Physics Homework - how to solve?
812983
I solved this as if the artist strikes the trampoline at 5 m/s downward.\nIs he standing on the trampoline and gets sent upward at 5 m/s, or is he standing on a platform some distance above the trampoline?\n\nFor striking the trampoline at 5m/s downward\nStart by considering the energy in the system:\nThe artist has kinetic energy of 1/2*m*v^2\nand potential energy of m*g*d\nwhere g is 9.8 m/s^2 and d is the net downward displacement of the spring\n\nThe spring will absorb the energy =1/2*k*d^2\nwhere k is the spring constant\nat equilibrium when the spring is fully compressed the energy in the spring will balance out the rest\nset all energy equal to zero for equilibrium\n0=1/2*kd^2-m*g*d-1/2*m*v^2\nmultiply by two\n0=k*d^2-2*m*g*d-m*v^2\n\nplug in the numbers\n0=6.2E+4*d^2-2*65*9.8*d-65*5^2\n0=6.2E+4*d^2-1274d-1625\nthe positive root, which is the displacement of the spring, \nset negative since I used positive motion downward\nd=-.172m\n\nSince this isn't the answer, I will assume that the artist was on a platform and jumped onto the trampoline.  I will compute the height of the platform above the uncompressed trampoline:\nIn this case the spring energy is computed from your answer:\n1/2(62000)*.31^2\n=2979\nthe potential energy lost due to compression is\n65*9.8*.31\n=197.5\nHis vertical velocity when he jumped from the platform\n=.5*65*25\n=812.5\nThe energy will sum to zero.\nThe energy in the spring came from three sources:\nHis vertical velocity in kinetic energy\nPotential energy by standing on a platform above the trampoline\nThe potential energy given to the spring due to his weight and travel during compression\nSo\n2979-197.5-812.5-65*9.8*h=0\nwhere h is the height of the platform\nh=1969/637\nh=3.1m\n\nj
289077
add all the equations, you get\n2(r+s+t) = 56\nso,\nr+s+t = 56/2 = 28\ndone!!
12692
you left out the number of seconds.  I'll give you the formulas you need though:\n\n(a)  a=v/t  so 6 divided by the time you forgot to put.\n(b)  F=ma  so .2 times the acceleration you just solved for in (a)\n(c)  same as (b), the puck exerts an equal force on the stick, the puck just goes flying farther because it's smaller.
899235
A) I think I got it...Use energy, except for g, use   Fg cos(19 degrees), found from a free-body diagram.\n\nso 1/2 mv^2 = mgh\n\n1/2 (4.7)^2 =  9.8(cos (19))  h    solve for h\n\nB)  now that you have h...maybe you can use   \n\nd y = V(sub 0 sub y) t + 1/2 (a x cos (19)) t^2\n\nyea, tell me how it comes out.
769928
T = 2pisqrt(l/g)\nl = g(T/(2pi))^2 = 9.80(1.0/(2pi))^2 = 0.25 m\n\n(The object's weight is irrelevant for small oscillations.)
404136
Calculate the total energy of the system before and after and make an energy balance.\n\nFor example, lets say you have an inelastic collision between two cars of equal mass m:\n\nInitial State:\nCar 1 moves at v1 to the right\nCar 2 moves at v2 to the left\n\nAssume v1 > v2\nThey hit and stick together completely.  Use conservation of momentum to find the total resultant velocity:\n\np1 + p2 = ptotal (p is the momentum of car 1 and car 2 respectively)\n\nm*v1 - m*v2 = (m+m)*vf\n\nvf is the final velocity of the clump\nthe minus sign is because I am choosing the direction "right" as positive (momentum is a vector)\n\nvf = (v1-v2)/2\n\nNow do an energy balance:\n\nEbeforecollision = 1/2*m*(v1^2) + 1/2*m*(v2)^2\nEaftercollision = 1/2*(m+m)*(vf)^2 + Efrictionlosses\n\nEbefore = Eafter (this is conservation of energy, and it is true as long as you account for all forms of energy...it wont be true if you forget about the frictional losses in Eafter)\n\nNow solve for Efrictionlosses\n\nTo get the total fractional energy loss, take the ratio Efrictionlosses/Ebefore
533021
there are some solved physics problems on \nhttp://www.dctech.com/physics/help/\nbut I don't know if it will help you out.
864532
Set the tension of the left (T1) times the cos 53 degress equal to the tension of the right (T2) times the cos 37 degrees. Solve for either T1 or T2 in terms of teh other variable. Then set the sum of T1 times the sin 53 degees and T2 times the sin 37 degrees equal to 23 kg times 9.8 meters per second. Substitute the term from teh first equation into the second one, and solve for the variable that is left. Then just plug that variable into the first equation to get the other variable.
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
929849
solve three body problems
835037
Let the two case be identified as having  moment-of-inertia I1 and I2. \nTorque T on to the each of the two bodies produces angular accelerations (alfa1 and alfa2) given  by the relation T= I x alpha \n\nalfa1 = T/ I1 and alfa2 = T/I2\n\nStarting form rest(wi =0), let the time taken to reach final angular velocity wf be t1\nusing:  wf =- wi + alfa x t  \n \nwf/alfa1 = t1 and wf/alfa2 = t2 \nleading to \n\nt2  = alfa1x t1/alfa2 \n\n     = I2 xt1/I1\n     = (I2/I1) x t1\n     =. ......Ans \n(Hoping you will put the values and get the answer you want)
29264
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.
926868
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.
917347
It sounds like you know what to do.  From your description, it looks like a basic statically determinate problem.\n\nYou have a known applied force P, and two unknown reaction forces, lets say A and B for the top and bottom reactions respectively.  You also have two equations of static equlibrium, specifically the sum of the forces in the "x" direction, and sum of the moments about any point you wish.\n\n(Ok, really you also have the force of gravity and the normal force, but those are the only forces in the "y" direction so that equation won't help you, especially since the mass or weight isn't given).\n\nAssuming to the right is positive, assume A is to the left and B is to the right (this is intuitive, but if directions are incorrectly assumed the answer will just have a negative sign, thats all).\n\n                    _ ------------> P\n                    |   |\n                    |   |\n                    |   |     10.5 m\n                    |   |\n                    |   |\n                    |   |\n                    |   |<----------A\n                    |   |\n                    |   |       1.5 m\n                    |_|--------------> B\n                    \n                 (FBD!)\n\nSum of forces in "x" gives: P - A + B = 0\nSum of moments about B: A(1.5) - P(12) = 0\n\nI took moments about an unknown reaction, B, to eliminate it.  Also, I assumed counter-clockwise to be positive.\n\nSubbing in the value of P = 5.40 N into the second equation:\n\nA = (5.40*12)/1.5 = 43.2 N\n\nThen, put that into the first equation:\n\nB = A - P = 43.2 - 5.40 = 37.8 N\n\nThat should do it.  Both come out positive because their assumed directions were correct, however on the Free-Body-Diagram they are in opposite directions necessary for equilibrium.\n\nCheers,
937830
T=m1*-9.8+m1*a\nT=m2*a--m2*-9.8\n\nset them equal and solve for a\n\ntrust me, i had a test on this today
136454
I will give you a direction to solve the problem...\n\nFirst I will do the problem from a rectilinear formulation(using force, acceleration, velocity) then transfer it to a curvilinear formulation(using Torque, angular accelaration, angular velocity).\n\ngiven F=ma (or F=m(v/t) or more specifically F=m(dv/dt) or even more specifically F=d(mv)/dt).\nWhere F=force, m=mass, v=velocity, and t=time.\n\ngives you Ft=mv (conservation of energy=>conservation of momentum or the impulse-momentum equation).\n\nThis states that it will take a certain amount of force for a certian period of time to produce a certain change in velocity of an object assuming that the mass remains constant.\n\nTransmitting to curvilinear motion force becomes torque, v becomes rv(radius time velocity or w), mass becomes moment of inertia (or I) and t stays the same.
1093339
No one on here is going to actually DO your homework for you.. but here is a place for you the get help -\n\nhttp://www.themathpage.com/aPreCalc/logarithms.htm\n\nhttp://en.wikipedia.org/wiki/Logarithm\n\nhttp://physics.mtsu.edu/~wmr/log_2.htm