Two physics question?
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let me guess. This is some kind of homework and you didn't read what you were supposed to........in time.\n     Just a thought.\nAnd if I'm right.....then shame, shame.
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a) 72.072 cm/sec^2\n\nb) 82.63 cm/sec^2
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KE = 1/2 mv^2\nfind the KE before the collision using the formula.\nfind the KE after the collision using the formula.\nsubtract the two values to find the change in kinetic energy
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(a) Gravity exerts a downward force on the block of 9.8 Newtons/kg * (0.25 kg) = 2.45 Newtons. Work done by gravity = force * distance = 2.45 Newtons * 0.12 m = 0.294 Joules\n\n(b) The spring is a bit tougher since the upward force increases as the spring is compressed. Force = 250 N * (C) where C is the distance the spring has been compressed (in meters). Work is then the integral of Force over distance, so Int(250 * C (dC)) from C = 0 to 0.12 M . The integral is 125 * C^2. Evaluation at 0 gives 0. Evaluation at 0.12 gives 125 (0.12)^2 =  1.8 Joules. So the work done by the spring is -1.8 Joules . Note that the work is negative, because the object is moving against the applied force.\n\n(c) When the block strikes the spring, it will have a kinetic energy of (1/2)mv^2 = (1/2)(0.25kg)v^2 = 0.125v^2.\nThis kinetic energy is absorbed by the spring as it compresses. Note that the work done by gravity is also absorbed by the spring.\nTherefore, we have the following equation:\n0.125v^2 + 0.294 = 1.8\nv^2 = 8(1.8-0.294) = 8(1.506) = 12.048\nSpeed of block just before it hits = SQRT(12.048) = 3.47 meters/sec downward.\n\n(d) I'll leave this as an exercise for you. Just apply the same techniques as above. Note that the block hits the spring with a velocity of 2(3.47) = 6.94 meters per second...
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Try these answers:\n\nFor the first one, shortest is 0.0086 m, longest is 8.6 m.\nFor the second one, it should be 2.95 Hz
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a) zero momentum, as velocity on top is zero. p=mv\n     p=0.100kg * Om/s\n\nb) Vf = Vi + at\n      0=15+(-10)t\n     t=1.5ec (to reach maximum height)\n\nDivide by 2 for halfway up .i.e 0.75ec\n\nVf = Vi + at\nO=15 + (-10)(0.75)\nVf = 7.5m/s\n\np=mv   (p = momemtum)\n\np = (0.100kg)(7.5m/s)\np = 0.75kg*m/s, upwards (Direction is important)
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a) vcenter = (m1*v1+m2+m2)/(m1+m2)\n\nb) v1 rel = v1 - v center\nv2 rel = v2 - v center\n\nremember velocity is a vector so you must keep track of magnitude and direction
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You can find a good summary here:\n\nhttp://www.sparknotes.com/testprep/books/sat2/physics/chapter12section2.rhtml
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So the first one can be solved easily using the energy method. Potential energy = mgh = 2*9.81*6\nKenetic energy  = 1/2mv^2 = 1/2*2 * v^2\nBecause there is no friction Potential Energy = Kenetic Energy:\n\n2(9.81)(6)=1/2(2)(v^2) \n\nNow you just need to solve for v, using your calculator or head.\n\nFor the second one you have to account for the amount of energy lost from friction. So, instead of KE=PE you get:\n\nKenetic Energy = Potential Energy - Energy lost by Friction\n1/2*m*v^2 = m*g*h - Friction energy \n\nSo you can do some algebra and\nm*g*h - 1/2*m*v^2 = Energy lost by friction \n\n\nThree is impossible without more information. You can't solve for u with just m, theta, and work. You need some information about the height, or the length of the ramp, or end velocity. Just think about it, you could have a ramp 100ft tall with little friction, or you could have one 1 ft tall with lots of friction. The work done would be the same even if you kept the masses and theta the same.
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1. The bouyant force of the water on the object is the difference in the reading, i.e. BF=(.0554-.0370)g\n                                   =0.0184gN\n\nBased on Archimedes' Principle, BF =Volume*Density of water displaced.\n \n0.018g=Volume*1000g\nVolume=.018/1000\n            =1.8*10^-5m3\n\n2. Density of Object=Mass/Volume\n                                =.0554/1.8*10^-5\n                                =3078kg/m3
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gravity and air resistance.
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cmon...dont u know basic physics...just calculate the resultant of these 2 forces\nanswer= [(132)^2 + (181)^2]^1/2\nif this makes sense!!
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Both the above answers are correct, provided the acceleration is horizontal. If it's not then it's a vectors and trig. question.
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1) KE = ½mv²\n1.1 x 10^9 = ½ x 8 x 10^4 v²\nv² = 2.75 x 10^4\nv ≈ 165.8 m/s\n\n2) another physics question: what is the speed of a 0.145kg baseball if its kinetic energy is 109J?\n\nDo as above\n\nKE = ½mv²\n109 = ½ x 0.145 x v²\nv² ≈ 1503.4\nv ≈ 38.8 m/s\n\n3) ANOTHER QUESTION: TWO 3.0 bullets are fired with speeds of 40.0 m/s and 80.0 m/s, respectively. what are their kinetic energies? which bullet has more kinetic energy? what is the ratio of their kinetic energies?\n\nAssuming 3.0 g bullets as they would NOT be 3.0 kg bullets\n\nSo 3 g = 0.003 kg\n\na. v = 40 m/s\nKE = ½mv²\n= ½ x 0.003 x 40²\n= 2.4 J\n\nb. v = 80 m/s\nKE = ½mv²\n= ½ x 0.003 x 80²\n= 9.6 J\n\nratio of energies (slower to faster) = 2.4:9.6 = 1:4 = (ratio of velocities)²\n\n4) another physics question: a car has a kinetic energy of 4.32 x 10^5J when traveling at a speed of 23 m/s. What is it mass?\n\nKE = ½mv²\n4.32 x 10^5 = ½m x 23²\nm ≈ 1633 kg
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you should try this site http://www.oberlin.edu/physics/dstyer/SolvingProblems.html
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i think u typed the question wrong, can u please correct it and if u have 2 unknowns, please give 2 equations, else u cant get a definite solution
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the first after collision moves in an opposite direction and the second ball moves towards the direction of collision\nfirst ball moves with a velocity of  16.66cm/s\nsecond ball with 66.66cm/s