Physics question below?
823386
Hi Alan the 1st!\nWe neglect the central aperture in CD, then inertia momentum is I=m*r^2/2 for any homogeneous disk, m=0.017kg, r=0.06m. Torq T=I*w’, w’ is angular acceleration, while w=w’*t, w=22 rad/s, t=0.88s, thus T=m*r^2*(w/t)/2, - plug in the proper numbers, be a good boy!
998322
Lets make an enormous assumption to fill in the gaps.\nYou have a thing on a string of length L and spin it round at a certain rate of rotation. Keeping the rate of roataion what would the tangential velocity of the thing be if the string was shortened to L/2?\nIn this case the tangential velocity (= the radius x the angular velocity) will also halve.
487670
None. It takes no power to maintain that speed (on level ground, no friction). The object has kinetic energy (0.5 * m * v^2). Power was required to put it in motion and power can be recovered by decelerating it. If you do the math, the energy is 3551.86 ft-lb, equivalent to 1 horsepower for 6.458 seconds. Answer 1 neglected the fact that weight must be converted to mass by dividing by g, and energy is proportional to v^2.
768228
a. V ice = pi(0.12^2-0.10^2)*3 = 0.041 m^3\nb. V ice = pi[(0.10 + x)^2 - 0.10^2]*3 = pi(0.10^2 + 0.20x + x^2 - 0.10^2)*3 = 3pi(0.20x + x^2)\ndV/dt = pi[0.60 + 6(0.02)]dx/dt = 0.000002pi\n= [0.60 + 6(0.02)]dx/dt = 0.000002\ndx/dt = 0.000002/[.60 + 6(.02)]
107382
v^2-u^2 = 2gh\nHere, v=1, g=-10m/s/s, h=0.6m\nHence, u^2 = 1+12=13\n=> u = 3.606 m/s
918781
Where is your freebody diagram? There are other forces acting on the block than just gravity...\n\nGood Luck
643953
depends on the direction of the force
912569
I think F is the answer. Taking acceleration due to gravity is constant at 9.81ms^-2, but this assumes that air resistance is neglected, and g is a constant.\n\nIf air resistance is NOT neglected, and acceleration due to gravity is not taken to be constant, H is the correct answer. The rest of the options are wrong.
751049
First find Vo:  Vo^2 = Rg/sin(28)  Where R = range & 8 = theta\n\nUsing your numbers, Vo = 41.523 m/s\n\nImpulse = change in momentum ----> Ft = mVo---> F = mVo/t\n\nSo, F = (.0431kg*41.523m/s)/.00438sec = 408.59 N\n\nVoila!
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
948228
D =M/V\n\nM=DV\n\nV=M/D
898375
I know the answer is in my basic physics book.  I would have to look it up.  You should too.
805324
Part a)\n\nVf^2 = Vi^2 + 2 * a * d\na = (Vf^2 - Vi^2) / (2 * d) = -4,489/162 = -26.7 m/s^2\nm = 220*10^3 / 9.81 = 22,426 kg\nF = m*a = -598,777 N\n\nW = F * d = 50,297,245 J\n\nPart b)\n\nF (from above) = 598,777 N\n\nI hope this helps!
826674
use conservation of angular momentum in these kind of problems\nNOT energy conservation\nif you dont know how to use the conservation equation, better read up on examples
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
484307
in air= 340m/s\nin water=1500m/s\nin solid=4500m/s
49314
Sorry, but with the time you spent typing these questions you could of looked them up in your physics book or googled them.  Sometimes getting others to do your leg work isn't always the best answer to a problem.  Good Luck in your new class!