Calculate the horizontal force that must be applied to a 1300 kg vehicle to give it an acceleration?
545558
Force = mass * acceleration\nForce = 1300 kg * 2.6 m/s2\nForce = 3380 Newtons
1001657
V= V(i) + AT Initial velocity plus acceleratin x Time\n\n\n   = V(i) + (F/m)T     Acceleration = Force/ Mass\n\nNote  use same system of units
412737
force = m*a + I*alpha\n\nthe first part is for the cart and the second is for the pulley\n\nThen relate the angular acceleration to the linear acceleration as follows:\nalpha=angular acceleration= w^2*r   in rad/s\nso a(accleration of cart)= w^2*r^2   m/s\n\nso force=m*a + I*(a/r)\n\nfrom which you can find a (acceleration)
857622
1) I think your 2nd c.f. should say c. acceleration; that's the units you're using------ the c force of a mass m with acceleration a is f = ma, so youhave 1.5 kg*18 m/s² = 27 N\n\n2) First find the velocity: v = d/t = 2πr/t = 2π*.75m/.8sec = 5.89 m/s\nThen, acceleration: a = v²/r = 5.89²/.75 = 46.256 m/s²\nThen, the tension: T = ma = .1kg*26.256m/s² = 2.626 N\n\n3)  v = 2πr/t = 2π*90/12.3sec = 45.974m/s\na = v²/r = 45.97²/90 = 23.485 m/s²\nF = ma = 900*23.485 = 21,136.5 N\nThe lateral friction of the tires on the track
856854
The acceleration of the system can be found using F=ma\nor a=F/m\na=65/(12+24+31)\n=.97 m/s^2\n\nfor the tensions, use the same equation considering the forces on each string.\n\nfor m1 the force is what is pulling the mass so that it accelerates at .97m/s^2\nF=12*.97\n=11.64 N\n\nfor the string between m2 and m3\nF=(12+24)*.97\n=34.92\n\nj
256104
Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.
1118371
double_nubbins has the right approach.  I will add a couple of hints: how long will it take the car to fall 19ft (remember the acceleration equation s=.5*g*t^2).  Assuming there is no air drag, the horizontal velocity of the car must be enough to cover the 52m width in that time.\n\nSpeed of the car? The horizontal speed? The falling speed?  The speed in the direction of motion?\n\nHorizontal speed: as calculated above\nFalling speed = g*t, where t is the time calculated above.\nSpeed  in the direction of motion = sqrt(horiz speed ^2 = falling speed ^2)
793496
The equation involved is \nT = I * alpha  where T = Torque; I = moment of inertia and alpha = angular acceleration.\nFor this system, \nI = m*L1^2 + m*L2^2\n = m*(20^2 + 80^2) * 10^-4 kgm^2\nInitial torque T = m * g * (80-20) *10^-2 J\nthis gives alpha = T / I = 8.65 rad/s^2.\nLinear acceleration = r * alpha\nHence a1 = 8.65 * 0.2 = 1.73 m/s^2\n          a2 = 8.65 * 0.8 = 6.92 m/s^2
738706
Both questions use the formula F=ma (Force = Mass * Acceleration):\n\n1) F = ma\n1310 = 214*a\na = 1310/214 = 6.1215 m/s^2\n\n2) F = ma\n3.6 = m*9.8\nm = 3.6/9.8 = 0.367 kg
883912
Let the banking angle be '$'.\nThe reaction force will be perpendicular to the banking surface.\nLet the reaction force be 'R'.\nLet 'm' be the mass of the car.\nvelocity of car (v)=27 mt/s\nradius(r)=60 mt.\n\nThe centrifugal force will be in horizontal direction. And this force will be balanced by a component of the reaction force 'R'.\nR sin$= mv^2/r ---------------------------(1) \n\nThe weight of the car will be vertically downward. This weight will be balanced by Rcos$.\nRcos$=mg  ----------------(2)\n\nNow dividing eqn(1) by eqn(2)\n\nRsin$/Rcos$=mv^2/r /mg\n=>tan$=v^2/rg = 27 x 27/60 x 9.8 = 729/588 = 1.239\n\n$= tan(inverse)(1.239)\n\nsolve above tan(inverse)  function. and find the value of the banking angle
955176
Force - Newtons\nMass - Kilograms\nAcceleration - metres per second
743343
Just like any other weight on an incline problem, draw a right triangle.  However, the lines aren't distances, they're force vectors.  The vertical vector = 17,000.  The horizontal line is irrelevant.  The hypotenuse is the force vector you want.  It represents the force of the truck sliding down the hill.  Basically, find this vector, and that's how much force of friction it takes to prevent sliding (so net force on the truck = 0).\n\nThe angle between the horizontal line of the triagle and the hypotenuse is 4 degrees, 20 minutes.  You want the other (non-right) angle.  SInce all three angles add up to 180 degrees, 180 - 4degs/20min (change that to decimals to save headaches) = the angle you need.\n\nNow, cos(that angle) = 17,000 / (the value of the vector your solving for, all it X).  Rearranging that gives you X = 17000/cos(that angle).\n\nHappy solving.
905915
You're going to have to resolve this two-dimensional problem into two one-dimensional problems.  I'd suggest using vertical and horizontal.  On the vertical side, we have gravity pulling down, and the normal force and one component of the tension from the rope pulling up.  This, by the way, is why the normal force is not equal to the weight - by pulling on the rope, you're partially lifting the sled off the snow.\n\nYou know that there is no acceleration vertically, so the net force vertically must be zero.  Knowing this, you can get the normal force as a function of the tension as the weight of the sled minus the vertical component of the tension.  Knowing the normal force and the coefficient of friction will give you the force necessary to pull the sled at constant velocity.  (If the velocity of the sled is constant, what can you say about the net force horizontally?)  Since the friction is only opposed by the horizontal component of the tension, you can find that.  Once you have that, you'll find that the only variable you have left is the tension in the rope, and you can solve for that.\n\nHope that helps.
949041
This problem relies on Newton's second law: F=ma\n\na. This question is not completely specified.  The mass of the rocket is then (F/a=m or 2.0E7/a).\n\nb. We first find the net force on the rocket and use the mass from the previous question. Now we compute the acceleration at liftoff (F/m=a or (25E6 - 2.0E7)/(2.0E7/a)).\n\nc. We derive a new formula by double integration on acceleration. \nintegrate(a dt) = at, integrate(at dt) = .5at^2. The sequence gives us acceleration -> speed -> displacement at time t. The rocket will be at velocity: +(25E6 - 2.0E7)/(2.0E7/a)(7*60) m/s and at displacement of: +.5((25E6 - 2.0E7)/(2.0E7/a))(7*60)^2 m.
35590
I concur with previous answerer. The acceleration is 1.5 m/s^2. Rearrange Newton's Second Law, to solve for the acceleration. Then, substitute what you know.
77860
hey, didnt i answer something very similar to this for you? you need to do F=ma so 43000000N/2100000kg then you get an acceleration, but then you have to subtract gravity from this because thats the acceleration produced by the rocket but there is still gravity pulling with 9.8 m/s^2 in the opposite direction. so just do (43000000/2100000) - 9.8, and that's your answer :)
939941
Yes you're on the right track, you know the momentum of the car along with the energy, both are conserved.  \n\nForce=Mass*Acceleration=Coefficient of Friction*Mass*Gravity\n\nYou can then solve for Acceleration=Coefficient of Friction*Gravity\nVelocity=Acceleration*t\nThen its simply a matter of filling t into this equation:\nDistance=1/2*Acceleration*t^2+Velocity*t