solve the uniform motion problem...?
831693
Let x be the distance he walked.\n\ntime = distance/speed\n\nso time walked + time in truck = 10 hours\n\nx/5 + x/20 = 10\n\nmultiply by 20\n\n4x + x = 200\n5x = 200\nx= 40 km\n\nCheck - he spent 40/5 = 8 hrs walking out, and 40/20 = 2 hours walking back for a total of 10 hrs.
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.
742316
Torqu equals force applied times radius, so the equation would be T=Fr    or    T=mar    so, T=(4)(9.8)(6), or 235.2 n/m\n\nMoment of inertia is not needed to solve this problem unless it later asks for another different answer.
1133022
It theoreticaly should not be too dificult to solve using vectors.  I am not going to solve it because I would prefer having exact values to work with, but basicaly, what you would need to do is find the complete distance, also make a graph where you find the time intersect of the torpedo and boat, and work with the x/y components to figure out what would be the ideal inclination.  It is a complex problem where you must work with more than one vector system, but it is doable if you work in parts.
1171207
Let us write coeff. of friction = k     \n                    acceleration due to gravity = g     m/ sq. s\n                    uniform velocity of motion = u      m/s\n                 \n                    Since k=F/R where R = weight of the box acting downward and F = Frictional force acting against the direction of motion.\n                   Hence, F = kR. Again F= ma, where a is the retardation on the box and R=mg.\n                    Therefore, a=kg. (m cancels out on both sides).\n\n                     By using the formula:\n                            v^2=u^2 - 2aS  ( here a is retardation) we have\n                      S= u^2/(2a) = u^2/(2kg).\n\n            Substituting the values:\n                         u=8 m/s\n                         k= 0.16\n                         g=9.8 m/sq. s\n                        The result is S= (8^2)/(2*0.16*9.8) = 20.41 m (approx.).
162313
Let V = speed (the magnitude of the initial velocity)\na = angle, t = time, \nVx = horizontal (x) component of velocity\nVy = vertical (y) component of velocity\nAx = acceleration in x direction\nAy = acceleration in y direction\nDx = net distance traveled in x direction\nDy = net distance traveled in y direction\n\nVx = Vcos(a)\nVy = Vsin(a)\n\n  Dx = 1/2 Ax*t^2 + Vx*t\nbut Ax = 0, so Dx = Vx*t, t = Dx/Vx\n\n  Dy = 1/2 Ay*t^2 + Vy*t\n(Ay = -g, the acceleration of gravity, negative for downward)\n -1/2*g*Dx^2/(Vcos(a))^2 + Vsin(a)*Dx/(Vcos(a)) = Dy\nMultiply by V^2, and collect terms with V^2 on the left:\n(Dx*tan(a)-Dy) V^2 = 1/2*g*(Dx)^2/(cos(a))^2\nV = sqrt(1/2*g*(Dx)^2/(cos(a))^2/(Dx*tan(a)-Dy))\nsubstitute a = 30 degrees, Dx = 6.05 m, Dy = 3.05-2.2 = 0.85 m, g = 9.8 m/s^2\nV = 9.51 m/s\nFor a = 60 degrees,\nV = 8.63 m/s
765864
I'm unclear with the way you stated the problem. Am I looking for a implicit derivative. I'll check back to solve it.
778679
I believe you will find the solution to your problem here:\n\nhttp://www.engineeringtoolbox.com/bernouilli-equation-d_183.html
872212
Use the certical component to find the time in the air and since the horizontal velocity does not change; Dx = vx * time\n\nDx = v^2 / g\n\nheight = v^2 / (2g)\n\nHope this helps without actually solving it for you.
776431
Integrate the acceleration equation with respect to time, twice.  Use, Xo = 500 and Vo = 10...solve for t.
127000
This problem is solved with the application of work and energy principle which states that total work done on a moving particle is equal to the change of its kinematics energy:\nW=T2-T1\nHere the only work done on the particle is dot to gravitation force and is equal to -m*g*h:\n\n-m*10*0.45 = (1/2)*m*(1)^2-(1/2)*m*(v)^2 --->\nm*v^2=(4.5+0.5)*m*2=10*m --->\nv=sqrt(10) m/s\n\nv is equal to sqare root of 10.(approximately 3.162 m/s)
321671
Ah, this brings me back!\nThis is conservation of momentum problem. That means that the initial momentum must equal the final. You must break up the values to x and y directions.\n\nThe 900kg mass is traveling 18mps * cos(30) = 15.6mps in +x\n                            18mps * sin(30) = 8.98mps in -y\nThe 750kg mass is traveling 0mps in x\n                            23mps in +y\n\nWith these values take the initial momentum as\n\ninit mom in x = 900kg(15.6mps) + 0 = final mom = 1650kg*(v)\nwhere v is final speed in x = 8.5mps\n\ninit mom in y = 900kg(-8.98mps) + 750kg(23mps) \n\nthat equals the final mom in y = 1650kg*(v)\n\nso v=+5.6\n\nIf you know the speed in the x and y directions, convert to vector notation by using s^2=vx^6+vy^2 = 10.2mps\nand the angle is atn(vy/vx)\n\nHope this helps
602635
For uniform motion:\n\ns/t is a steadily increasing line [straight diagonal across your page]\nv/t is a flat line at a non-zero level [horizontal]\na/t is a flat line at zero [horizontal]\n\nFor uniformly accelerated motion:\n\ns/t is an increasing curve [like the right-hand side of y=x^2]\nv/t is a steadily increasing line [straight diagonal]\na/t is a flat line at a non-zero level [horizontal]
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.
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
758142
a) There is an equation derived from the motion in x and y that results like:\n\ny = (tanθ )x - g x²/ [2(Vo cosθ)² ] we know y, x and θ we need to solve for Vo:\n\nVo =  (x / cosθ) √[ g / (2 (x tanθ  - y )]\nIf we put  the plane at the initial hit at 1 m, y is 21 m -1m = 20 m\nAnd as we know that x = 130 m and θ= 35°\nSubstituting:\n\nVo = 41.7 m/s <<<<<<<<\n\nb) t = x / ( Vo cosθ )  Substituting: t = 3.81 s <<<<<<<\n\nc) Vfy = Vo senθ - gt   (Vfy is the final velocity of y after the time t)\nSubstituting, its components are:\nVfy = -13.46 m/s      <<<\nVox = Vfx = Vo cosθ = 34.16 m/s   <<<\n\nand its magnitude:\n\nVf = √(Vfx² + Vfy²) = 36.71 m/s <<<<<
855447
if it is equations of motion then \nv=u+ft\ns=ut+1/2ft^2\nv^2=u^2+2fs\nhope this is what you wanted