper. I think a 50hp is plenty big enough for a pontoon boat. You could probably water ski with that actually I know you can because my uncle has a 50 evrude had it for years hauls butt. but the 30hp is too small unless your just staying in about a 5 mile area then that would be fine. Good luck and good boating.
21955
well - the displacement of a nimitz-class aircraft carrier is 100,000 tons.  that means when the carrier is sitting in water, the underwater part of the aircraft carrier takes up a volume that, if filled by water, would weigh 100,000 tons, or 200 million pounds.  the density of water is 62 lbs per cubic foot, so that means the carrier displaces around 3.2 million cubic feet of water, or 24 million gallons.\n\nhowever, if you had a container that was the same shape as the hull of the carrier, but just a tiny bit larger, you would only need to fill the container with a very tiny bit of water.  then, when you move the aircraft carrier into the container, the water will spread out and form a thin layer between the hull and the container.  the ship is still floating on this thin layer, even if it's only like a millimeter thick!  i don't really know how to calculate the volume of this thin layer without knowing the actual surface area of the submerged part of the ship.\n\nhowever you could model the underside of the carrier as a cylinder.  the length of the carrier is 317 meters and the beam is 41 meters, so the surface area of a half-cylinder with these dimensions is the length of the ship times half the circumference of the cylinder, so A = 317 * pi * 41 m = 40,000 square meters.  so the volume of water covering this area in a 1-mm thick layer is 40 cubic meters, or about 10,500 gallons.  still quite a bit of water.\n\nof course, it is pretty impractical to suggest that you could build a container that is exactly one millimeter larger than the outside of a ship.  if you had any places that stuck out just a tiny bit, the ship would no longer be floating and just be resting on the container.  so a more reasonable estimate might be a container that is one meter larger all around, which means you'd need about 10 million gallons of water to fill the container.
251348
After 2.5 hrs, Boat A has gone 50 km, and Boat B 80 km. Let the distances of A and B be two sides of a triangle, a 70 degree angle connecting them.\n\nThus we have a triangle with two known sides and one known angle.\n  \nLabel it A, B and C starting from the bottom left angle. Label each angle's opposite side its respective lowercase (a,b, and c). Fill in for b and c, and angle A. We don't know if this is a right triangle or not, so we're going to need to use the law of cosines:\n\na^2 = b^2 + c^2 - 2*b*c*cos(a)\nSubstitute for each known variable:\na^2 = (50)^2 + (80)^2 - 2(50)(80)cos(70 degrees)\na^2 = 2500 + 6400 - 8000 * cos (70 degrees)\na^2 = 8900 - 8000 * .34202\na^2 = 6163.84\na = 78.5101\n\nThus, using the law of cosines, the distance (a) from the two boats is approximately 78.5101 km. It is not exact as there is no exact value for cos (70 degrees).
353792
tan7.5° = 3.25/L\nThus L = 3.5/tan7.5°\n≈ 26.59 ft
283876
2.13m*1.52m*.38m = 1.23m^3 of water.\n\nAt 4°C pure water has a density of about 1000 kg/m^3\n\n1000 kg/m^3* 1.23m^3 = 1230kg (mass of water)\n\n2.13m*1.52m= area of bed = 3.2376m^2\n\n(1230kg + 91kg)*9.81m/s^2= 12959.01N = weight of frame and water\n\nweight of bed / area of bed = 12959.01N/3.2376m^2\n\n= 4002.66 N/m^2 = (0.581 pounds per square inch)\n\nThis assumes that the frame has a contact area with the floor the same size as the bed dimensions.
623418
As long as your boat is under 10ft. wide, you don't need a permit,, I had a Bertram 25 hardtop the beam on it was 10ft.2in. I towed it all over in Texas, without a problem !!!! and did not have a permit !!!
104166
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
415762
Assuming:\n1. that the ships area does not tapper as it goes down (which of course they do in real life)\n2. that the density of water is 1x10^3 kg/m^3\n\nwe know that the new arrangement is a ballanced force arrangement.\ntherefore the change in weight must equal the change in upthrust.\n  delta W = delta upthrust\n\nupthrust before:\nF upb = densityxgxVolume b\nFupb = densityxgxAreaxh    (h=depth of ship)\n\nupthrust end:\nFupe = densityxgxVolume e\nFupe = densityxgxAreax(h-10.5) \n\nchange in upthrust:\ndelta Fup = denistyxg(Ah -A(h-10.5))\ndelta Fup = densityxgx(10.5A)\n\nchange in weight:\ndelta W = mg\n\nsubs in:\nmg = densityxgx(10.5A)\nm   = densityx(10.5A)\nm/density = 10.5A\nA = m/(10.5xdensity)\n\nm = 1507480kg,  density = 1000kg/m^3\n\nA = 1507480/(10500)\nA = 143.57 m^2\n\nsmall in comparisson to flight deck, however as i said this is assuming a uniform area of hull ie floating cuboid, aship would be much more like a triangular prism with one point sunk into the water. i'll be dammed if i'm going through the calculas to work out that one!!!!!
633811
The unfortunate circumstance that caused the loss of your tires may well have made your canoe rise up in the front and if you are unable to push the bananas through the screen door you'll have to eat them and become fat.  You might take in water and if you are just fat enough you will even out and be able to continue down river.  So, I would say that you can push about 20 bananas through the screen door so you still have enough to eat or just leave in the canoe.  See 1lbs of bananas is 5  or so.  Two tires would weigh around 100 lbs.  If you are in the back of the canoe the bananas in the middle and the tires of course up front and the screen door on one side or the other then it would just work out that  you have 120 lbs of banas and only need 100lbs to equalize the weight once the tires were gone.  Of course I wouldn't eat them just to get fat and take on water. I'd push them to the front of the canoe and push the rest through the screen door.  The peels will make it very dificult but if I had time to go down river with tires, bananas and a screen door I'd have time to push 20 bananas throgh the screen.
637819
You must already understand the displacement description so here's an attempt to physically understand the process.\n\nGravity pulls everything with mass  towards the center of the Earth.  This includes water and even air.  This is what creates air pressure and water pressure within a column of the fluid.  The force towards the center of the Earth is called the Weight.  Weight is simply the mass of an object multiplied by the acceleration of gravity (32.2 feet per second).\n\nMolecules in fluids can move around very easily.  They are constantly being pulled downward by gravity.  Generally, they are blocked by other fluid molecules already below them but they pile up and apply a pressure to the poor suckers below them.  The molecules at the bottom of the stack are under tremendous pressure in very deep portions of the ocean because they have to support the weight of the molecules above them.  (This also creates a sea level air pressure of 14.7 pounds per square inch)\n\nShips weigh less than the water they displace.  Gravity is pulling that ship down, trying to get it to the center of the Earth.  Once the weight of the water equals the weight of the ship, the water pushes back with enough force to keep it afloat.  Water molecules can flow easily.  When they begin to weigh more than the effective weight of the ship, they will flow beneath it and push the ship upward.
134718
water will get deeper
552534
Fluid Pressure= weight density x (times) depth\n\nFl. p.= 10,000 kg/m^3(thats the weight density of water) x 15 m\n\nFl. p.= 150,000 m^2 (because one m cancels)\nHope this helps!