If a boat going 6 mph goes into a cross current of 4 mph, at what angle will the boat veer?
466510
33.7 degrees\n\nIt is calculated using vectors  6 straight ahead and 4 to one side.  The hypotenuse is sqrrt(4^2 + 6^2) = sqrrt(52).\n\nSo the angle the boat will veer is the angle that has a sin equal to 4/sqrt(52) = .5547 = 33.7 degrees
947129
a) Force on the box due to max acceleration witout moving the box = the maximum frictional force = 0.24* Normal force = 0.24*Weightof box = 0.24*mg\n\nSo max acceleration = 0.24*g = 2.352m/s^2\ns = .5at^2 = 10.584\n\nb)\nthis is a vector problem\nthe boat needs to be driven towards west inorder to cross exactly to a point at north.\nresultant speed of boat = 140/20 = 7m/s\nso using pythagorian theorem, U^2 = 7^2 +4^2, or U=sqrt(65) = 8.0623\n\nc)displacement = (11i - 5j) - (7i - 8j)  = (4i + 3j) \nforce = (3i + 4j)\nwork done = Fs = (3i + 4j)(4i + 3j)  = 12 + 12=24\n\nSo change in Kinectic energy  = work done  = 24J
536856
as you know that\na1v1=a2v2 \nso \npi r1 square v1= pi r2 square v2\nv2= 1600 ft/sec\n\ni hope it helps
695797
Assuming that 0.5% means 0.5% of the way straight down, or 0.5% of 90 degrees = 0.45 degrees, and that gravity is neglected due to its insignificance over that short a time, and that 90 mph = 132 fps,\n\nx = 132t cos 0.45\ny = 5 -132t sin 0.45\n\nwhen x = 56, t is approximately 0.425 sec and y is approx. 4.56 ft
1005012
Maybe it has something to do with grounding. If a ship is in the ocean and not touch anything then there is no reason for the current to go up right?
257654
Say the anchor weighs 15 pounds in air and displaces 2 pounds of water.\nBefore the drop, the boat displaces its weight plus the 15 lb weight in air of the anchor. This is the maximum displacement situation, so the water is at maximum height.\nAfter the drop, if the anchor hits bottom, the boat displaces its weight only and the anchor displaces 2 lb of water, the minimum displacement situation, so the water is at minimum height.\nAfter the drop, if the anchor doesn't reach the bottom, the boat displaces its weight plus the 13 lb weight in water of the anchor. The anchor displaces 2 lb water. The total displacement is the same as the maximum displacement situation, so the water is at maximum height.
980357
The answer has to be a speed!  It involves seeing that dy/dt = 1/2 x^(-1/2) dx/dt in this case.  Also, the distance from the origin is given by s^2 = x^2 + y^2; to get ds/dt (the required result), differentiate that wrt "t".  Put this all together yourself; it's the only way to learn rather than simply parrot.\n\nWithout a cast-iron guarantee, I get the value 27/(4 x sqrt 5) \ncm/s.\n\nLive long and prosper.
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.
284176
ok well his initial velocity is 0 and the distance is 402m and time is 4.913. you have to find the acceleration. well the only way i can think of doing this  without using these basic equations that are derived from newtons law (i forgot the derivation) is this.     you find the average velocity needed to go 402m in 4.913s. which is just 402m/4.913s = 81.823....\nthis is just the average velocity and because acceleration is constant, this means you can just find the average by doing  Vi + Vf / 2. you have Vi =0    so Vf/2 = 81.823........ so Vf = 163.6474.........m/s\nacceleration is  change in velocity /time. you have change in velocity  (Vf-Vi)   so divide it by time.   \n163.6474-0/4.913   =  33.3 m/s^2\n\n2. so the acceleration due to gravity is 9.8m/s^2. you know the distance is 10m and initial velocity is 0. so Vf = a x t\nso Vf = 9.8t. and Vf+Vi/2 = d/t    same things basically. so Vf/2 = 10/t  which means  Vf x t = 20\nVf = 9.8t\nVf x t = 20\nso then that means that (9.8t) x t = 20\n 9.8t^2 = 20\nt^2 = 2.04....\nt = 1.4285.....  so it takes her that long\nand now you just find Vf by putting this time in the equation  Vf = 9.8t = 14m/s. thats final velocity\n\nb. ok so this question confused me at first but it snot hard. once she hits the water (with speed of 14m/s), she travels for 2m before coming to a stop. so distance is 2m. Vf = 0. Vi = 14m/s and you have to find acceleration. \nso a = (Vf-Vi) x t = -14t\nso you need time now. you do average velocity  again\nVi + Vf / 2 = d/t\n14/2 = 2/t\n7 = 2/t\nt = 0.2857.... s\nthen put it in the other equation and you get a = -14t\nso a = -4m/s^2.\nforce = mass x acceleration\nmass is 61kg so you do 61kg x -4 = -244 N\nthe water exerted 244N against the movement (thts why its negative).\n\nso hope this helped. let me know if you need more help or whatever. laterz
962720
Lots of unnecessary complications here.\n\nAirspeed is the hypotenuse. 20.4 is the side opposite the angle we are looking for.\n\nThe heading, east of north, must be at angle sin^-1(20.4/241).\n\nSpeed relative to the ground is the third side of the triangle, and is \n\nsqrt(241^2 - 20.4^2)\n\nZidane has got the angles all wrong. The wind is from the east. We want to fly due north. That's where the right angle is. 241 is the hypotenuse of the triangle. The wind serves to make the ground speed less than 241, not greater.
652370
Okay, I've thought about this and it's not possible to determine the rate of acceleration from the given criteria.  It is an "average" of 60 miles per second, BUT how many miles in the first second compared to the number of miles in the final, fifth second?  If that is known, you can project MPH from seconds 6 to 10 or the additional 1/4 mile.  How do you calculate the rate of acceleration in your theory?  How fast is the vehicle traveling at each point of the 1/4 mile distance?
976821
Well, these are just opposite vectors, so you subtract the 50 m/s of the wind from the 100 m/s of the plane.\n\nActually it's : 100 + (-50) = 50 m/s at 90°\n\nE-mail the other other questions to me so I can send you a better response with graphics.\n\nI'll be happy to help :)
467716
When you sail, you cut across the wind at an angle. The more you turn into the wind or sheet in the sail, the faster you go. To get back where you started from you 'tack' into the wind at a sharp angle, then turn the other way in a zig-zag pattern. The sail tries to push the boat back but the hull and keel cut into the water so you actually move into the wind
371345
d=vt\n\nWhere d is distance, v is velocity, and t is time.\nV here is the speed of sound through water, which you'd have to look up, and t=2.9 seconds.  Remember when you solve that your "d" actually covers the distance from the ship to the ocean floor AND BACK, so you'll have to divide it in half to find the depth.
823160
first problem: let current=c and the speed of the kayak=k\n\nk+c=4.7\nk-c=3.4        add the two equations.\nso, 2k=8.1    divide by two.\nso, k=4.05    plug this into one of the equations.\n4.05+c=4.7   subtract by 4.05\nc=.65          \n \nboth answers in mph.\n\nuse this same logic for problem two.
303745
54 mi in 3 hours is a speed of 18mph (54/3=18),\n54 mi in 3.6 hours is a speed of 15mph (54/3.6=15),\nTHe current is half the difference between the two - it helps as much as it stops you....\nTherefore current is 1.5mph ((18-15)/2=1.5) and the boatspeed in calm water is 16.5mph (15+1.5 or 18-1.5).\n\nThe equations are, with boat rate=x and current=c;\nx+c=18      (1)\nx-c=15      (2)\n(1) ==> c=18-x\ninsert c into (2) ==> x-18+x=15\n==>2x=33 ==>x=33/2=16.5\nInsert x into (1) or (2) to get c=1.5   \n\nHope it helps,\nFredrik, MSc
987748
1.Compute initial velocity V0\nUsing equation S=V0*t-(g*t^2)/2 we find that the initial velocity V0=35/3 + 3g/2\n2.What ever goes up must come down V0*t=-(g*t^2)/2 t=2*V0/g\n3.Total time travel = 2t \nYou to the math\n\nBest\n\nHint:  the arrow will stop midair in about 5.3 seconds