how to convert rev/min to rad/s. please help?
828009
One revolution = 2pi radians.  One minute = 60 seconds.  So 1 rev/min = 2pi/60 rad/s.  Multiply 11.2 rev/min by 2pi/60 to get 1.17286 rad/s.
475880
baud rate = bit rate/N\nbaud rate = 3000/4\nbaud rate = 750baud/s
1185159
I think it is \nf(s)=√(pi/s)\n\nBy the way how do you do the sign for the sqrt?\nI still cant figure it out.\nI can do it in MS Word but when I copy into the the Firefox I get anything, but what I desire.\n\nIt is getting late, very late\n\nI'm running 2000. \nThank you Scott\nand \nnwohiosteve  thank you for the hint. My memory is refreshed.\npress ALT, ASCII  number , reliese ALT works
378935
adiabatic lapse rate?
812026
a) There are 2pi radians in a revolutuion\nThe angular speed in radians pers sec is simply 2pi xRPS\nwhere RPS is the number of revolutions per sec (1RPS = 60RPM) \n\n2400x2pi /60 sec = 251.33 radians per sec\n\nb) The circumference = pi Dia =  23.562 in = 23.562/12 =1.9635ft\n The speed of the saw teeth is 2400/60 = 40 RPS (x 19635 ft) =78.54 ft/sec\n\nJust a note about angular speed:\nThe angular speed is the answer to the question: how many times does it rotate in a given amount of time? This can be expressed in radians per sec, RPS or RPM etc.\nThe speed of the teeth (or of some other point on the blade) is the answer to the question: What distance do the teeth travel in a given amount of time, for a given RPS, RPM etc.?\nIf you take a disk, or circular blade, and draw a radial line on it and then give the disk a certain RPM then the points on your line will have the same angular speed, since different points on the radial line will all travel 360 degrees(or 2 pi radians) every revolution. But the points will be travelling at different speeds. Points at the edge of the disk will have the greatest speed since they travel a greater distance (i.e. a greater circumference) per revolution than points which are closer to the center of the disk. So the teeth of the saw will have a greater speed than points on the blade which are closer to the center, but they all have the same angular speed.
148738
Homework, huh?  Badly copied from the source.  Make it an answerable question, clean up the gibberish and try again.  For instance, 90 degrees phase lag at what frequency?  Sorry to be curmudgeonly, but this shouldn't be just about getting answers without understanding.  Anyway, here's a hint.  At 90 degrees, Xr=Xc, and RC=1/(2*pi*f).  For phase lag the R is in series and the C in shunt.
952712
Since the angular acceleration is constant, you can take the average of 48 and 120 = 83 rads/sec.\n\nThus 7*83=581 revs.
892957
Try this converter. Its simple and free.\n\nhttp://www.snapfiles.com/get/svideoconverter.html
819170
7 pi radians/second = 3.5 revolutions per second.\nCircumference =  20 ft * pi.\n\nLinear velocity = 3.5 * 20pi  = 70pi ft/sec.
302941
translated in assembler this addition and store statement would be written as follows:\nMOVE J, R0 (1 Instr. fetch + 1 Operand fetch)\nADD  K, R0 (1 Instr. fetch + 1 Operand fetch + 1 Addition)\nMOVE R0, I (1 Instr. fetch + 1 Operand store)\n\nwith a speed of 200MHz 1 cycle is 5ns (nano seconds)\nlets assume an Instr. fetch is 1 cycle, an Operand fetch/store is 3 cycles and the Addtion is 1 cycle then we get 3 + 9 + 1 = 13 cycles, 13 * 5 ns = 65ns approx.
140474
Use the formula Ac = v^2/r = r w^2, where w = angular speed.\n\nSince you already have the angular speed, you can use the second form of the equation above.\n\nAc= r w^2 =(1.0 m) (2.0 rad/s)^2
756404
a = rω² → ω = √(a/r) = √(9.8/1100) = .09438 rad/sec\n\n*86400sec/day*1rev/2π rad = 1297.928 rev/day
275054
1 radian= 57.2957795 degrees\nGo to refdesk.com and look at references for calculators
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
929521
V=(4/3)pir^3\ndV/dt=4pir^2*dr/dt\n10=4pir^2*dr/dt\nA=4pir^2\ndA/dt=8pir*dr/dt\n=4pir^2*dr/dt*/r\n=10/r\nwhen r=8 dA/dt\n=10/8cm^2/sec
913766
The angular acceleration is 10/6 = 1.66667 rad/sec^2\n\na) I = Ta/(alpha)a = 36/1.6667 = 21.6 kg-m^2\n\nSince it loses 10 r/s in 60 sec, the deceleration is 10/60 = -1/6 r/s^2\nb) Td = I*(alpha)d = 21.6*(-1/6) = -3.6 N-m\n\nTheta (acc) = .5(alpha)a t^2 = .5*(10/6)*10^2 = 83.333 rad\nTheta (decel) = .5(alpha)d t^2 = .5*(1/6)*60^2 = 300 rad\n\nc) Theta total = 383.33 rad = 61 rev
522220
try it :\n\nhttp://www.rdrop.com/~cary/html/endian_faq.html