pHySiCs pLeaSe...heLp anYoNe =(?
930430
The center of gravity of the cone lies at  one-third of the height from the bottom of the cone.\n\nThe angular elevation of this point from the circular base is given by tan A = h/(3r) where  h is the height of the cone and r is radius of the circular base.\n\nFor the cone to topple the vertical line from the center of gravity must fall outside the circular base.\n\nThat is after tilting the cone the vertical line from the center of gravity must be just more than 90 degree from the ground.\n\nTherefore the angle through which the cone has to be tilted\n is (90 - A.) \n\nA can be found from tan A = h/(3r).
695472
Okay,\n\nQuick method:\n\nShe picks up a certain K.E. (kinetic energy) before she hits the water and loses this K.E. 3m under. \n\nWe have KE = Force x distance = mad\n\nso we have m x a1 x d1 = m x a2 x d2\n\nm =65kg, a1= 9.8m/sec, d1=13m, d2= 3m\n\nF(ave)= m x a2= m x a1 x d1/d2 = 2760 1/3 N \n\nThe force exerted on her by the water from when she hits until she stops 3m under will not be constant but the average force stops her in a distance of 3m and has the above value
735831
Did you put your entire homework assignment on here? I noticed you posted a bunch of science questions just now.\nIf you don't have your textbook handy [you really should be looking up these answers so that you can remember and understand them when you have your test/exam], go to the links below and look them up. The 1st is a down loadable physics textbook. The 2nd has animations of physics.
1119730
ahh, good theory problem...\nlet's see, there's energy, friction, force (normal, tension, as well as the other ones mentioned), gravitation, mass, pendulum motion, gotta be more that I've missed, but no doubt someone'll be here to correct me.
443144
you should try this site http://www.oberlin.edu/physics/dstyer/SolvingProblems.html
732533
D = M/V\n\nDensity= Mass/Volume
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
35590
I concur with previous answerer. The acceleration is 1.5 m/s^2. Rearrange Newton's Second Law, to solve for the acceleration. Then, substitute what you know.
823034
You can find a good summary here:\n\nhttp://www.sparknotes.com/testprep/books/sat2/physics/chapter12section2.rhtml
899593
Hello,\n\nIn this link you will have what you need:\n\nhttp://www.bsu.edu/web/fseec/pie/Lessons2/Physics/Potato%20Battery.doc\n\nI believe now will be OK.
858861
1) The gravitational potential energy of water at the top of a waterfall is first converted to kinetic energy (falling) and then converted to both acoustic (sound) and thermal (temperature) energy when it hits the pool of water at the bottom. The temperature of the water in the pool will be slightly higher than the temperature of the water at the top of the waterfall.\n\n2) The thing to remember for this problem is that the energy in a system before some action = the energy in that system after some action:\n\nP0 + K0 + Th0 = P1 + K1 + Th1\nWhere P is gravitational potential energy (mgh), K is kinetic energy (0.5mv^2), and Th is thermal energy (leftover).\n\n22*9.81*4.5 + 0 + 0 = 0 + 0.5*22*2.5^2 + Th1\nSolve for Th1:\nTh1 = 22*9.81*4.5 - 0.5*22*2.5^2 =  902.44 Joules of thermal energy were generated due to friction.\n\n3) Split this problem into two cases, the first 10m and the second 10m. Find the velocity of the crate after the first 10m:\na = F/m = 200/70 = 2.86 m/s^2\nv = sqrt(2*a*d) = sqrt(2*2.86*10) = 7.56 m/s\n\nFor the second 10m, the force of friction on the crate is:\nFf = 0.3*70*9.81 = 206 N\n\nThe net force on the crate is 6N in the direction opposite of movement. The acceleration for the second 10m is:\na = F/m = 6/70 = 0.086 m/s^2\nvf = sqrt(vi^2 + 2*a*d) = sqrt(7.56^2 + 2*-0.086*10) = 7.45 m/s\n\n5) Since watts (W) are measured in J/s and joules (J) are measured in N*m, a watt is the same as N*m/s. Convert lb to N and ft to m to go from horsepower to watts:\n\n550 ft*lb/s * 0.3048 m/ft * 4.45 N/lb = 746 N*m/x = 746 W\n\nPS - You have in your problem description that horsepower is equal to 764 W but that should be 746 W.\n\nI hope this helps!
829880
the equation is:\nF=G*m1*m2/(r*r);\nso, F=6.67*10E-11*0.14*0.17/(0.65*0.65);\n\nthis is the force that either will exert on the other
239996
Work done = force x distance moved by object i thinkin a straight line ,so ithink some conversion maybe needed.
923403
Start off by drawing the picture. You have a right triangle, with a hypotenuse of 25, a base (b), and height (h). \n\nDon't get jumbled down by numbers yet, ignore them. Visualize what you are calculating.\n\nVelocity / speed is the derivitive of position. (v = delta x / delta t). So all you need to do is find an eqation relating the geometry using known and unknown quantities and then differentiate with respect to time. Sounds confusing but all you have to do is....\n\nYou know, a^2+b^2 = c^2 for a right triangle. In our problem,\n\nb^2+h^2 = 25^2\n\nThis is our position equation. Now differentiate with respect to time.  DO NOT forget the CHAIN RULE! <------ VERY important\n\n2*b*db/dt +2*h*dh/dt = 0.\n\nFrom here just use algebra; where,\n\nb=7 ft, and db/dt = 2 ft/sc.\n\nWhich leaves two equations and two unknowns.\n\nThe important thing to remember is to take this step by step and try to visualize what you are calculating. Also, looking at the units of the given quatitities can give you some idea of what you will be required to do. \n\nAlso, just remeber that speed / velocity is the derivative of position with respect to time (the units reflect this).
762251
If 2 billiard balls of mass m (ma and mb) are perfectly elastic and ball A approaches from the left with velocity 2 m/s and hits ball B which has velocity 1 m/s, and is also going to the right, what's the status after the collision. \n\nA is going to the right with velocity 1 m/s and B is going to the right with velocity 2 m/s.\n\nma*2 m/s + mb*1 m/s = ma*1m/s + mb*2m/s\nmomentum is conserved
295799
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
843122
good start.\n\ntotal inital momentum = 480-397.5\n\nthis is an inelasitc collision, and aferwards you have \nboth masses travelling at velocity v:\n\n(75+120)v = 480-397.5\n\nsolve for v.  This is the velocity at which they both slide.\nThe inital energy now is 1/2(75+120)v^2\n\nThis energy is converted to frictional energy:\n\ninital kinetic energy = total work of friction = Fr(distance)\n\nFr = uN = u(120+75)g = .3(195)(9.8)\n\nso we have\n\n.3(195)(9.8)(distance slided) = 1/2(195)v^2 \n\nYou know v form above.  So find distance slided.