A person in a wheelchair approaches a 0.70-m-high ramp at a speed of 4.1 m/s. Calc.  the final speed?
149886
Hope this helps,  first remember that energy must be conserved so E initial will equal E final so you use the equations for potential and kinetic energy.  mgh for potential and 1/2mv^2 for kinetic.  You know the initial velocity, gravity and the height of the ramp so you simply break it into components and get the answer of 3.15 m/s.  HOpe this helps
620893
The best way to work this out would be to plot a speed time graph for the information.\n\nOr you could express it as a simultaneous equation.
680534
Use the equation from the link below.  You will find that \nv/c = 0.14365, or\nv = 0.14365 c\nor\nv = 43,100 km/s\n\nThe object is moving away from the observer at this rate.\n\nBy the way, you probably won't get credit unless you can work out the math steps I left out.
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).
952952
one third of the elastic limit is 8x10^7 Pa and the total area is 3x10^(-4) m² so the total force is 2.4x10^4 N.  It takes 12,740 N to hold the elevator stationary agains the acceleration of gravity which leaves 11,260 N available to accelerate the elevator upwards, so the maximum acceleration would be about 8.66 m/s².\n\n\nDoug
311411
Math model of problem:\n\nD2S/Dt^2 = 2.1  \n\nIntegrating:\n\nDS/Dt = 2.1*t + c1\n\nIntegrating again:\n\nS = 2.1*t^2/2 + c1*t + c2\n\nFrom the statement of the problem:\nWhen t=0 DS/Dt=0 and S=0\nTherefore c1 and c2 both are zero\n\nSo S = 1.05*t^2\nSo t = square root of S/1.05\nThat is t = square root of 70/1.05\nthus t = 8.165 secs.\n\nSo boat accelerates for 8.165 secs in travelling from rest over a distance of 70 metres
414568
.75 seconds, and Acceleration
924813
(288^2)/(2*32.2) is the maximum height.\nat time 288/32.2 seconds\nit hits ground at time time   t+576/32.2, where t is obtained from\n45=288+.5*32.2*t^2
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
951151
Weight the person has to lift = mass of the bucket * accelaration due to gravvity\nWeight the person has to lift = 24 * 9.8 J \nWork done = weight lifted * DIstace\n8*1000 J = 24 * 9.8 * Distance\nDistance = 24*9.8/8*1000 meters\nDistance = 3*9.8/1000 meters\nWell this is the answer. Right now i am feeling quite lazy to calculate all this. Hope you can manage with this answer. \nThanks
685470
the speed = distance/time
872577
He can accelerate at 0.9g for the whole 50 m. The relevant equation is D=at^2/2, thus t=sqrt(2D/a). Solving, t=sqrt(100/(.9*9.80665)) = 3.366 s, and he reaches a speed of a*t = 29.71 m/s.
888459
v(t)= -gt - ve(ln((m - rt)/m))\n\nx(t) = ∫v(t) dt\n\n= ∫[-gt - ve(ln((m - rt)/m))] dt (Note ∫lnu du = ulnu - u + k)\n\nLet u = (m - rt)/m Thus du = -r/m dt ie dt = -m/r du\n\nSo x(t) = -½gt² - ve*-m/r (u(lnu - 1) + c\n= mve/r [ (m - rt)/m * (ln((m - rt)/m)  - 1] - ½gt²\n= ve(m - rt)/r * (ln((m - rt)/m) - 1) - ½gt²\n\nIf g = 9.8 m/s^2, m = 30000 kg, r = 155 kg/s, and ve = 3000 m/s and t = 1 minute = 60 s\n\nx(60) = 3000 * (30000 - 155 * 60)/155 * (ln (30000 - 155 * 60)/155 - 1) - ½ * 9.8 * 60²\n\n≈ 138390 m\n\n= 138.39km
946074
avg. speed = distance/time = 75/11 = 6.82 m/s
725445
You can't without the radius of the wheel...m/s is a tangential speed.
484307
in air= 340m/s\nin water=1500m/s\nin solid=4500m/s
927736
Everyone above assumed a constant speed.  the more general formula is \n\nD = ∫v dt