physic .I have answer I need the way of solution(answer1-  5/1m/s   2-   14/9 m/s)?
340116
a) d= 0.5 * 9.8 * t^2 + Vi * t = 10 m\nFor t=1 sec:\nVi= 10 - 9.8*0.5 = 5.1 m/s \n\nb) g= (Vf-Vi)/ t\nFor t=1sec and Vi=5.1\nVf= 9.8+5.1= 14.9 m/s\n\nHope it is clear enough.
666681
1/2 * mass * velocity^2.
1128176
work = change of kinetic enegy\nW = 1/2 m (vf² - vi²)\nW = 0.5 * (0.250 kg) * (16² - 20²) m²/s²\nW = -18 J
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.
288780
From the following link, look for "Chapt22, Problem-24".\n\nThis problem is similar, except the speed is 10% instead of 11%. I'm sure you can use the same formulae though.\n\nhttp://www.jca.umbc.edu/~george/html/courses/phys112/2004fall/homework/04fall_phys112_hk8.pdf
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
557508
T = 2 * pi * sqrt(l/g)\n2 = 2 * pi * sqrt(0.9942/g)\n--- taking your length to be in metres\n\ng = 9.8123607 m/s2
940115
341m/s x 3600s/h x 1km/1000m = 1227.6km/h
705605
formula\n\nv^2 = u^2 + 2gh\n\nv = final velocity\nu = initial velocity\n\nso v^2 = 0 + 2*9.81*0.041 = 0.80442 meter\n\nthus v = 0.8968 meter / second\n\nhere mass and angle is not required.. unless some conditions are given ... which is none here ... so 0.8968 meter/second is the answer
923403
Start off by drawing the picture. You have a right triangle, with a hypotenuse of 25, a base (b), and height (h). \n\nDon't get jumbled down by numbers yet, ignore them. Visualize what you are calculating.\n\nVelocity / speed is the derivitive of position. (v = delta x / delta t). So all you need to do is find an eqation relating the geometry using known and unknown quantities and then differentiate with respect to time. Sounds confusing but all you have to do is....\n\nYou know, a^2+b^2 = c^2 for a right triangle. In our problem,\n\nb^2+h^2 = 25^2\n\nThis is our position equation. Now differentiate with respect to time.  DO NOT forget the CHAIN RULE! <------ VERY important\n\n2*b*db/dt +2*h*dh/dt = 0.\n\nFrom here just use algebra; where,\n\nb=7 ft, and db/dt = 2 ft/sc.\n\nWhich leaves two equations and two unknowns.\n\nThe important thing to remember is to take this step by step and try to visualize what you are calculating. Also, looking at the units of the given quatitities can give you some idea of what you will be required to do. \n\nAlso, just remeber that speed / velocity is the derivative of position with respect to time (the units reflect this).
1134148
Since sheriefhalawa didn't give you an answer, but merely told you how to find it, I expect this will be of more interest:\n\nImpulse(1) = 26 * .056 = 1.456 Ns\nDv(1) = I(1)/m  =  9.1 m/s\nv(1) = 3.1 + 9.1  =  12.2 m/s\n\nI(2) = -26*.056 = -.6608 Ns\nDv(2) = -.6608/.16 = -4.13 m/s\nv(2) = 3.1 - 4.13 = -1.03 m/s
378804
Vf=Vi+At\n5=12+A*2.3\nA= -3.04348 m/s^2
124298
Answer to #2\nt=1.565 seg\nV= 2.556 m/seg\nt=1.565 seg, the same.
759333
all object fall 4.9m after 1 sec\n\nX=1/2(9.8m/s^2)(2s)^2\nx=19.6m\n\nx=(1/2)(9.8m/s^2)(3s)^2\nx=44.1m
359140
Use the equation:\n\n    v = u + at\n\nwhere:\n      v = 0 (because the dog comes to a stop)\n      u = 1.5 m/s\n      a = -.35 m/s^2\n\nSubstitute into the above and solve for t (answer is between 4.0 and 4.5 s)
723976
Question is: why on earth would you want to make that calculation?  m/s^2 is the standard form for acceleration.
962403
2M = wx(L-x)\n\n(2M)/w = x(L-x)\n\n2(50N/m) / (5N/m) = x(20m-x)\n\n(100N/m) / (5N/m) =\n\n(20 N/m) = \n\nsorry i couldn't get an answer\nthat is as far as i can get\nit might help a little bit though