What is the crate's acceleration?
920572
coefficient of kinetic friction = Friction Force / Weight\n\nFriction Force = (coefficient of kinetic friction)x(55x9.81)\n                       = 188.84 N\n\nSo Net Force = 220 -188.84 = 31.16 N\n\nF=mxa,  a=F/m=31.16/55=0.5665 m/s2
863271
mass doesn't matter here, just the acceleration\nv=u+at\na=(v-u)/t\na=(9-0)/0.15=60 m/s^2
822312
wish I could solve em for you but ill try and help still.\n\n1) The acceleration should be the derivative of the equation for the car's velocity.  The original distance/time equation would be the anti-derivative of the velocity.  \n\n2 The equation to find out how long it will take to get there is h+ax-16x^2 in feet.  (a is the original speed and x is number of seconds and h can be the height of the building.)\n\nHope this helps
804445
First you want to know the acceleration due to gravity of the planet Glob. At the surface the gravitational acceleration (g) can be determined by equating the weight of an object with the gravitational force law\n\nm*g = G*M*m/R^2\n\nwhere R is the radius of Glob, M is Glob's mass, G is the universal gravitational constant, and m is the mass of the object. Solving for g yields\n\ng = G*M/R^2\n\nNow, throwing an object into the air, and neglecting air resistance, we can consider conservation of energy. The kinetic energy of the object when it is released by the astronaut is completely converted into potential energy when the object reaches the top of its travel.\n\n0.5*m*V^2 = m*g*H\n\nwhere V is the initial speed of the object and H is the height to which it traveled, and g is the gravitational acceleration solved for above. Solving for V we find\n\nV = sqrt(2*g*H).\n\nThe satellite traveling in a circular orbit of radius D above the planet has its centrifugal force balanced by the gravitational force of the planet.\n\nFc = m*v^2/D = G*M*m/D^2 = Fg\n\nSolving for the speed of the satellite (v) needed in order to maintain orbit at radius D yields\n\nv = sqrt(G*M/D).
921575
9.8 m/s^2 is the acceleration due to earths gravity. calculate the volume of the paint bucket. multiply it by 7.65 to get the mass.
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)
865118
Force acting along the inclined plane[30.1]\n=mgsin30.1\nForce acting normal to the inclined plane\n=mg cos30.1\nForce of kinetic friction opposing the motion\n=.242*mgcos30.1\nWhat is the net acceleration?\n=gsin30.1-.242*gcos30.1\n=g[.5015-.242*.8651]\n=.2922g\nA]If we let the crate roll, it will accelerate due to gravity with out any external force being applied. To make it move in uniform motion we have to apply force in opposite direction\n=.2922*9.81*30.3\n=86.85 newtons\nB]work done by the force\n=86.85*4.75\n=412.55 newton-meters\n=412.55 joules\nC]Force of friction*4.75\n=30.3*9.81*cos30.1*.242*4.75\n=295.60 joules\nD]zero, since it acts perpendicular to direction of motion\nE]Work done by gravity\n=mgsin30.1*4.75\n=30.3*9.81*.5015*4.75\n=708.06 joules\nF]as in B]
813872
The applied torque is Fapp*R=70 Nm.\nThe tension on the rope supporting the 30 kg box is m*(g+0.8) = 318.2 N. \nThe torque producing this tension is 318.2*r = 63.64 Nm.\nThe remaining torque, 6.36 Nm, accelerates the wheel at 0.8/r = 4 rad/s^2.\nThe inertia of the wheel is torque/ang. accel. = 6.36/4 = 1.59 kg-m^2.
702361
ok biker bike, here we go.\n\nacc.= g = 9.8 m/s2, (any object moving in air will have this acc. or retardation)\n\nb) we apply formula\ns=ut + 1/2at^2\nu=20\nt=4\ns=20*4 + 1/2 *9.8 *4*4\n  =158.4 m\n\nc) v=u + at\n  therefore, t= v-u/a\n                   =50-20/9.8\n                   =3 seconds (approx)
288909
well, acceleration is usually refernced as bing positive, therefore deacceleration, being the opposite of acceleration, should be negative.  but, however, if you say that something is deaccelerating at a rate of x units/time, then it should be positive.\nhope that helps!!
652894
the amount of pressure exerted on an object
853525
Both the above answers are correct, provided the acceleration is horizontal. If it's not then it's a vectors and trig. question.
860005
I finished the problem.  The answer is s = 0.48517m.  It was hard, although now that I've worked it, it doesn't seem as difficult anymore.  Had to do some serious thinking though.\n\nThe collar A will tend to slide <down> the bar CD.  The vector CD (from C to D) is [-0.2, -0.3, 0.25] with magnitude 0.43875 and direction cosines [-0.45584, -0.68376, 0.56980].  The weight of collar A, directed downward, is 8g = 78.4N and is represented as [0, -78.4, 0].\n\nWe don't have to worry about the collar being supported by bar CD (the normal force).  The component of the collar directed down the bar is the inner (scalar, dot) product of the weight vector and the direction cosines.  This force is 78.4 x .68376 = 53.60706N.\n\nLet E be the point on the bar CD such that BE is perpendicular to CD.  Then BEA is a right triangle, and by the Pythagorean Theorem, the force along BE is sqrt(150^2 - 53.60706^2) = 140.09384N.  (I'm using numbers stored in the calculator and showing five decimal digits.  All this can be rounded off at the end.)\n\nSince the force triangle BEA is similar to the distance triangle BEA, we just have to get one of the legs to find our distance s = AB.  We will get the distance BE.\n\nTriangle BEA is in the plane defined by the points B, C, and D.  Vector CB = [-0.4, 0.2, 0.15].  The vector cross product CB x CD (see above) is normal to the plane defined by points B, C, and D.  CB x CD = [0.095, 0.07, 0.16].\n\nNow, the cross product CD x (CB x CD) lies in the plane defined by B, C, and D; is perpendicular to CD; and is parallel to the line segment BE.\n\nCD x (CB x CD) = [-0.0655, 0.05575, 0.0145] with magnitude 0.08723 and direction cosines [-0.75091, 0.63914, 0.16623].\n\nNow we can get parametric equations of the lines CEAD and BE, and find the point of intersection E.  Starting with point C and using vector CD, the first line is\n\nx = 0.4 - 0.2t\ny = 0.3 - 0.3t\nz = 0 + 0.25t\n\nStarting from point B and using the direction cosines from vector BE, the second line is\n\nx = 0 + 0.75091u\ny = 0.5 - 0.63914u\nz = 0.15 - 0.16623u\n\nSetting like terms equal to each other, we have three equations in two unknowns.  The three equations are not linearly independent.  Solving the x and y equations for t and u, we get\n\nt = 0.29871 and u = 0.45313\n\n(and these results are confirmed in the z equations).  Plugging these results into the x, y, and z equations, we find that the intersection point E is (0.34026, 0.21039, 0.07468).\n\nWe know the coordinates of point B, so by the Pythagorean Theorem, the distance BE is 0.45313m.\n\nComparing the force and distance vectors in similar triangles BEA, we have\n\ns/150 = 0.45313 / 140.09384\n\nand the length s of the cable AB is 0.48517m.  (Answer)\n\nA difficult, but fun, problem.
840333
You need to consider conservation of angular momentum\n\nI0*w0 = If*wf\n\nI0 is the initial moment of inertia, If is the final moment of inertia, w0 is the initial angular speed, and wf is the final angular speed.\n\nI = Iss + Io\n\nwhere Iss is the (constant) moment of inertia of the student and the stool, and Io is the moment of inertia of the objects. This general relation applies to both the initial and final moments of inertia. Io can be computed from\n\nIo = 2*m*r^2\n\nwhere m is the mass of the object and r is the distance from the axis of rotation. The factor of 2 accounts for the two different objects. The first equation then becomes\n\n(Iss + 2*m*r0^2)*w0 = (Iss + 2*m*rf^2)*wf\n\nwhere r0 is the initial radius (1.0 m) and rf is the final radius (0.40 m). Solving for the final angular speed you have\n\nwf = (Iss + 2*m*r0^2)*w0/(Iss + 2*m*rf^2)\n\nUsing this expression it is easy to compute the final angular speed.
465426
When an object is thrown from a top of a tower in a horizontal direction, it is having a horizontal uniform motion and an accelerated downward motion due to the gravitational force. The path is a parabola.\n\nSimilarly, the charged particle is having a uniform motion and an accelerated motion due to the electrical force perpendicular to the uniform motion.  \n\nTherefore, the path is a parabola.
456850
Do your home work. \n\nStill I tempt to answer this question.\n\nW= weight of the box\nmu= coefficient of friction.\n\nWe assume there is no acceleration. So force to upward movement is Fup = Wcos(20)*mu + Wsin(20),\n\nsimilarly Fdown = Wcos(20)*mu - wsin(20),\n\nso\nWcos(20)*mu + wsin(20) = 4Wcos(20)*mu - 4wsin(20)\n5Wsin(20) = 3Wcos(20)*mu\nmu = 5sin(20)/3cos(20)\nmu= 5/3 tan (20)\nmu = 0.606
372574
If the vehicle is moving at a constant speed, then the object is moving forward at that same speed before being tossed. After it is tossed, it continues to move forward at the same speed (as you and the vehicle) because there is nothing to slow it down.\n\nTry tossing the object when the vehicle is accelerating. The object, after being tossed, does not enjoy the acceleration that you and the vehicle do, but instead continues to move forward at the speed it had when tossed, and will land "behind" you. This is the principle behind a famous result of Einstein. If the vehicle has no windows, you can't tell whether the vehicle is accelerating, or standing still with "gravity" pulling tossed objects toward the back of the vehicle. Even if you try to use a beam of light to determine acceleration or gravity, you will get the same bending of the light beam.