help with newtons laws please1?
931143
Basically the law you are talking about is the theory that an object in motion will stay in motion until another force acts upon it.\n\nMomentum is calculated by multiplying the mass by the velocity of the object.  There are many applications for these caclulations like if you wanted to find out what would happen in a head on collision between two objects.  You would be able to calculate in which direction the two objects would travel after the collision.\n\nAnyway, here is a website that I found for you that explains it completely.
539746
F = ma\n\nF - Force\nm - mass\na - acceleration
987931
you are making a mistake with the units\n\ndid you mean kg/s^2? if it is so the answer is 10300 kg/s^2\nif did you mean kgf/m it is 1050 kgf/m \nif not your units are wrong. \nremember that 1 N (newton) is 1 kgxm/s^2\nand 1 kgf = 9.8065 kgxm/s
874727
You need to sum the components. Let Fr, Fl, Fu, and Fd be to right, left, upward, and downward directed forces, respectively. Then\n\nFx = Fr - Fl\n\nand\n\nFy = Fu - Fd.\n\nThe right and upward directions are taken as the positive direction so that the minus sign indicates that the forces (Fl and Fd) are directed to the left and downward.\n\nThe magnitude of the resulting force is then computed as\n\nF = sqrt(Fx^2 + Fy^2)\n\nand the direction can be found from\n\ncos(q) = Fx/F\n\nUse some reasoning to be sure that the direction you compute is consistent with the orientation of the force components.
928343
Torque.\n\nThe weight of each particle is mg. The torque caused by particle 1 is mg * 20, counterclockwise, and the torque from particle 2 is mg * 80, clockwise. Therefore the net torque is mg * 60, clockwise.\n\nSo the force on the left particle is (mg * 60) / 20 = 3 mg, and the force on the right particle is (mg * 60) / 80 = 3/4 mg. Both of these are clockwise, so the left particle goes up and the right particle goes down. To get acceleration, just cancel m.
851079
You do not have enough numerical values to solve, so I can only help you with variables.  Drawing a free-body diagram, you can see that these are the following forces:\n\nApplied Force (F), parallel to incline moving down\nStatic Friction (f), moving opposite to F\nWeight (W), moving down\nNormal (N), perpendicular to incline.\n\nLet's define x-axis parallel to incline and y-axis perpendicular to incline.  We only have to break up W into components parallel to these two axes.  Using Newton's 2nd law, we have:\n\ny-axis -----------> N - mgcos(theta) = 0 [block does not move in this direction]\n\nThus, we have N = mgcos(theta)\n\nx-axis ---------> F + mgsin(theta) - f = 0 [since the friction is static, the block is just about to move]\n\nThus, we have F+mgsin(theta)=f, where f = uN = umgcos(theta)\n\nUsing these equations, you can solve for u.  Thus:\n\nu = [2000 + mgsin(theta)]/[mgcos(theta)]\n\n---------------\n\nHope this helps
935341
1 sum the forces and set equal to zero. 2 forces, gravity and electric.\n\n2. conservation of energy. initially all kinetic, finally all potential. KE1=U2. U2 is electric potential. solve.\n\ngood luck
210438
the newton's three laws of motion are as follows: law of inertia ( an object at rest will stay at rest unless acted by an outside force, a moving will continue on ots motion unless acted by an outside force), law of acceleration ( thw acceleration of a body is directly propotional the force and inversely proportional to its mass: a=f/m) and the law of interaction ( in every action there is always an equal and opossite rreaction)
733729
http://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation
217280
According to Newton's second law,\nF = m*a \nF = Force\nm = mass\na = acceleration\n\nnow, also,\n\na = (v-u)/t \nv = final velocity\nu = initial velocity\nt = time\n\nput this expression into first equation, you get\n\nF = m (v - u)/t\n= (mv - mu)/t\n\nalso, momentum = mass * velocity\n\nso,\nF = (final momentum - initial momentum)/t\n\nwe can also say that\n\nF = Rate of change of momentum
903133
From what I learned in physics class, the force you are speaking of is measured in Newtons.
21991
http://www.newton.dep.anl.gov/askasci/math99/math99009.htm
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 :)
1045430
Momentum - P = mv (P=momentum, m=mass, v = velocity)\nThree Laws of Newton - F = ma (F=Force; a=acceleration)\nEquations of motion - 1) v = u + at\n                                    2) S = (1/2)ut + at^2\n                                    3) v^2 = u^2 + 2aS\n(v=final velocity; u = initial velocity; S=distance)\n\nIf this can help you please mail me back !\nBEST OF LUCK !!!!
966154
The magnitude of a force is measured in newtons(N)
612059
Consider two objects having masses 1kg and 5 kg are in a horizontal plane.\n\nWe apply a force of 10 newton to the 1 kg mass and 50 newton to the 5kg mass. \n\nThat is, we apply less force for less mass and greater force for greater mass.\n\nNow we find the acceleration of the two bodies.\n\nThe1 kg mass will have an acceleration of 10/1 = 10 m/ s^2. That is its speed will be going on increasing by 10meter / second in every second. \n\nSimilarly the 5 kg mass will have an acceleration of 50/5 = 10 m/s^2. That is its speed will be going on increasing by 10meter / second in every second. \n\nBoth have the same acceleration and both have the same speed at any time after start.\n\nNow returning to gravitational pull, let us place the 1kg mass in one pan of the balance and the 5kg mass in another pan. \n\nThe pan having 5kg mass will come down and the pan having 1kg mass will go up.\n\nThis is because the earth is pulling the 5 kg mass with a force of 50 newton and 1kg mass with a force of 10 newton.\n\nIf  we drop the two objects from the same place, as we have seen already the 1 kg mass will fall down with an acceleration of 10 m/ s^2 and the 5kg mass will also fall down with 10 m/ s^2. But the forces acting on them are different.\n\nThe Earth’s pull is such that it gives less force for less mass and greater force for greater mass.\n\nTo be more specific it gives a force of 10 newton for every kilogram of mass and hence the acceleration is for any mass is the same as approximately 10m/s^2.
121693
cmon...dont u know basic physics...just calculate the resultant of these 2 forces\nanswer= [(132)^2 + (181)^2]^1/2\nif this makes sense!!