Help me with one physics problem, show steps.?
371345
d=vt\n\nWhere d is distance, v is velocity, and t is time.\nV here is the speed of sound through water, which you'd have to look up, and t=2.9 seconds.  Remember when you solve that your "d" actually covers the distance from the ship to the ocean floor AND BACK, so you'll have to divide it in half to find the depth.
596087
This is very straightforward. Which step are you having trouble with? The same goes for your other question.
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
210676
Use free body diagram to balance the forces and u can derive the equation of motion \nIf u want i can give u full derivation.\n"::
1163903
Set m1 = 0.145 kg, v1i = 0.720 m/s\nand m2 = 0.295 kg, v2i = 2.22 m/s\n\nConservation of linear momentum:\nm1v1i + m2v2i = m1v1f  + m2v2f\nSimplified\nm1(v1i-v1f) = -m2(v2i-v2f)\n\nConservation of energy (kinetic only):\n(1/2)m1v1i^2 + (1/2)m2v2i^2 = (1/2)m1v1f^2 + (1/2)m2v2f^2\nSimplified:\nm1(v1i - v1f)(v1i + v1f) = -m2(v2i - v2f)(v2i + v2f) \n\nSolving the two simplified equations:\nv1f =((m1-m2)/(m1+m2))v1i + ((2m2)/(m1+m2))v2i\nand\nv2f =((m2-m1)/(m1+m2))v2i + ((2m1)/(m1+m2))v1i\n\nTherefore,\nv1f = 3.315 m/s\nv2f = 1.23 m/s
898375
I know the answer is in my basic physics book.  I would have to look it up.  You should too.
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...
1144696
gravity and air resistance.
353265
Hess's law \n\n\n +142KJ  that is all you want "right" is the answer
966112
u can consult ur physics book
855600
u can also solve this by takin s= -100m\nn by the formula S= ut + 1/2 gt^2\ng = 9.8m/s^2. u = -6m/s .......4 the 2nd part\nand \n\nfind time of the 1st half by the formula \n=> v = u + at....[ where u = 6m/s, v = -6m/s(velocity in oposit direction)]\n\nALSO REMEMBER THAT WEN A TRAJECTORY AGAIN REACHES ITS STARTIN PT.  IT HAS THE SAME MAGNITUDE OF VELOCITY\n\n-ve sign is given to all dat is below the level or in the direction below the startin point\ncombine these 2................ n voila..\nwith time u can master this easy.... n logical method
737967
Draw a free body diagram as shown below: \n\n                [m1]------> T<---[m2]------->89N\n              <--------            <---------\n                f1 =19.6N        f2=29.4N\n\n\nFirst solve for f1 and f2:\n\nf=ukR is the general formula where f is the force of friction, uk is the coefficient of friction, and R is the force normal to the surface. In this problem R is the weight of the block=mg. Now substitute known values\n\nf1=0.2*10*9.8\n   =19.6N to the left\nf2=0.1*30*9.8\n   =29.4N to the left\n\nLooking at the free body diagram:\n\nfor m1:\n\nnet force, f1=T-19.6\n                      \nFrom the formula, f=ma  where f is the net force, m the mass, and a is the acceleration.:\n\nf1=10a\n\nThus,\n\nT-19.6=10a\n         a=(T-19.6)/10\n\nfor m2:\n\nNet force, f2=89-29.4-T\n                  =59.6-T\n\nSimilarly as for m1:\n\n59.6-T=30a\na=(59.6-T)/30\n\nNow a for m1 is also the same a as for m2, therefore:\n\n(T-19.6)10=(59.6-T)/30\nThe zero on each side cancels out. \n3T-58.8=59.6-T\n4T=59.6+58.8\n4T=118.4\n  T=29.6N in the direction shown in the above free body diagram.
756774
P=P (momentum before = momentum after)\nmv=mv\n(55kg)(4.6m/s) = (230kg)v\nv= ((55kg)(4.6m/s))230kg\nv=1.1m/s
12996
I can't draw a free-body-diagram here, so it's going to be somewhat more difficult to explain (a picture is worth 1000 words and all)...\n\nSince the block is moving at a constant velocity, we know that all the Forces are balanced.  this means that the Force of friction is equal to the force of gravity pulling the block down the incline.\n\nI'll write this as \nFf = uN  \n\nSince we know the angle of the slope, 25 degrees, we can break the Force of gravity into 2 components,  one that points down the incline (i'll refer to it as Fx) and one points the opposite direction (I'll refer to is as Fy) of the Normal vector N.\n\nIf you draw that triangle you notice the the hypotenuse has magnitude(length) of mg (mass times the gravitational constant.\n\nnow, lets use the sin & cos relationships to determine the length of the 2 components: Fx & Fy\n\nsin = opposite / hypotenuse\ncos= adjacent / hypotenuse.\n\nIf you look at the diagram, the opposite is Fx so\nsin 25 = Fx / mg \nFx = sin25 * mg \n\nand the adjacent is Fy so\ncos 25 = Fy / mg\nFy = cos25 * mg  \n\nagain since all force are in balance, \nFy = N \n\nand Fx = uN \nsin25 * mg = u * Fy \nsin25 * mg = u * cos25 * mg \nwhich reduces to\n\nsin25 / cos25 = u \nand \nfinally\nu = tan25 = ~.466
800064
For these problems, it would be helpful to draw a free-body diagram to represent all the forces on this problem.  Here you have 4:\n\n1) Weight in the vertical direction going down, let's denote it W\n2) Force, moving in the x-direction, denote it F_1\n3) frictional force, moving opposite of Force, denote it F_2\n4) Normal force, perpendicular place of contact (opposite direction of W), denote it N\n\nNow use Newton's 2nd law to write equations in each direction:\n\ny-direction: \nN - W =0 (because the block does not move in the y-direction).  Therefore:\n\nN = W, where W = m*g\n\nx-direction:\nF_1 - F_2 = 0 [because the block is just about to move when F_1 is applied].  Therefore:\n\nF_1 = F_2, where F_2 = u*N\n\nMaking the necessary substitutions for F_1 and N, you get:\n\n30 = u*(4.4)(9.8).  Solving for u, you get --------->\n\nu = 0.7\n\n--------\n\nHope this helps
1008231
Set the equation of motion.\nTaking upward direction as positive\ns=vt-1/2gt^2\nat maximum height s=10,v=0\ng=-g since upward direction is positive\nyou get t= sqrt(2s/g)\n=sqrt(2*10/9.8)\n=1.428 secs
105143
Draw a force diagram to help you figure this out.\n\nRemember, perpendicular vectors are independent.\n\nF=ma