Can someone please help me with physics?
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Piston 1 (in the figure linked) has a diameter of 3.0 in. Piston 2 has a diammeter of 1.5 in. In the absence of friction, determine the force necessary to support the 500 lb weight.\n\nAnswer: The force required at piston 1 = 125 lb. This is because the pressure in piston 1 = pressure in piston 2\ni.e. 125 /one quarter area = 500/ area.\nArea of piston 2 = pi x [3.0/2]^2  sq. in.       Area of piston 1 = pi x [1.5/2]^2  sq. in\nUse the principle of moments to find force F:\n\nie. 125 lb x 2.0   =   F x (2 + 10)\nHence F = [125 x 2] / (2 + 10)\n              =  125 / 6    lb force
955176
Force - Newtons\nMass - Kilograms\nAcceleration - metres per second
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hey, didnt i answer something very similar to this for you? you need to do F=ma so 43000000N/2100000kg then you get an acceleration, but then you have to subtract gravity from this because thats the acceleration produced by the rocket but there is still gravity pulling with 9.8 m/s^2 in the opposite direction. so just do (43000000/2100000) - 9.8, and that's your answer :)
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I hope I still have the equations memorized. I will try to give you a helping hand, but please check the answers.\n\nMomentum is = mv.... mass times velocity\n\nM= mv = 1.24*10^5 * 75.5 meter per second = 9362000 kg.m/sec\n\nHorizontal component of the momentum is equal to Moment cos(x), where x is the angle of approach which is 11.5. The equations are:\n\nVert component is = mv*sin(11.5)=1,797,504 kg.meter/second\n \nHoriz component = mv cos(x) = 9,174,760 kg.meter/second
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Try: allaboutphysics.com I think that might help.
1166211
I do not know you that well yet...............\n\n\nHow about a physics question or something challenging?
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The last poster needs to have his dynamic physics retaught, or stop trying to sound like an engineer.\n\nThe first mistake, is identifying newtons = force x distance.  This is actually a measure of work or energy and measured in joules.  The second mistake identifying force as being measured in joules.  Force is measured in Newtons.  Since it was Isaac Newton in the second law of motion that defined Force.\n\nYour question cannot really be answered without more information.  What is the direction of motion, horizontally or vertically?  \n\nVerically?  Then 150 lbs of force.  Pounds were once used to measure force.  If you want to use metric, convert the 150 lbs to kilograms, multiply by 9.8 and then you must exert at least this much force in Newtons to overcome the force of gravity.  150 lbs is 68.04 kg.\n\nHorizontally, what type of surface?  Determine the coefficient of friction, and multiply it by the mass in kilograms.  Then this is the amount of force you need to exert to overcome the force of friction.  There are two types of friction, static and dynamic.  In other words it takes more force to move an object at rest, but once the object is in motion it takes less force to keep it move (opposing the dynamic force of friction).\n\nWatts by the way is a measurement of power, which is energy derived by time, Power = dE/dt.  Joules per second.
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Try this:\nhttp://kr.cs.ait.ac.th/~radok/physics/f3.htm
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Alright... physics is fun if you get the right problems :) .  \n\nThe collision is elastic... your book should have a conservation of momentum equation.  It states that p1+p2 = Ptotal\n\nYou have to use component vectors for the velocity after the collision.  A good thing to do would be to draw the problem.  It would make it a lot easier.  So use conservation of momentum.\n\nAnd there are two equations that would really help you:\n\nFinal velocity of block A= ((MB-MA)/(MA+MB))*VAinitial\n\nFinal velocity of block B= ((2MA)/(MA+MB))*VAinitial\n\nHelp with physics:\n\nALWAYS draw pictures epecially on your exams.  Get on the good side of the professor, go to a tutoring center if you have one, do many problems and memorize how to do the basic ones if you want an A on the test.
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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.
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For your first problem...\n\nYou know that T = Fc = (mv^2) / r\n\nWhere T is the Tension, Fc is the Centripetal Force, m is the mass, v is the velocity, and r is the radius.\n\nYou're given m = 80 kg, r = 0.90 m, and T = 600N.\n\nSo...\n\n600 N = (80kg * v^2) / (0.90 m)\nSolve for v...\nv = +/- 2.598 m/s\n\nSo since -2.598 m/s doesn't make sense (a negative speed), the solution would be ~ 2.6 m/s.\n\nFor your second problem...\n\nIts been a long time since I've done circular motion.  This site has the equation you need to use (its too messy to type here): http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/carbank.html if you can figure out the angle.  So if you know what the angle should be, that equation will get you the coefficient of friction.\n\nIf you can't figure out the angle, at least attempting to solve it using that equation and some reasonable angle should get you some points I'd imagine.\n\nIf I can think of the correct angle I'll post an edit to this.
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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.
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Where is your freebody diagram? There are other forces acting on the block than just gravity...\n\nGood Luck
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Solve for momentum.\n\nBefore collision total momentum  = mv\n\nDefine your coordinate system so that the particle was traveling along the x axis in the positive direction, and the collision happens at the origin. Then all of the momentum before the collision is in the x direction. the initial y component of momentum is zero.\n\nparticle 1 goes flying off at v1 at angle a1\n\nparticle 2 goes flying off at v2 at angle a2\n\nThe total momentum afterwards must be equal to the momentum before. So the y component of p1 plus the y component of p2 must equal the y component of the momentum before collision (zero).\n\nSo m*v1*sin(a1) = m*v2*sin(a2)\n\nSimilarly the x components must add up so \n\nm*v1*cos(a1) + m*v2*cos(a2) = mv\n\nNow just solve the first equation for V1 and substitute in the second equation. \n\nThen solve the second equation with substitutions to find the value of v2 in terms of v, and the sin and cos of the two angles. \n\nFinally substitute the value of v2 back into your equation for v1 to find the value of v1.\n\nI leave the simple algebra of solving the equations and the calculation of plugging the numbers into the equations to the student.
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1) I think your 2nd c.f. should say c. acceleration; that's the units you're using------ the c force of a mass m with acceleration a is f = ma, so youhave 1.5 kg*18 m/s² = 27 N\n\n2) First find the velocity: v = d/t = 2πr/t = 2π*.75m/.8sec = 5.89 m/s\nThen, acceleration: a = v²/r = 5.89²/.75 = 46.256 m/s²\nThen, the tension: T = ma = .1kg*26.256m/s² = 2.626 N\n\n3)  v = 2πr/t = 2π*90/12.3sec = 45.974m/s\na = v²/r = 45.97²/90 = 23.485 m/s²\nF = ma = 900*23.485 = 21,136.5 N\nThe lateral friction of the tires on the track
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Very simple. start out asking yourself what information do i know about the problem. Anything from masses and forces to symmetries of the problem. Next Ask yourself what additional information you need to solve the problem, values, formulas, etc. from there it should all fall out with ease.
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http://en.wikipedia.org/wiki/Physics