Recent content by Mqqses
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Power for accelerating a ski lift
Thanks for your insight into this problem, I think I'll leave this alone for now.- Mqqses
- Post #7
- Forum: Introductory Physics Homework Help
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M
Power for accelerating a ski lift
So would I use an integral from 0 to 6.4s? ∫06.4 ∂PE/∂t + ∂KE/∂t where PE= mgh and KE= 1/2mv2?- Mqqses
- Post #5
- Forum: Introductory Physics Homework Help
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M
Power for accelerating a ski lift
I'm not sure how to approach post #2, but I think that would be some sort of ∂PE/∂t + ∂KE/∂t for total energy required to move the lift. The sin θ may be a wrong idea. Am I getting the gist that an integral might be used for this?- Mqqses
- Post #4
- Forum: Introductory Physics Homework Help
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M
Power for accelerating a ski lift
Homework Statement A ski-lift has a one-way length of 987 m and a vertical rise of 275 m. The chairs are spaced 21 m apart, and each chair can seat three people. The lift is operating at a steady speed of 14 km/hr. Neglecting friction and air drag and assuming the average mass of each loaded...- Mqqses
- Thread
- Lift Power
- Replies: 7
- Forum: Introductory Physics Homework Help
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M
Projectile motion submarine missle problem
R=sqrt((R*g*.001)/(sin(2*45)) is the formula for range. Edit: the .001 is to convert the gravitational constant from m/s^2 to km/s^2.- Mqqses
- Post #14
- Forum: Introductory Physics Homework Help
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Projectile motion submarine missle problem
The question he is referring to neither describes the Earth as "flat" nor "round" so it is hard to grasp these questions in the sense of the Physical world. I am also working on this homework problem, and I've gotten question b) right, but not a). I also have not a clue what question c) is...- Mqqses
- Post #11
- Forum: Introductory Physics Homework Help