Please Help Me with Physics?
931404
Gravity is proportional to the mass of the planet.  I'd say the acceleration on this planet is 0.1045 * 9.81 m/s^2.  \n\nFor b) use x =  1/2at^2\n\nt = time  \na = acceleration due to local gravity
734736
first one is easy\n\nfirst find the normal force\n\nFf (friction) = Fn(normal force ) (.35)\n0(motionless ) = Fn .35\nFn = -.35\n\nfrom there its easy\n\nFn = mgsin(theta)\n\nsolve for theta\n\n-.35 = (200)(9.8) sin(theta)\n\nand solve it ur self\n\nsecond one, tell us wat u want us to find and we can tell u, but here are the things u can find, ill give u the formulas\n\nacceleration = 0 (constant speed)\ntheta = 15 degrees\nvelocity = constant\n\nthats pretty much it, from what u gave me
179131
Try this:\nhttp://kr.cs.ait.ac.th/~radok/physics/f3.htm
391655
force = constant * distance\nF=kx\nforce = mass* acceleration\nF=ma\n\nx=F/k\nx=m*a/k = 4.0*9.8/800= about .05m, or 5cm
7127
here's a site with pics that might give you some ideas:\n\n\nhttp://www.physics.usyd.edu.au/competition/comp1998/comp98.html
653943
gravity and inertia\n\nread more below. It is really fascinating.
574012
I've alread answered two of your questiions in my last answer....\n\nIn this one, the guys above are correct - however, I'd try working it out for myself if I was you. Your tutor will soon realise something is wrong if you are giving answers to a subject you don't understand. Read the chapter, ask you teacher if you need more help. Learn Newtonian mechanics for yourself.
969486
I just helped someone else with their physics question, so I'm alittle tired of it.  But, I suggest you make an equation sheet...physics is all about equations.  So, if you have a piece of paper with your equations, all you have to do is look at the sheet and plug things in...
901641
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
898375
I know the answer is in my basic physics book.  I would have to look it up.  You should too.
481598
There are three objects; 5kg, 9kg, and a cable of negligible mass.\n\nWhen the three move, they must have the SAME acceleration.\n\nTherefore, consider the three as a single object of mass 5 + 9 = 14 kg, with acceleration ‘a’.\n\nThe only force that is pulling the three is the force given by the 9kg object. (The gravitational pull is only on this 9kg.)\n\nThis force is 9 x 9.8 N = F.\n\nSince F = M a,   9 x 9.8 = 14 x a.\n\nFrom this a = 6.3 m/s^2.\n\nAll the three objects are moving with an acceleration 6.3 m/s^2.\n\nIf the body 5kg is moving with an acceleration of 6.3 m/s^2, then what is the force acting on it?\n\nWe know that f = ma; hence, f = 5 x 6.3 = 31.5 Newton.\n\nSince the tension of the rope is the pulling force, tension =31.5 Newton.\n\nThe tension can be calculated using the forces acting on the other end of the cable also.\n\nThe downward force on 9kg is 9 x 9.8 N; the tension is the upward force T. \n\nThe net force acting on 9kg is (9 x 9.8 - 9 x 6.3) = 9 x 3.5 = 31.5 N = T.
979735
If we just give you the answer, you'll never learn properly.\n\nAs has been suggested you need to get everything in the right units first!!\n\nNewtons 2nd law F= Ma\n\nForce = Mass X Acceleration\nForce is in Newtons\nMass in Kilograms\nAcceleration in Metres per second\n\nYou already have the mass - 38,000kg\n\n\nYou need to figure out the acceleration:\n\n\n60 kilometres/hr = 60,000metres/hr\n60,000metres/hr divided by 60 = 1,000 meters per minute\n1,000 metres per minute divided by 60 will give you the needed finishing velocity of metres per second\n\nAcceleration = finishing velocity / time ( 20 seconds )\n\nForce therefore = 38,000 X your answer for acceleration!
696750
try the following. they are really helpful.\n\nwww.tutor4physics.com\nwww.physicsclassroom.com
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.
3541
all you have to do is calculate the work that the engine should have done.  W=Fx meaning work equals force times distance.  The force is mass times acceleration, so it would be 10 x 9.8 (9.8 meters per second squared is the acceleration due to gravity) giving you 98 (lets round to 100 so i dont have to get a calculator or think too much because im tired, and your other data only has one significant figure anyway, if you think you need it more precise you can go get a calculator and do the work im going to show you with 98 on a calculator...)\n\nanyway, so to find the work (W=Fx) you multiply the force we just found (100N) by the distance so you get 100N x 4m = 400J.\n\nNow you know that the engine should have done 400J of work, but instead it did 500J, meaning that it did extra work, so it's not 100% efficient, so to find the percentage of its efficiency, just divide 400/500 and multiply that by 100 to get the percentage.  So its .8 x 100 so 80%.\n\nAnd if i may ask, why do you keep asking very basic questions on completely random physics topics that you would clearly not be doing all at once, and if you already did all of it, you should already know these answers?
170255
It is much easier to just give you a link instead of explaining it:\n\nhttp://www.shef.ac.uk/physics/teaching/phy107/solvegray.html
698592
The acceleration due to gravity is independent of the mass of the object.\n\n The speed of any mass thrown up will decreases by an amount 9.8 m/s per second. \n\nIf the initial speed is u, then its kinetic energy is ½ m u u. Its potential energy is zero.\n\nThe total energy is½ m u u + zero = ½ m u u.\n\nAs it goes up its kinetic energy decreases whereas its potential energy increases keeping the total energy same.\n\nWhen it goes up through ‘h’ meter,  its total energy is [ ½ m u u + mgh].\n\n[ ½ m u u + mgh] = ½ m v v.\n\nm u u + 2mgh = m v v. \n\nCancelling m on both sides,\n\nu u + 2gh = v v\n\nThus we have the familiar equation vv = uu + 2gh.\n\nWhen the ball reaches 9 m its speed becomes zero.\n\nHence,  v v = 2 x 9.8 x 9\n\nSolving we get v = 13. 3 m/s . \n\nThe speed with which it is thrown up is 13. 3 m/s.\n\nWhen the height is 8 m, now having known the initial speed we use the same equation for solving for uu.\nvv = uu + 2gh\n\n13.3 x 13.3 = uu + 2 x 9.8 x 8 .\n\nSolving for u, we get  u u = 196 or u = 4.43 m/s.\n\nThe problem can be viewed in other way.\n\nWhen the ball is at its heighest point namely at 9m its potential energy is mgh. where h is 9m\n\nWhen it falls through one meter its kinetic enery increases and potetial energy decreases , keeping the total energy as the same.\n\nmg x 9 = ½  m vv + mg x 8.\n\nOr mg ( 9-8) = ½  mvv \n\ng x 1 = v  v/2\n\n2 g = v v. \n\nv v = 2 x 9.8 = 19.6 (m/s)^2. \n\nv = 4.43 m/s.\n\n Note that the mass gets cancelled and hence we need not know the mass of the ball.