anonymous
  • anonymous
An elevator starts from rest with a constant upward acceleration andmoves 1m in the first 1.8 s. A passenger in the elevator is holding a 8.7 kg bundle at the end of a vertical cord. What is the tension in the cord as the ele- vator accelerates? The acceleration of gravity is 9.8 m/s2 . Answer in units of N
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Ok so we know Mass=M=8.7kg Distance=d=1 meter Time=t=1.8 seconds Gravity=g=9.8m/s^2 And the formula for Acceleration I get by the formula for Distance=D=1/2at^2 so moving things around I get Acceleration=A=2d/t^2 =(2*1m)/(1.8s)^2 = 2m/2.34s^2 = .62m/s^2 So to find then plug in to T=m(g+a) 8.7kg(9.8m/s^2+.62m/s^2)=8.7(10.42)=90.63 kg*m/s^2 Which is 90.63N ??? Correct?
anonymous
  • anonymous
right wrong???? any thoughtS?
phi
  • phi
looks good

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anonymous
  • anonymous
k. its probably wrong somehow but thanks....
anonymous
  • anonymous
ok it was right. someone needs to give ME a medal...
anonymous
  • anonymous
hvn't u studied about pseudo force.?
phi
  • phi
If you insist that you have made a mistake, perhaps you are referring to this line Acceleration=A=2d/t^2 =(2*1m)/(1.8s)^2 = 2m/2.34s^2 = .62m/s^2 where you have 1.8^2 = 2.34 Of course 1.8*1.8= 3.24. However, 2.34 is clearly a typo because you get 0.62 which is 2/3.24 (not to mention the obvious swap of digits 2 and 3)

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