what is the effective duration of a carried motion in a meeting?
140359
That sounds like a "Robert's Rule" question.  Try http://www.robertsrules.com/.  They have a forum specifically for this type question.\n\nMy understanding would be that once a motion carries its permanent unless and until another motion to reverse or suspend was passed.
362084
45 mins plus extra time
103569
Displacement.  If you mean its change in position *divided* by the time it took, that would be its average velocity.
700467
It would state the effective date in the question.  \n\nIt would take effect on that date, unless there is a court challenge to the constitutionality of the question.\n\nSee this wikipedia site for details.
338704
Good things to include (besides good spelling of the words 'write', 'minutes', etc.):\n\nInvited attendees, actual Attendees.\n\nThe agenda\nOld business from the prior meeting\nNew business from the new meeting\n\nEach item that was discussed and who said what\n\nFinally summarize with the list of action items (what needs to get done as follow-up, by when and by whom).\n\nAnd if there will be another meeting, note that.\n\nIf you use Microsoft Office, you can say 'New' and choose from templates... type 'Meeting Minutes' and it will give you a sample.  I've given a link below.  That should help immensely and will even make it look professional.
518470
six and a half minutes?
753766
First we determine the total moment of inertia of the system (with everyone on the rim).\n\nPeople: I = m r^2 = (65*150) (96^2) = 8.986 * 10^7\nSpace Station: I = 5.08 * 10^8 (it says 5.08 108 in your question, so I'm gonna assume that this is what you meant)\nSum = 5.679 * 10^8 kg*m^2\n\nIf the apparent acceleration = 1g\nThen, using a = v^2 / r , we can get the linear velocity of the edge\n\n9.8 = v^2 / 96\nv = 30.672 m/s\n\nTo get the angular velocity, use w = v / r \n(w is angular vel.)\n\nw = 30.672 / 96 = 0.3195 radians/sec\n\nSince angular momentum is conserved, we can equate the following:\nI * w (before meeting) = I * w (during meeting)\n\nObtain I (during meeting)\n\nPeople: I = m * r^2 = (65*50) * (96^2) = 2.995 * 10^7\nSpace Station: I = remains the same\nSum = 5.380 * 10^8 kg*m^2\n\nConservation of angular momentum:\nBefore = During\n(5.679*10^8) * (0.3195) = (5.380*10^8 ) * w\nw = 0.337 radians/sec\n\nFind the linear velocity:\nv = w * r = 0.337 * 96 = 32.377 m/s\n\nAcceleration to maintain circular motion:\na = v^2 / r\na = 32.377^2 / 96 = 10.92 m/s^2\n\n(that's 1.11g, if you're wondering)\n\nTada!
147775
2 quick: 5 min\n2 long: 2 Days
8725
a) eleven minutes.\nb) eight minutes.
1175600
They will meet when:\n\n4mph x Time = 3miles - 5mph x Time\n\nSo the answer is 0.33333 hours \n\nOr just 20 minutes.\n\n(not 27)
938340
Minutes for Extraordinary Meetings are kept by the corporate secretary.  They are not available to the general public and are only available to those who are shareholders.  If you are a shareholder, check with the corporate secretary.
866278
time taken divided by number of words (usually four letters plus a space = word) so time divided by 5\n\ntry this link for a full explanation\n\nhttp://www.talenttech.com.au/pdfs%5Cwhitepapers%5CComputationTypingSpeed.pdf
1065911
If i remember right it is divided into 3 segmients and they are roughtly about 45 minutes, 35 minutes, and 45 minutes. It  sometimes varies.
1151062
What is the standard way to describe motion ?
63833
50 feet or about 2-3 seconds
628570
velcoity is change in displacement/time
295237
20 minutes.\n\nAt minute 0, persons C (13) and D (17) begin crossing.  At minute 13, person C is finished and person B (7) starts.  At minute 17, D is finished and A (3) can start.  At minute 20, both A and B finish.