physics problem .... I need help please ?
961939
The weights of the liquids inside 2 branches must be equal, therefore:\n\na) m_water + m_oil = m_mercury (the amount of mercury that raised in the other branch to keep the balance)\n\nm_water = D_water * V_water = D_water * h_water * Area\nm_oil = D_oil * V_oil = D_oil * h_oil * Area\nm_mercury = D_mercury * V_mercury = D_mercury * h_mercury * Area\n\nSubstitute this in the balance of weights above, we have\nD_water * h_water * Area + D_oil * h_oil * Area  = D_mercury * h_mercury * Area\n<-> h_mercury = (D_water * h_water + D_oil * h_oil) / D_mercury\n\nb) To bring mercury's surface two the same level again, the weight of the to-be-added water must be the same with the weight of both oil and water in the other branch.\n\nYou can do the same as above to find:\nh_water2 = (D_water * h_water + D_oil * h_oil) / D_water
1014316
would you like to supply me with some formulas? im out of grade 12 physics and i forgot most of the formulas. trust me from experience, just plug numbers into a formula and things will work\n\nF=mg = ma\nFk=μkN\na= vf^2 - vi^2 / 2d
40451
well figure out the net force with F=ma and just subtract. like just say 93 x .26 is the net force, the force thats acually being used for motion. so you subtract this from 36 to see what force is opposing this. so the total equation is:\n36 - (93 x .26)
858739
Use Bernouillis equation and you will have your answer. You can use the equivalent lengths method to find the friction.
981852
You can solve this problem by resolving the forces along the ramp for part a) and normal to the ramp for part b).\n\na) The net force along the ramp must be zero for the crate to move at constant velocity (ie no acceleration), so we have:\n\nFcos30 = mgsin30\nF=mgtan30\n\nb) the reaction of the ramp on the block will be normal to the ramp and equal to the sum of the imposed forces resolved normal to the ramp, so:\n\nR = Fsin30 + mgcos30
1120781
do your own homework\n\nits probably 0.26 kg M/s ^2 since its a velocity\n\nyou also probably need to convert it to rad/s whatever that is
680534
Use the equation from the link below.  You will find that \nv/c = 0.14365, or\nv = 0.14365 c\nor\nv = 43,100 km/s\n\nThe object is moving away from the observer at this rate.\n\nBy the way, you probably won't get credit unless you can work out the math steps I left out.
734736
first one is easy\n\nfirst find the normal force\n\nFf (friction) = Fn(normal force ) (.35)\n0(motionless ) = Fn .35\nFn = -.35\n\nfrom there its easy\n\nFn = mgsin(theta)\n\nsolve for theta\n\n-.35 = (200)(9.8) sin(theta)\n\nand solve it ur self\n\nsecond one, tell us wat u want us to find and we can tell u, but here are the things u can find, ill give u the formulas\n\nacceleration = 0 (constant speed)\ntheta = 15 degrees\nvelocity = constant\n\nthats pretty much it, from what u gave me
901641
Someone posted the exact same set of questions a few days back, either here or on the physics page - you ought to be able to find the methods in the responses. Have a look.\n\n\nBest of Luck - Mike
748368
first get the time from the horizontal motion  t = .51 / 1.2 s (uniform motion, no acceleration here)\n\nthen use h = 1/2gt^2 for the vertical motion (with initial velocity zero)
179131
Try this:\nhttp://kr.cs.ait.ac.th/~radok/physics/f3.htm
796334
Consider the forces and torque in this problem. Setting up so that the x-axis is positioned parallel to the incline\n\nSFx = m*a = m*g*sin(q) - Fs\n\nwhere m is the mass of the cylinder, a is the acceleration, g is the acceleration due to gravity, q is the angle of the incline and Fs is the statis friction force. For the torque\n\nST = I*a/r = r*Fs\n\nso that we have for the statice friction force\n\nFs = I*a/r^2.\n\nPlugging this into the first equation\n\nm*a = m*g*sin(q) - I*a/r^2\n\nand solving for the acceleration\n\na = m*g*sin(q)/(m + I/r^2)\n\nNow, finding the time to travel a length L can be done in several ways. The first, and perhaps simplest is\n\nL = 0.5*a*t^2\n\nso that\n\nt = sqrt(2*L/a)\n\nAlso, you can consider how many revolutions of the cylinder is takes to achieve the same distance. This mean dividing the distance L by the circumference of the cylinder. Multiplying the result by 2*pi gives the angular travel so that the time can be determined by\n\n2*pi*L/(2*pi*r) = 0.5*alpha*t^2\n\nwhere alpha is the angular acceleration and is equal to a/r. \n\nConservation of energy can be employed as well\n\nm*g*h = 0.5*m*v^2 + I*w^2\n\nwhere h (= L*sin(q)) is the height, and w (= v/r) is the angular speed. That is to say that the potential energy at the start is equal to the kinetic energy after traveling through the distance L along the incline. Solving for v yields\n\nv = sqrt(2*m*g*L*sin(q)/(m + I/r^2))\n\nwhich is equal to the acceleration times the time traveled. From this \n\nt = sqrt(2*m*g*L*sin(q)/(m + I/r^2))/a
797830
Keep in mind these equations\n\nx(t) = x-initial + (v-initial)t + (1/2)(a-gravity)t^2\nv(t) = v-initial + (a-gravity)t\n\n(v-final)^2 = (v-initial)^2 + 2g(x-final - x-initial)\n\na-gravity = -9.8 m/s = -32 ft/s (i think..)\n\nTherefore\n\nx(t) = 0 + (64)t - (16)t^2\nx(t) = -16t^2 + 64t\n\n(v-final)^2 = (v-initial)^2 + 2g(0)\n(v-final)^2 = (v-initial)^2\n\nsince it is going in the opposite direction,\nv-final = -(v-initial)
950937
You're in luck.  I teach physics and I know exactly how to answer this question (it's based on the work-energy principle).  It's pretty simple, really.  But it would be best if we discussed it by messenger.  Yahoo Messenger me, my ID is\nfortitudinousskeptic\n\n- Kevin
948754
we juss need to take vector sum of the two forces. assuming the wind is directly in horizontal direction....... sqrt(9.25sq+1.05sq)=9.3N
170255
It is much easier to just give you a link instead of explaining it:\n\nhttp://www.shef.ac.uk/physics/teaching/phy107/solvegray.html
783086
Gravity is pulling down on the object, air is resisting it (pushing up, if you will) and causing friction, and if there's any wind it's pushing sideways on the object. I doubt you'd have wind in your problem, though. Wind is complicated :)