Physics Problem ☺ ☻ ☺?
965191
This isn't very hard - you just need to do each of the little pieces. I'll give some hints.\n1) When the block is at the extreme from the rest position all the system energy is stored in the spring. The stored energy is spring force (k*x) times spring deflection distance (x). So it's k*x^2 where x is 8.5. \n\nThe block has maximum speed when the block is in the center of the oscillation range (the spring is fully relaxed at that point), so all the energy is in kinetic energy. Energy is conserved, so\nMax spring energy = Max KE\nk*x^2 = (1/2)*m*v^2\n\n2) Best hint here: think, no math req'd.\n\n3) Spring force F = k*x. Acceleration = F/m\n\n4) Acceleration is max (see #3) at x = 8.5 and zero at x = 0. In the formula\nx = (1/2)*a*t^2,\nyou can use the average of the accelerations those 2 points for a.
660156
http://www.physlink.com/education/askexperts/ae25.cfm
1022451
I got 85% in physics last term.I am pretty good at the subject.I will try my best to help you if only you can frame the question.
929849
solve three body problems
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.
1120781
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
948476
I won't show in details since it's hard to type out the work, but you'll get the rest, here it goes:\n\na)\nWork = mgd  *d=change in height\nW = (10kg) (9.8m/ss) (sin15) (-7.5m) = -190J\n\nb)\nW=(friction coef)(natural force)(distance traveled)\nnatural force = (cos15)(10kg)(9.8m/ss)\nW= (0.4) (cos15) (10kg) (9.8m/ss) (-7.5m) = -280J\n\nc) \nW=Force x distance\nW=(100N)(7.5m)=750J\n\nd)\n(initial KE)+ (puller work)=(gravity work) + (frictional work) + (final KE)\n(final KE) - (initial KE) = 750J - 280J - 190J = 280J\n\ne)\n(final KE) - (initial KE) = 280J\n*solve for final v\nv = squareroot (2 x [280j+ (1/2)(10kg)(1.5squared)] / 10kg) =7.63m/s
256104
Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.
172516
Sorry I'm not good in physics.\n\nAre you from the Philippines? For sure you are because you know Sandara Park.
510692
Try: allaboutphysics.com I think that might help.
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.
730564
1. friction\n2. W=mg  , he has a mass of 100Kg. Weight is the force=newtons, g=pull of gravity N=100 x 9.8m/sec^\n98Newtons\n3. correct
877088
E = hc/λ \nλ = 1.2 x 10^-10 m\nc = 3 x 10^8 m/s\nh = 6.626068 × 10^-34 m2 kg / s\n\nE = 6.626068 × 10-34 * 3 x 10^8 / 1.2 x 10^-10\n\nE = 1.7 x 10^-15 J.
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
795982
∫_0_2 g(t) dt = 5\n\n(a)\ny= ∫_0_4 g(t/2) dt\nlet t/2 =x ; so dt = 2dx; and the range becomes x:0 to 2\nso, y= ∫_0_2 g(x) 2 dx = 2  ∫_0_2 g(x) dx = 2 * 5 = 10\n\n(b)\ny= ∫_0_2 g(2-t) dt\nlet 2-t=x; so dt = -dx; range becomes x:2 to 0\ny= - - ∫_0_2 g(x) dx = 5
443144
you should try this site http://www.oberlin.edu/physics/dstyer/SolvingProblems.html
617628
Ignore the answerer above. Learning physics can be tough.\nSolve as below:\n\nu=30m/s\nv=0\na = -6m/s^2\n\nusing v=u+at\nt=(v-u)/a\nt=(0-30)/(-6)\nt=5 seconds is the answer