Hard Physics Problem?
831615
R being half way on the circle then:\n\nb)\nave velocity on x axis is v*Integral(sin(a)) (delta a) /(45deg) from 0deg to 45deg, on y the integral is v*Integral(cos(a)) (delta a) /(45deg) from 0deg to 45deg, the x integral evaluates to -0.292, divided by pi/4 is -0.373, the y integral is 0.7071, divided by pi/4 is 0.9\n\nv=w*r, which is 1.25m*4*pi/s=15.7m/s\n\nso v_ave PQ = 0.9*15.7m/s (j) - 0.373*15.7m/s (i)=\n14.13 (j) - 5.86 (i),, the magnitude is correct, i've checked\n\na) over PR this is very simple since integrals both evaluate to 1, so you have +/- 2/pi in both cases, sign is +for y - for x\nthe ave velocity is 10m/s (j) -10m/s (i)\n\nc) at P, it is -15.7m/s (i) +0 (j)\n\n\nNote: the magnitude of ave velocity can be any number <= 15.7m/s. consider the case for full revolution. the ave velocity vector is zero vector\n\nPS ok im not modifying this anymore
971746
This problem involves kinetic energy, force of kinetic friction, work, and conservation of energy.\n\nGiven: weight of girl + sled=645N\n           mass of girl +sled=645/g\n                                        =6.58N\n           speed of girl +sled at bottom=7m/s\n           coefficient of kinetic friction=0.050\n\nKinetic Energy, KE at the bottom of the hill=1/2mv^2\nKE=1/2*(6.58)7^2\n     =161.2 Joules\n\nForce of kinetic friction, f=uR, where u=coefficient of kinetic friction, and R is the force normal to the surface equal to weight of girl+sled.\n\nf=0.050*645\n =32.25N\n\nWork=f*s, where f is the force of kinetic friction and s is the distance the sled travels. Thus\n\nWork=32.25s\n\nBased on principle of conservation of energy:\n\nKE=Work\n161.2=32.25s\ns=161.2/32.25\n  =5.0m
179131
Try this:\nhttp://kr.cs.ait.ac.th/~radok/physics/f3.htm
782037
I solved the problem by stating that P would travel distance d in time t\nd=20*t+.5*.3*t*t\n\nand Q would travel d+20 in time t\nt is the time to collision\n\nd+20=30*t\nd=30*t-20\nSubtract the equations and solve for t:\n\n0=.15*t*t-10t+20\nt=2.064 seconds\n(the second root is superfluous)\nd=41.9m\n\nWhich proves collision\n\nj
943651
I am pretty sure that if you looked in your textbook, it would give examples of this kind of thing, and from that you should be able to figure it out on your own. You're a smart kid, you are taking physics instead of some easy A class, now act like it, stop asking others to do your work, and try to solve it on your own.\n\n\nLong Live Jambi
610467
Go to your local physics professor. If you are correct, (s)he will no doubt help you to get it published. If you are incorrect, (s)he should be willing to show you where your math is off.
757274
READ THIS CAREFULLY\nAs the block just slides on the table the net force just equals the limiting friction acting on the body\n\nhence,\nF=coeff. of friction * Nor,mal rxn\n\nnormal rxn= weight of body on a plane\n\nhence F= 20*9.8*0.4\n              =78.4N\n\nACC \nit is just begg to slide and the given data is just unsufficient 2 find the acc. of the box
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.
760426
Figure out the centripetal force.  After that, this becomes a sliding problem turned on its side.
206468
In elastic collision, energy is not lost.\n\nmv2 + MV2 = mu2 + MU2\n(1055 + 690) * (2.15)2 = 1055u2 + 690U2\n\nalso mv+MV = mu + MU\n(1055 + 690) * 2.15 = 1055u + 690U\n\nYou have 2 equations with 2 variables, solve them.
898375
I know the answer is in my basic physics book.  I would have to look it up.  You should too.
866141
You have to calculate p=massXspeed before the "collision".  The quantity remains constant, so if the mass increases, the speed must decreases by the same ratio for the p to stay the same.  (First, you have to convert the 8N weight to a mass using the 9,8N/kg gravity constant)
757859
haha, k  Draw a picture and draw a free-body diagram.\n\nI don't feel like busting out a paper/pencil  so I'll try it in my mind\n\nF(sub g) cos Theta  =  Mg cos(20 degrees)\n\nbecause the Normal force is equal to the y-component of gravity.\n\nTwo similar triangles, the incline and one which has the Force of gravity straight down, and the y compentn equal opposite of the normal force, and X component equal opposite to F\n\n\nlots of physics i'm skipping in my explanation tho
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
703117
here u go!\n a):\nuse the principal of F=ma to find the frictional force.  \nIn order to do this, you need to determine acceleration, so use the kinematic formula: v^2=vo^2+2a(x-xo).  \nwith this equation, you should find a to be equal to -1.5m/s^2.  \nPlug this answer into the F=ma equation, with m equal to .090kg.  Since Friction is the only force in the horizontal direction, F is equal to the frictional force.\nSolving for F=(.090)(-1.5)\nF=-.135N, so its magnitude is .135N\n\nb.):\nOne physics principal states that a frictional force is equal to mew (the coefficient of kinetic friction) times the normal force.\nSince you now know the frictional froce, just solve that equation.\nThe normal force in this problem is equal to the opposite of the gravitational force, which is (9.8)(.090)=.882\nSo, when you have the equation \n(.882)(coefficient of friction)=.135\njust divide .135 by .882 and you will get the coffecient of friction to be .153!!!!!!! yay!!!
1144696
gravity and air resistance.
355325
Newtons, or sometimes kilograms