i need some help find the answer to this for my physics class
206079
rockets were not invented yet, they used canons
355495
The potential energy at the top is mgR = 2 x10 x R = 20 R.\n\nKinetic energy at the bottom of the curve = 0.5 m U ^2 = 0.5 x 2 x U^2. = U^2.\n\nAs there is no friction  K.E = P.E\n\nU^2 = 20 R.\n\nOn the horizontal path, \n\nInitial speed is U.\n\nFinal speed is V = 0.\n\nDistance traveled = 8 m.\n\nAcceleration = Fricttional Force / mass = -8/ 2 = - 4 m/s^2. ( Negative sign since the force acts opposite to the motion.\n\nUsing V^2 - U^2 = 2as, and also noting that V=0,\n\nU^2 = 2 x 4 x 8 = 64.\n\nU = 8m/s.\n\nThe time of travel is found using V= U+ at.\n\nt = -8/ (-4) = 2 second.\n\nUsing U^2 = 20 R.\n\nR = 64/ 20 = 3.2 m.
122992
k ill try to help u  even though i dont condone cheating. i just know how it is to not understand something and its due in a few hours and it sucks. just remember, nobody will be able to help u on the exam.\n\nso 1. \na. acceleration is change in velocity divided by time. \nsince it started from rest, initial velocity is 0 and then final velocity is given as 25.3 m/s. and the time is given as 0.59s. so to get acceleration   you would do  \na =  Vf-Vi / t\n  =  25.3 - 0 / 0.59\n  = 42.88......\n\nb. the force is equal to mass x acceleration. since you know the mass and acceleration, you just multiply them and you get F = 38421.69... N. \n\nc. you have to find the normal force in this case, which is the force that opposes gravity. like upward force. so in this case, gravity is acting on the driver and force of gravity is equal to mass x 9.8 = 666.4 N. and that means the normal force is 666.4N (since he is not moving up or down the total vertical force = 0)\n\n2. ok well in this case, you know the initial velocity is 0, the distance is 404m and the time is 4.944s. if its constant acceleration, you could assume that the average velocity is just Vi+Vf/2. so if thatz average velocity, then distance = average velocity x time\n     404 = 4.944s  x  (0+Vf)/2\n 404 x 2 / 4.944 = Vf\nVf = 163.43........m/s\n\nthe acceleration would just be Vf-Vi/t. and since you know Vf now  you just do   a = 163.43-0/4.944\na = 33.056....m/s^2\n\n3. ok well acceleration is 9.8m/s^2 (gravity)     distance is 9.7m. initial velocity is 0m/s. so this time, you have to solve 2 equations. the first one is the acceleration equation. a = Vf-Vi / t = Vf / t = 9.8\n9.8 = Vf/t\nVf = 9.8t\nand the other one is the distance = t  x  Vi+Vf/2   average velocity one. \n9.7 = t x Vf / 2\n19.4 = Vf x t\n19.4 = (9.8t) x t\n19.4 = 9.8t^2\nt^2 = 1.979...\nt = 1.406....s\n\nnow that you know the time, you find the Vf by putting it in any of those 2 equations. Vf = 9.8t = 9.8 (1.406)\nVf = 13.788... m/s\n\npart b. so in this case, you use the initial velocity as the one u just found (the velocity of the swimmer at the surface of the water)   so Vi = 13.788m/s\nand Vf = 0    since they come to a stop. and d = 2m\nso you have to find the acceleration. first you need to find the time by using the average velocity equation.\nd = t x Vi+Vf/2\n2m = t x (0+13.788)/2\n4 = 13.788t\nt = 0.29s\nnow that you know the time, you have to find acceleration\na = Vf-Vi/t\n  = 0-13.788 / 0.29\na = -47.5286m/s^2    its negative because its downwards and up is positive in our case. so the force = mass x acceleration and you just got acceleration and her mass is 65kg so you just multiply by 65kg to get F = 3089.36N\n\n\n4. well F = ma and you have the force. when they say it weighs this much newtons, then it means  use gravitational acceleration which is 9.8. so 2350N = m x 9.8m/s^2\nm = 2350/9.8\nm = 239.79kg\n\ni hope you actually understand this stuff. try to do your homework by yourself   dont take advantage of this site. youll just end up screwing yourself over. and if u dont get anythin, let me know and ill try to help
504825
Where is your freebody diagram? There are other forces acting on the block than just gravity...\n\nGood Luck
511220
1) The total rocket, with mass 3M travels 260m over the first "half" of the parabola. On exploding, the fragment with mass M falls straight down - i.e., all the momentum in the horizontal direction remains with the larger fragment. Therefore its speed in the horizontal direction will increase:\n\nP = constant\nBefore explosion, P = (3M)(u)\nAfter explosion, P = (2M)(v)\n\nTherefore the speed in the horizontal direction will increase to 3/2 times the original speed.\n\nThe vertical momentum of the larger fragment will not increase, as the smaller fragment falls as from rest i.e. no original momentum. Therefore the time taken for the larger fragment (indeed, either fragment) to reach the ground from the apex of the motion will be the same as it was to reach the apex from the starting point. Since the horizontal distance was 260m, the extra distance covered will be that multiplied by 3/2, since the speed increased by that much, so the total distance from the launch point is 260 + 3/2(260) = 650m.\n\nYou do it by considering the horizontal and vertical motions independently.\n\n2) If the bobs collapse elastically, then no energy is lost in the collision. The energy of the larger bob on contact is entirely its kinetic energy - this comes entirely from the potential energy it had at the top of its swing. If the angle is 15deg and the length of the cord is 50cm, then the bob is drawn back until it is at a height of 1.7cm. The formula for potential energy is U = mgh, so the potential energy of the bob at that point is \n\nU1 = (0.39kg)(9.8m/s2)(0.017m) = 0.06497 joules\n\nconverting all units to SI.\nThis is the kinetic energy of the bob as it strikes the second one - all of this energy is transferred to the second bob, as kinetic - the second bob rises until it stops, when all the kinetic energy is converted to potential - and so use the same formula to get the height it reached.\n\nU2 = 0.06497 J = (0.14kg)(9.8m/s2)(H)\n\nwhere H is the maximum height of the lighter bob. It works out as\n\nH = 0.047 m\n\nor 4.7 cm. Using simple trigonometry, this corresponds to an angle of 25.1 deg.\n\nHope this helps
386616
See my answer to this question in "General-Science and Mathematics."  :)
300228
V = (v + u) / (1 + uv/c²)\n\nTry that. The velocities are quite high so relativistic. Haven't tried it yet though.\n\nNevermind, the velocites aren't nearly high enough to be relativistic. If it was km/s it would work.
636658
do your own homework\n\nits probably 0.26 kg M/s ^2 since its a velocity\n\nyou also probably need to convert it to rad/s whatever that is
59521
yes try posting question under \n\nHome > Science & Mathematics > Physics \n\nYou will get an answer there.
515446
it is a much easier problem than it seems..\nthe key point is the fact that the body is moving at constant speed. meaning that the net force on it is zero --> no acceleration\n\nthere are two forces. friction and the horizontal component of the force that is exerted at 40 degrees.\n\nso simply ...\n\n\nFx = Friction==>\n976 cos40 = coeff. *N *g\n\n\n now there is equilibrium at the vertical axis as well.\n\nmeaning that \n\nN+Fy = m*g =>\n\nN = (mg- fy)\n\nso we mix them together and hve\n\n976cos40 = 0.667 (mg- 976sin40) =>\nm=50.31kg\n\nin this problem you use that there is equilibrium in both of the x and y axis.
491970
Someone posted the exact same set of questions a few days back, either here or on the physics page - you ought to be able to find the methods in the responses. Have a look.\n\n\nBest of Luck - Mike
531381
Hi Yo-Yo!\nEnergy conservation law: m*g*h=E1+E2, m=0.12kg, h=1.2m, g=9.8m/s^2, E1 is translation kinetic energy, E2 is rotational kinetic energy.\nE1=m*v^2/2, v is translation speed at full length h, E2=I*w^2/2, w=angular speed at full length h.\nNow v=w*r, r=3.2*10^(-3) m, thus m*g*h=m*(w*r)^2/2+I*w^2/2, hence w=sqrt(2*m*g*h/(m*r^2+I)). Plug numbers in and find w & v (be a good boy!).\nMeanwhile v=a*t & h=a*t^2/2, ‘a’ is linear acceleration.\nThus t=v/a & h=a*(v/a)^2/2, so a=v^2/(2*h). Plug numbers in and find a & t. Angular acceleration being w’=a/r. My regards!
474473
We place the weight of the rod at the center of its length or 0.90m from one end. Its Potential Energy, PE at the horizontal position with respect to the particle, is 2.4g*1.40.  This energy is converted to a lesser PE at theta=35 degrees.  PE of the rod at this point is          2.4g*(1.40-0.90cos35). The particle had initial PE equal to zero, but at 35 degrees from its original position, it gained a PE equal to mg(1.40-1.40cos35). The PE gained by the particle is equal to the dissipated energy  during the collision. \n\nBased on the principle of conservation of energy:\n\nPE of system, initial=PE of system, final\n\na)Substitute the above values:\n\n2.4g*1.4=2.4g(1.40-0.90cos35)\n+mg(1.40-1.40cos35)\n\ng cancels out.\n\n2.4*1.4=2.4*0.67+0.26m\nm=(3.36-1.60)/0.26\n   =6.76kg\n\nb)Dissipated energy=6.76g*0.26 \n                           =17.2 Joules\n\nPls draw a diagram showing the position of the rod and the particle, before and after collision. You'll then find my explanation easier to understand.
292295
Consider two objects having masses 1kg and 5 kg are in a horizontal plane.\n\nWe apply a force of 10 newton to the 1 kg mass and 50 newton to the 5kg mass. \n\nThat is, we apply less force for less mass and greater force for greater mass.\n\nNow we find the acceleration of the two bodies.\n\nThe1 kg mass will have an acceleration of 10/1 = 10 m/ s^2. That is its speed will be going on increasing by 10meter / second in every second. \n\nSimilarly the 5 kg mass will have an acceleration of 50/5 = 10 m/s^2. That is its speed will be going on increasing by 10meter / second in every second. \n\nBoth have the same acceleration and both have the same speed at any time after start.\n\nNow returning to gravitational pull, let us place the 1kg mass in one pan of the balance and the 5kg mass in another pan. \n\nThe pan having 5kg mass will come down and the pan having 1kg mass will go up.\n\nThis is because the earth is pulling the 5 kg mass with a force of 50 newton and 1kg mass with a force of 10 newton.\n\nIf  we drop the two objects from the same place, as we have seen already the 1 kg mass will fall down with an acceleration of 10 m/ s^2 and the 5kg mass will also fall down with 10 m/ s^2. But the forces acting on them are different.\n\nThe Earth’s pull is such that it gives less force for less mass and greater force for greater mass.\n\nTo be more specific it gives a force of 10 newton for every kilogram of mass and hence the acceleration is for any mass is the same as approximately 10m/s^2.
91871
see if this helps.....\n\n\n\nAnalysis of a simple gravity pendulum\n \nFigure 2. Force diagram of a simple gravity pendulum.To begin, we shall make three assumptions about the simple pendulum\n\nThe rod/string/cable on which the bob is swinging is massless and always remains taut; \nThe bob is a point mass; \nMotion occurs in a 2 dimensional plane, i.e. pendulum does not swing in and out of the page. \nConsider Figure 2. The blue arrow is the gravitational force acting on the bob, violet arrows are that same force resolved into components parallel and perpendicular to the bob's instantaneous motion, the motion along the red axis, which is always perpendicular to the cable/rod. Newton's second law\n\n \nwhere F is the force acting on mass m, causing it to accelerate at a meters per second2. Because the bob is constrained to move on the green circular arc, there is no need to consider any force other than the one responsible for instantaneous acceleration parallel to the known path, the short violet arrow in our case\n\n \n \nLinear acceleration a along the red axis can be related to the change in angle θ by the arc length formula\n\n \n \n \nThis, however, is not the acceleration we seek because the gravitational force on the bob causes a decrease in angle θ. One accounts for that by placing a negative sign in front of a, thus:\n\n \n \n \nFigure 3. Trigonometry of a simple gravity pendulum.This is the differential equation which, when solved for θ(t), will yield the motion of the pendulum. It can also be obtained via the conservation of mechanical energy principle: any given object which fell a vertical distance h starting from rest, would have acquired kinetic energy equal to that which it lost to the fall. In other words, gravitational potential energy is converted into kinetic energy. Change in potential energy is given by\n\n \nchange in kinetic energy (body started from rest) is given by\n\n \nSince no energy is lost, those two must be equal\n\n \n \nUsing the arc length formula above, this equation can be re-written in favor of \n\n \nbut what is h? It is the vertical distance the pendulum fell. Consider Figure 3. If the pendulum starts its swing from some initial angle θ0, then y0, the vertical distance from the screw, is given by\n\n \nsimilarly, for y1, we have\n\n \nthen h is the difference of the two\n\n \nsubstituting this into the equation for  gives\n\n \nThis equation is known as the first integral of motion, it gives the velocity in terms of the location and includes an integration constant related to the initial displacement (θ0). We can differentiate, by applying the chain rule, with respect to time to get the acceleration\n\n \n \n \nsame as obtained through force analysis.\n\n[edit]\nSmall angle approximation\nThe problem with the equations developed in the previous section is that they are unintegrable. To shed some light on the behavior of the pendulum we shall make another approximation. Namely, we restrict the motion of the pendulum to a relatively small amplitude, that is, relatively small θ. How small? Small enough that the following approximation is true within some desirable tolerance\n\n \nif and only if\n\n \nSubstituting this approximation into (1) yields\n\n \nSolution to this equation is a well-known, and quite expected, oscillatory function\n\n \nwhere θ0 is the semi-amplitude of the oscillation, that is the maximum angle between the rod of the pendulum and the vertical.\n\nThe term  is a pulsation, which is equal to ,\n\nwhere T0 is the period of a complete oscillation (outward and return).\n\nSince ,\n\nthe period of a complete oscillation can be easily found, and we have obtained Huygens's law:\n\n \n \n[edit]\nArbitrary-amplitude period\nFor amplitudes beyond the small angle approximation, one can compute the exact period by inverting equation (2)\n\n \nand integrating over one complete cycle,\n\n \nor twice the half-cycle\n\n \nor 4 times the quarter-cycle\n\n \nwhich
37063
A quick guess without having reviewed your math, try adding g = 9.8 m/sec^2 to your answer and see if that is the answer that is needed.
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Hi!  You are right.\nIf they have equal masses and are elastic then  b) it is going to happen. Why? well Vf= 2*Vcm-Vi. Where all are vectors and :\nVf= final speed;   Vcm=center of mass speed ;  Vi=initial speed.\nThus 0=2*Vcm-Vi => Vcm=V/2=(m1V1+m2V2)/(m1+m2)=\n=m1*Vi/(m1+m2)  => m1=m2.    \n  " a)" isn't possible because the total momentum must be preserved.\n          Hope it helps!