Is the slope of the line equal to the weight of the car plus the weight bars?
82803
Yes acceleration and force should have a linear correspondence in this experiment (noise excluded). Obviously Force = Mass * Acceleration.\n\nIt is difficult to give much more qualitatively without a better description though.
547508
No. \nThe system is static and the sum of the forces equal zero, but there definately are forces flowing through the bar.  Replace the bar with a piece of plastic and then what happens.  The system is no longer in equilbruim, becomes dynamic and the plastic gets torn to shreads.  The same would be true if say you used a piece of wire.  Just because something is not moving does not mean there are no forces flowing through it.
939535
Sorry, but you'll need to give a little more information. How are the pulleys and weights arranged? Just a little picture would help.\n\nTypically, pulleys, weights, tensions and accelerations are extensions of the block on an incline problem except the pulleys are providing the force vector direction change.\n\nIt might help to think of the problem as a sum of vectors and look for where their vector sum lies.
842997
No... though some my say M+Y=L... but that's not realistic. Plus this is under math, and it should be science. Ask your friends about it. You're not going to recieve much comfort on this site with a question like that.
701538
yup you can do this with momentum conservation laws.  I wont do it for you though.\n\nBut I will help you out.\n\nResolve in the direction of the slide after the crash.  Work out the angle each car was travelling relative to this line (a diagram will help). Then use momentum of the collected impact slide.  From that you can work out the velocities of the two separate vehicles using cos*angle*mass*velocity=momentum
15317
The “slope” of a line is the change in one variable with respect to the change in the other variable.  For example: rise over run.  \n\nThe change in Y with respect to X is indeed -2.  As X increases by one Y will decrease by two.\n\nHowever you may also be interested to know the change in X with respect to Y is: -1/2.  As Y increases by one X will decrease by ½.
696645
Nice trick, combining multiple questions into one. I'll answer one q, no. 3.\nThe 300N force along the slope doesn't contribute to the normal force so has no effect on friction. The normal force is m*g*sin(slope) = 100 * 9.81 * 5/20. Your picture\nhttp://i34.photobucket.com/albums/d130/MATTRAINES/q4.jpg?t=1164668264\ndoesn't give a value for coefficient of friction, so if we assume it's 1, then Ff = the normal force.
454204
b-although gradient is deals with curves(like quadratic graphs) and slopes with straight lines
787000
slope is equivalent to dy/dx = y'\n y = x -1\ndy/dx = dx/dx -0\n dy/dx = 1\n   y' = 1\nTherefore the slope of the given equation is 1.
861510
The way to solve this problem is to decompose the forces acting on the car, but not in the x and y direction. Instead, decompose the forces in the direction of the incline and the normal force.\n\nHence, we have:\n\nF(friction) = μN, moving up the incline, and μ is the coefficient of static friction\n\nThen we have gravity acting on the car, and the component that's acting down the incline is = mgsinθ, where θ is the incline of 15°.\n\nFor the car to be safe in making this turn, the forces along the incline must be equal:\n\n(a) F(friction) = μN = mgsinθ = (1450)(9.8)sin(15°) = 3677.82 N\n\nBut the forces acting perpendicular to the incline on the car must also be zero:\n\nN = mgcosθ\n\nNow substituting this into the previous equation, we get:\n\nμN = μ(mgcosθ) = mgsinθ, this implies that the coefficient of static friction is:\n\n(b) μ = sinθ/cosθ = tanθ = tan(15°) = 0.268
532726
CW= PW / (.5)(A)(Gr)(vcubed)\n\nCW=drag coefficient\nPW= aerodynamic power (kilowats)\n.5= constant\nA= total front area of the vehicle in meters squared.\nGr= gradient of the terrain.\nv3= velocity cubed.\n\nGr can be found with the equation:\nGR=(mass)(gravity)(velocity)(sin of the angle of the gradient)\n\nyour total answer should be  no more than 1
902316
Let y be required distance from the far end.\n\nThe weight is acting at the C.G i.e., at 2 m from the far end.\n\n2/3 of the weight w is acting upward at y and \n\n1/3 of the weight w is acting upward at 4m from the far end.\n\nTaking moments about the far end, \n\n2/3 W y + 1/3 W 4 = W x 2. \n\n Dividing both sides by 2W and multiplying by 3 on both sides\n\ny + 2 = 3 \n\ny = 1m.
729746
you solved it right, since the line BC never goes up/down(the y-coordinate never changed, graph it and you'll see) the slope is 0. and 0/7=0, so you're right.\nit would be impossible if you got 7/0, because you can't divide by 0.
984341
I think you actually need to use Tangent for this problem. \n\nThe wheelbase of the truck is 1.8m. Assuming it is perfectly balanced, the center of gravity would be directly over the center of that; 0.9m from one wheel. That's the base of the triangle.\n\nThe vertical side rises up to the center of gravity; which is unknown, or X.\n\nA 30 degree tilt causes the truck to roll over. This means that the angle of our triangle is 90 degrees minus 30, or 60deg.\n\nNow we have, Tan60=X/0.9\n1.73(0.9)=X\nX=1.56
298529
If you know the density of the material, you can measure it's volume.  D = M / V.  Therefore, the mass equals the density of the object times the volume.  The weight of the material is its mass times gravity.
757648
Friction force:\nF = m g c\n\nm = mass of car\ng = gravitational accel\nc = coeff. of friction\n\nCentripetal force:\nF = m v^2 / r\n\nm = mass of car\nv = velocity\nr = radius of curve\n\nSet the two forces equal to each other:\nm g c = m v^2 / r\n\nCancel m, plug in given values:\n9.8 m/s^2 * c = (12 m/s)^2 / 52 m\nc = 0.28257\n\nSo, c has to be at least 0.28257 to keep the car from slipping.
173295
It refers to the slope of a line and the y-intercept.