When the ice cube is released, what distance will it travel up the slope before reversing direction?
814294
Since no energy is lost to friction or any other forces, this problem can be solved using energy methods. The energy in the compressed spring is simply:\n\nEs = 1/2 * k * x²\n\nAs the spring expands, it does work on the ice cube to speed it up. However, not all of the energy in the compressed spring is converted into kinetic energy of the ice cube. Note that gravity will pull the ice cube down the ramp according to the equation:\n\nF = m*g*sin(theta)\n\nThe energy imparted from the spring to the cube will be equal to the spring energy MINUS the work done against gravity. YOU HAVE TO TAKE INTO ACCOUNT THE NEGATIVE WORK DONE BY GRAVITY! This work is:\n\nW = F * x\n\nW = m*g*sin(theta) * x\n\nTherefore, the net energy imparted to the ice cube by the spring is simply:\n\nKE = Es - W\n\nKE = 1/2 * k * x² - m*g*sin(theta) * x\n\nThis is the kinetic energy of the cube after leaving the spring. Since no other forces are present, once the ice cube reaches its maximum height, its potential energy will be equal to its initial kinetic energy:\n\nPE = KE = 1/2 * k * x² - m*g*sin(theta) * x\n\nPotential energy is simply:\n\nPE = m*g*h\n\nm*g*h = 1/2 * k * x² - m*g*sin(theta) * x\n\nwhereh is the height of the block above the surface. Using simple trigonometry:\n\nh = L sin (theta)\n\nwhere L is the distance the ice cube travels up the slope after leaving the spring. Substituting this expression into the potential energy relation provides:\n\nm*g*L sin (theta) = 1/2 * k * x² - m*g*sin(theta) * x\n\nSolving for L:\n\nL = (1/2 * k * x² - m*g*sin(theta) * x) / (m*g*sin(theta))\n\nFinally, substituting all of the known values yields the answer:\n\nL = [(.5)(23 N/m)(.12 m)² - (.053 kg)(.12 m)(9.81 m/s²)sin(35°)] / [(.053 kg)(9.81 m/s²)sin(35°)]\n\nL = .4353 m\n\nThe ice cube will travel 43.53 cm up the ramp after leaving the spring. Hope that makes it clear! :)
1057502
There are actually an infinite number of answers to this old puzzle.  The traditional answer is of course, "You are at the North Pole."\n\nThis assumes that the walking "1 mile due east" portion is defined as "keeping the north pole directly left of you as you walk", or inversely, "keeping the south pole directly right of you as you walk".  Thus when you walk to the east, you simply make a 1-mile arc about the north pole.\n\nAnother set of infinite answers is "any point 1 + 1 / (2 * pi) miles away from the soth pole".  Following the directions, you will walk one mile toward the south pole, then walk one mile east which will take you *in a complete circle around the south pole*, then one mile back north to your starting position.\n\nFollowing this logic, you can find an infinite number of circles as a solution.  For example, "any point 1 + 1 / (4 * pi) miles away from the south pole" will have the easterly portion of the walk take you completely around the south pole *twice*.  And "any point 1 + 1 / (6 * pi)" will walk you around the south pole three times, etc.
970380
Let rock be thrown at an angle of θ at a velocity of Vo\n\nThen vertical component = Vo sinθ\nand horizontal component = Vo cosθ\n\nAlso acceleration in horizontal direction = 0 \nand acceleration in vertical direction = -g = -9.8 m/s²\n\nVertical direction\n\na = -9.8\nv = u + at\nSo Vt = Vosinθ - 9.8t\nand s = ut + ½at²\nSo h = Vosinθ*t - 4.9t²\n= 0 when it is at ground level\nie t(Vosinθ - 4.9t) = 0\nSo t = 0 or Vosinθ/4.9 ... ****\nSo it will take Vosinθ/4.9 s to cross the river.\n\nHorizontally\n\nVt = Vocosθ + 0 (as a = 0)\n= Vocosθ\n\ns = ut + ½at²\n= Vocosθ * t\n\nSo 75 = Vocosθ *Vosinθ/4.9\n= Vo².sinθcosθ / 4.9\n= Vo² * 2sinθcosθ / 2*4.9\n= Vo² * sin2θ/9.8 \nThus Vo² * sin2θ = 75*9.8 = 735\nThus Vo² = 735/sin2θ\nNow least value of Vo occurs when Vo² is minimum and this will occur when sin2θ is maximum (=1)\n\nSo Min possible Vo = √735 \n≈ 27.11 m/s (projected at 2θ = 90° ie θ = 45°)\n\nMinimum time = Vo*sin45°/4.9 (from ****)\n≈ 3.91 s
839427
Assuming you mean minimum force required to move the mass up the slope, the answer is 100N.\n\nI also assumed you meant distance parallel to the slope was 200, meaning the angle between the slope and the horizontal would have sin=20/200 = 0.1. I will call this angle "&".\n\nDraw a free body diagram with a force (parallel to the slope) acting on a block, and a reaction force (perpendicular to the slope) under the block. The weight of the block acts straight downwards with a force of 100*10= 1000N\n\nJust as the block is about to move upwards (i.e not going up or down), all horizontal forces are equal 0\n\n=> Fcos&=Rsin&\n\nAll vertical forces are also equal 0\n\n=> Fsin&+Rcos&=1000\n\nRearranging first equation:\nR=Fcos&/sin&\nsub R into second equation:\n=>Fsin&+Fcos^2&/sin&=1000\nMultiply all terms by sin&\nFsin^2& +Fcos^2&=1000sin&\nTake out F\nF(sin^2&+cos^2&)=1000sin&\nLaw: sin^2+cos^2=1\n=>F=1000sin&\n=>F=100N
550744
If I'm reading your description right, I think your first boat is heading up and to the right ("direction N.68 degrees E") making an angle of 68 degrees with north, or 22 degrees with east.  And I think your second boat is headed directly east ("S.90 degrees E").  So I'm assuming that they form a triangle, with one angle being 22 degrees and another one being a 90 degree angle, since we're looking for the point where one boat will be directly south of the other.\n\nThe first boat traveled 12t miles so far.  The second boat left an hour later and traveled 16(t-1) miles.\n\nThe first boat's path turns out to be the hypotenuse of the triangle, and the second boat's path will be one of the adjacent legs to the 22 degree angle.  So we can set up an equation with trig identities:\n\ncos 22 = 16(t-1)/12t\n0.9272 = 4(t-1)/3t\n2.7816t = 4t - 4\n4 = 1.2184t\nt = 3.3 hours\n\nSo after 3.3 hours, one boat will be directly south of the other.\n\nFinding that distance can be done one of several ways: using the Pythagorean theorem, sine relationships, law of cosines, etc.  I'll use sine, since we're already doing trig.  After 3.3 hours, the northernmost boat will have traveled 39.6 miles.  The distance between the boats becomes our side opposite the 22 degree angle.  So,\n\nsin 22 = distance between/39.6\ndistance = 39.6sin 22\n= 14.83 miles\n\nNow if I read the angle wrong, then simply go back and change all the "22" angles to the correct one and recalculate.
391989
Ever heard the old saying about throwing a penny to the ground from the top of the Empire State Building?\n\nThe mathematical equation below will explain it all...\n\nGravity + Large ball of ice = OUCH!\n\nEver wonder what Newton would have done if hit in the head with hail rather than an apple? LOL\n\n:)
808463
I assume that the total travel along the incline including the spring compression is L = 3m.  The potential energy lost in this slide is m*g*h, where h = vertical distance covered.  This is L*sinø, where ø is the 35º angle.  So the total energy gained by the block and then absorbed by the spring is m*g*L*sinø.   The energy absorbed by the spring is given by E = ∫F*ds, where F is the force from the spring.  If the spring constant is k, then F=ks, and E = k∫s*ds [0 to ∆s] = .5*k*∆s^2.   ∆s is the compression of the spring.  Equate this to the  energy gained by the block to get\n\nm*g*L*sinø = .5*k*∆s^2\n\n∆s^2 = 2*[m*g*L*sinø]/k\n\n∆s = √[2*(m*g*L*sinø)/k]
778096
The height of the ramp give potential enery in the following format\n\nE = mgh      where E is potential energy, m is the mass of the car in Kgs, g is the gravitational acceleration of the planet you're on (in meters per second per second (9.8 on earth)) and h is the height of the ramp.\n\nThe angle or slope of the ramp will not affect the potential energy, but will have an effect on the rate of change of potential energy.
728081
i'm going to assume that the x axis is marked time and the y axis is marked displacement (that's a posh vector way of saying distance)\n\nWhen the particle is above the x axis (y is +ve) the particle is forward from it's starting point\n\nConversely for moving backward it is below the x axis (y is -ve)\n\nIt speeds up when the line has +ve gradient (from bottom left to top right)\n\nconversly it slows down when gradient is -ve (top left to bottom right)\n\nyou get these last 2 by differentiation.\n\nnow double differentiate to get the acceleration\n\nVelocity is at a max when there is a peak (maximum where gradient = 0) \n\nVelocity is at a min where there is a trough (minimum where gradient is =0)\n\nThese max and min are like Vs and inverted Vs, so the +ve and - just to the left and right of these determins which is which.\n\nOn a graph, you can tell by looking.\n\nIt stands still, when the line is horizontal in a line rather than a max or min
942655
Assuming that I understand the geometry correctly, the combine package-rocket will move off the cliff with only a horizontal velocity component. The package will then fall under gravity traversing the height of the cliff (h) in a time T given by\n\nh = 0.5*g*T^2\n\nHowever, the speed of the package-rocket system must be sufficient to cross the river in that time\n\nv2 = L/T\n\nConservation of momentum says that\n\nm1*v1 = (m1 + m2)*v2\n\nwhere m1 is the mass of the rocket, v1 is the speed of the rocket, m2 is the mass of the package, and v2 is the speed of the package-rocket system.\n\nExpressing v2 in terms of v1\n\nv2 = m1*v1/(m1 + m2)\n\nand then expressing the time in terms of v1\n\nT = (m1 + m2)*L/(m1*v1)\n\nsubstituting T in the first expression\n\nh = 0.5*g*(m1 + m2)^2*L^2/(m1*v1)^2\n\nsolving for v1, the speed before impact is given by\n\nv1 = sqrt(0.5*g/h)*(m1 + m2)*L/m1
642567
rise over run tells you how far you move up/down and right/left.\nIf the slope is 4 = 4/1\nthen you can move up 4 and to the right by 1 OR you could move down 4 and to the left by 1.\nIf the slope is -5/8, then you would move up 5 and to the left by 8 OR down 5 and to the right by 8.
181237
It depends on the type of compass.  If it relies on satellite technology then there's no difference.  If it's a traditional needle type magnetic compass then the principle is still exactly the same in that the needle will point to the magnetic north pole it's just that the relative direction of travel will be different in the US to in Egypt.\n\nThe magnetic north pole is in Canada (just north of Hudson Bay) so from the US you'd effectively be walking upwards whereas setting off from Egypt you'd be walking on something of a diagonal course.  Bear in mind that these directions are relative to one another because in reality wherever you set off from you'd be heading north (assuming of course that you followed the direction of the needle).\n\nAlyssa's answer is wrong.  Imagine you're on the equator and set off walking east (sailing on the wet bits) and kept heading east for a very long time, you'd eventually come back to where you started - there wouldn't be a point where east suddenly became west or vice versa.\n\nOne thing that would differ is magnetic deviation (also known as magnetic declination) - this is the angle between magnetic north and true north.  There's about 1000 miles difference between the locations of the true north pole and the magnetic north pole so your position on earth would affect the angle of magnetic deviation.\n\nThe principle of using a compass will be exactly the same in Egypt as in the US.  The only thing that varies is magnetic deviation - which in itself varies because magnetic north isn't a fixed point.
1088871
50 because if you rearrange it in the format of y=mx+b with m being the slope it would equal to 50 in your case
796334
Consider the forces and torque in this problem. Setting up so that the x-axis is positioned parallel to the incline\n\nSFx = m*a = m*g*sin(q) - Fs\n\nwhere m is the mass of the cylinder, a is the acceleration, g is the acceleration due to gravity, q is the angle of the incline and Fs is the statis friction force. For the torque\n\nST = I*a/r = r*Fs\n\nso that we have for the statice friction force\n\nFs = I*a/r^2.\n\nPlugging this into the first equation\n\nm*a = m*g*sin(q) - I*a/r^2\n\nand solving for the acceleration\n\na = m*g*sin(q)/(m + I/r^2)\n\nNow, finding the time to travel a length L can be done in several ways. The first, and perhaps simplest is\n\nL = 0.5*a*t^2\n\nso that\n\nt = sqrt(2*L/a)\n\nAlso, you can consider how many revolutions of the cylinder is takes to achieve the same distance. This mean dividing the distance L by the circumference of the cylinder. Multiplying the result by 2*pi gives the angular travel so that the time can be determined by\n\n2*pi*L/(2*pi*r) = 0.5*alpha*t^2\n\nwhere alpha is the angular acceleration and is equal to a/r. \n\nConservation of energy can be employed as well\n\nm*g*h = 0.5*m*v^2 + I*w^2\n\nwhere h (= L*sin(q)) is the height, and w (= v/r) is the angular speed. That is to say that the potential energy at the start is equal to the kinetic energy after traveling through the distance L along the incline. Solving for v yields\n\nv = sqrt(2*m*g*L*sin(q)/(m + I/r^2))\n\nwhich is equal to the acceleration times the time traveled. From this \n\nt = sqrt(2*m*g*L*sin(q)/(m + I/r^2))/a
824568
We can use trigonometry to determine the vertical and horizontal components of the stone's velocity.  This will let us determine how far it goes.\n\nImagine a right triangle with a hypotenuse of 14 units and one angle of 26 degrees.  The side opposite the 26 degree angle will be as long as the stone's vertical velocity.  The side adjacent to the 26 degree angle will be as long as the stone's horizontal velocity.\n\nSince:\n\nsinA = opposite / hypotenuse... \nsin26 = vertical velocity / 14...\nvertical velocity = 6.14m/s\n\ncosA = adjacent / hypotenuse;\ncos26 = horizontal velocity / 14...\nhorizontal velocity = 12.56m/s\n\nIf the object had an initial vertical velocity of 6.14m/s, we know how long it stayed in the air because:\nacceleration = change in velocity / time\nWhich is the same as saying:\ntime = change in velocity / acceleration\nThis applies in any situation where the acceleration is constant.\nSo, if we want to know how long it took for the stone to reach 0m/s vertical velocity (the point at which it stops rising and starts falling), the change in velocity will be 6.14m/s.  If we just divide this number by acceleration due to Earth's gravity (9.81m/s^2), we get 0.63 seconds.  The stone traveled upward for 0.63 seconds.  This means that it must have also traveled downward for 0.63 seconds before hitting the ground, for a total flight time of 1.26 seconds.\n\nNow, we have a simple velocity equation.  Since we know that:\ndistance = velocity * time \nSince we're only worried about how far the stone traveled horizontally, we use the stone's horizontal velocity: 12.56m/s.  Time will be the total time that the stone was moving: 1.26sec.\n\nSo...\ndistance = 12.56 * 1.26 = 15.83 meters.\n\nHope that helped!
130805
(a)\nDy = 1/2A*t^2 + Vy*t, but Vy = 0, so t = sqrt(Dy/(1/2 A))\nDy = -1013 m, A = -g  (both negative since downward)\nt = sqrt(1013/4.9) = 14.38 s\n(b) Dx = Vx*t \nVx = 100 km/h = 27.78 m/s\nDx = 27.78*14.38 = 399.4 m\n\n6.  Dy = 1/2A*t^2 + Vy*t\nA = -g, Dy = -60 m, Vy is not specified.  If they jump up when they dive, they have more time to clear the rocks.  But we have to assume they just run straight off the edge, so Vy = 0, and t = sqrt(60 m/(1/2*g)) = 3.5 s.\nTo go > 25 m, Dx = Vx*t > 25 m, so Vx > 25 m/3.5 s = 7.14 m/s\n\n7. Dy = 1/2 A*t^2 + Vy*t, Vy = 0 (thrown horizontally),\nso t = sqrt(Dy/(1/2*g)) = sqrt(0.32/4.9) = 0.256 s\n\n   Dx = Vx*t = 12.3 m/s*0.256 s = 3.14 m\n\n8. Dy = 1/2 A*t^2 + Vy*t.\nBy now you should see that the motion in the y (vertical) direction depends only on the initial velocity in the y direction, the distance to fall in the y direction, and the acceleration in the y direction.  The speed in the x direction has nothing to do with it.
873659
First one's pretty easy.  Since you have a frictionless track (and we're assuming there is no air resistance here, either), the speed is due solely to the change in potential and kinetic energies.  At the start, the cube has a potential energy of mgh, or 0.530 kgX0.387mX9,8 m/s^2 = 2.01 J.  By the time it reaches the bottom of the loop, it has converted all of that potential energy into kinetic energy (since we're calling the bottom of the track to be zero altitude).  As it travels up the loop, it gains potential energy again as it fights against gravity, slowing it down.  At the top of the track, therefore, it has a potential energy of 0.30 J, meaning the kinetic energy must be the difference, or 1.71 J.  Solving for velocity (KE = 1/2 mv^2), you get 2.54 m/s.\n\nThe second case is a little more tricky, in that it involves rolling friction.  However, friction only plays a role in starting and stopping motion -- if the cylinder achieves pure rolling motion, the frictional component goes to zero because there will be no relative motion between the points of contact on the ramp and the cylinder (no sliding).  If we assume that there is never any slipping along its path, then it should behave just like the cube sliding along a frictionless track, and the speed will still be 2.54 m/s.  I could be wrong, but I think that's the answer.