The question tells me that the velocity of a 960 kg (with riders) amusement park ride is 37.07 m/s [down] after free falling for a height of 70.1 m (this is one of those dropping rides). I now have to figure out the power of a braking system that stops the ride in 0.3 s. I also have to figure out the average force the braking system applies on the ride.Please show work and explain. Thanks in advance.
Would you have preferred that he let these companies fail and the recession become a full depression?
YES!!! It has failed in every case it has been tried, unless you consider Castro's little island jail to be a success! This is because power left in the hands of a few will always lead to Tyranny by them. Read Mark Levin's book - Liberty and Tyranny.
Yeah. Also, Bush and the TARP/Bailout mess was similar. I wish these Marxists/Socialists, like Bush and Obama, would just go away. RON PAUL 2012 No more Socialism from the RINOs and Demos.
maximum present day trains use disc brakes, utilized utilising air tension in a brake cylinder. finished service braking is usually approximately 3.5 to 4 bar. Diesel contraptions can stop very effectively, yet can nonetheless take over a mile (from one hundred mph contained relating to a one hundred seventy). besides the shown fact that, if the railhead is contaminated, this distance would be plenty greater suitable, as the two the wheels will lock up, or a device call WSP (Wheel Slip secure practices) will activate, this is comparable to anti-lock brakes on a vehicle. automatic sanders could additionally help in any such situation. Sorry, i'm no longer able to respond to approximately HSTs or loco hauled inventory, i'm no longer experienced in those.
Checking the velocity after falling 70.1 metre involves equating potential energy (Ep) with kinetic energy (Ek): mgh (mv^2) / 2 gh v^2 /2 v √2gh √(2 x 9.8 x 70.1) 37.07 m/s Power dissipated in stopping vehicle kinetic energy / time P (mv^2 / 2) / t (960 x 37.07^2 / 2) / 0.3 2,198,696 watts Calculating the average force involved the impulse equation: Ft ?p m(v-u) where: F average force; t time; ?p momentum change; m mass; v final velocity; u initial velocity Ft ?p m(v-u) F (m(v-u)) / t 960(0 – 37.07) / 0.3 -118,624 Newton It’s always good to check things. So, another way of calculating the average force would be to determine how far the vehicle travels as it is brought to a stop and then apply the work formula (W Fs). First determine the deceleration rate: a (v – u) / t (0 – 37.07) / 0.3 -123.6 m/s/s Now the distance travelled in stopping: s (v^2 – u^2) / 2a (0^2 – 37.07^2) / (2 x -123.6) 5.56 m Now apply work formula: F W / s Ek / s (mv^2 / 2) / s (960 x 37.07^2 / 2) / 5.56 118,634 Newton Good, the two force values are essentially the same !
