Asked by Geee
Thank you for the respond!! I appreciate it. But it says on my paper
360 rev/min = 360 (2pi/60) rad/s = 12pi rad/s and so
D= 2r = 2(s/theta) = 2(40/12pi) ft = 20/3pi ft = 2.12 ft should be the answer :(
I have 2 more if that's alright.
A point on the rim of a turbine wheel of diameter 10 ft moves with a linear speed 45 ft/s. Find the rate at which wheel turn (angular speed) in rad/s and in rev/s.
360 rev/min = 360 (2pi/60) rad/s = 12pi rad/s and so
D= 2r = 2(s/theta) = 2(40/12pi) ft = 20/3pi ft = 2.12 ft should be the answer :(
I have 2 more if that's alright.
A point on the rim of a turbine wheel of diameter 10 ft moves with a linear speed 45 ft/s. Find the rate at which wheel turn (angular speed) in rad/s and in rev/s.
Answers
Answered by
Henry
Circumference = pi*D = 3.14 * 10 = 31.4 ft.
Va = 1rev/31.4ft. * 6.28rad/rev * 45ft./s = 9 rad/s.
Va = 9rad/s * 1rev/6.28rad =
Va = 1rev/31.4ft. * 6.28rad/rev * 45ft./s = 9 rad/s.
Va = 9rad/s * 1rev/6.28rad =
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