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Elina [12.6K]
2 years ago
5

A horse on the merry-go-round moves according to the equations r = 8 ft, u = (0.6t) rad, and z = (1.5 sin u) ft, where t is in s

econds. Determine the cylindrical components of the velocity and acceleration of the horse when t = 4 s.
Engineering
1 answer:
Luba_88 [7]2 years ago
5 0

Answer:

velocity = 0.664 ft/s, acceleration= 0.365 ft/s²

Explanation:

z= 1.5 sin 0.6t

angular velocity= dz/dt= d(1.5sin0.6t)/dt

                       = 0.9cos0.6t

                    at t=4

                       =0.9cos(0.6×4)

                        = -0.664 ft/s

Acceleration= dv/dt

                      = d(0.9cos0.6t)/dt

                      = -0.54sin0.6t

             at t=4

                   acceleration= -0.54sin(0.6××4)

                                           = 0.365 ft/s²

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Answer:

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Explanation:

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and

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\frac{Q_{square}}{Q_{circle}} = \frac{AV^2}{AV^2}  

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A surveyor knows an elevation at Catch Basin to be elev=2156.77 ft. The surveyor takes a BS=2.67 ft on a rod at BM Catch Basin a
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Answer:

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elev = 2156.77 ft

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put here value

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