Suppose that the elevator is broken in the building in which you work and you have to climb 9 stories?
111317
The external work is equal to the potential energy you gain.\n\nPE = mgh = (58kg)(9.8 m/s/s)(33 m)\n= 18757 J\n\nYour body only works at 25% effeciency though, so\n\n.25 = Output Work/Input Work = 18757 J/Input Work\n\nInput Work = 18757J /.25 = 75028 J\n\nWe need to convert to kcal.  Joule found that \n\n1 cal = 4.186 J,\n\nTherefore, Input Work = 75028 J / 4.186J/cal = 17924 cal\n\nRounding to 2 significant digits, and converting cal to kcal, we get\n\nInput Work = 18 kcal\n\nHence, climbing 9 stories doesn't even work off half an Oreo (An Oreo has about 50 food Calories, or 50 kcal)\n\n10 kcal/ minute is a reasonable rate of calorie consumption above basic metabolism for a typical person while exercising.  So a consumption of 18 kcal while climbing 9 stories doesn't seem unreasonable.
922568
if you fall down the stairs again maybe it will come to you.
693885
Your problem cannot be solved if there's no other given. Lots of factors should be considered in fainding the probability to be stacked in an elevator such as the probability that the elevator will malfunction, probability that the elvator will take you to the second floor, third floor, fourth floor and so on and so forth.
344505
Not if you work in a medical building and some drugs came up missing...otherwise I don't see why not.
431808
What you just said would be good enough reason. If there are unsafe work condition, then that gives the potential for people to get hurt. That is a lawsuit waiting to happen. How bad are the work conditions? If I were I would take it up with the Better Business Bureau if it is serious enough..but if you have already quit, I don't understand why you need another reason?
599097
depends on the magnitude of the earthquake, how tall the building, the architecture.. many things\n\nthe safest place in a building is under things that wont come down, like big supporting arcs \n\nI would recomend getting out of the building
100100
Maybe. If your failure to climb the stairs drastically adversely affects the operation of the company, then maybe you will be encouraged to "retire". I am putting that word in quotation marks because this suggests that you deserve to be given benefits, in case. I don't believe that this condition can be a valid reason to terminate your services without giving you remuneration in return...However, don't take my word for it. For official advice, you may go to the Department of Labor or the Labor Relations Commission or their equivalent agencies in your country. We have these agencies here in the Philippines...Ok?
711446
As the elevator moves up with an increase in the weight of the student from 849N to 942N, there was a net force upward acting on the student. This net force equals 942-849=93N. If the scale read 849 originally when the elevator was at rest then the mass of the student must have been 849/9.8=86.6kg. Now solve for the acceleration using the formula F=ma, where F is the net force, m the mass of the student, and a the acceleration.\n\nF=ma\n93=86.6a\na=93/86.6\n  =1.07m/s^2\n\nTo bring the reading back to 849, will need a net downward force of 942-849=93N. m remains the same. With the same net force as in the above, but in the downward direction, the acceleration must be the same in magnitude, i.e, 1.07m/s^2. But this time it is deceleration because the elevator would be slowing down, the net force being in the downward direction.\n\nThe elevator decelerates as it approaches the 74th floor because of the lower reading of 774N. The net force is 849-774=75N downward.\n\nF=ma\n75=86.6a\n a=75/86.6\n   =0.86m/s^2.
1068049
NO.\nSHE IS NOW AT DOWNWARDS JOURNEY.\nSHE CANT GO UP.
271774
in a lift/elevator full of people, especially your boss and other workmates.
61458
Your answer is correct, and may be fine for your professor, but for readers who are not already familiar with the problem, both the question and the solution are confusingly worded.  I suggest that in the future, you copy the original question here exactly.\n\n\nI would precede your solution with the following:\n\n\nYou can find the floor in a maximum of 14 attempts.\n\nDrop a bulb from the floors in this order:\n14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100\n\nIf the bulb breaks, stop and test starting from just above the last floor that it didn't break on, and go up one floor at a time.  If the bulb breaks on the first attempt, resume the test from the first floor.\n\nExamples:\n14 (breaks), 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13\n\n14, 27, 39 (breaks), 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38\n\n14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99 (breaks), 96, 97, 98\n\n14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100 (breaks)
385467
Hummmm, you could start at one and let us know how that goes.  Then you could go to second and again let us know and finally when we don't hear from you we will know what floor was not safe for you :)
503065
4 cubicals but one time i got it in from 9
875545
Come on, you don't know the difference between an elevator and an escalator?\n\nhttp://science.howstuffworks.com/escalator.htm
434539
You have to go to platform 9 and 3/4
942786
The tension is the weight + the force Fb required to accelerate the bundle along with the rest of the elevator. \nNeed to calculate the elevator's acceleration a\ny = (1/2)*a*t^2 where y = 1m and t = 1.6s. Solve for a.\n\nNow calculate Fb\nFb = m*a. Solve for Fb.\n\nT = W + Fb
611132
unless you break a window