Problem#1 The flip flap railway had a 25 feet (7.62 m) tall loop and had zero normal force at thetop, a) calculate the velocity at the top of the ride. b) using the work energy equation, calculatethe velocity at the bottom of the ride. Solve for the acceleration and divide by 9.8 m/s? and thenadd 1g because on the earth's surface the acceleration due to gravity is 1 g.

Name: Problem#1 The flip flap railway had a 25 feet (7.62 m) tall loop and
Uploaded: 2024-03-20T06:57:25.000Z
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Description: Problem#1 The flip flap railway had a 25 feet (7.62 m) tall loop and had zero normal force at thetop, a) calculate the velocity at the top of the ride. b) using the work energy equation, calculatethe velocity at the bottom of the ride. Solve for the

Given, The height of the railway track (diameter of the loop): d=25 ft =7.62 m g = 9.8 m/s2 We need…

Answered: Problem#1 The flip flap railway had a…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Problem#1 The flip flap railway had a 25 feet (7.62 m) tall loop and had zero normal force at the
top, a) calculate the velocity at the top of the ride. b) using the work energy equation, calculate
the velocity at the bottom of the ride. Solve for the acceleration and divide by 9.8 m/s? and then
add 1g because on the earth's surface the acceleration due to gravity is 1 g.

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what about the last part? how many g's does the coaster experience? I calculate 6, but every story I've read about the Flip-Flap says 12 g's.

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