A 75.0 kg person is on a Ferris Wheel that has a radius R = 16 m. The tangential velocity of the Ferris Wheel is 8.25 m/s. Calcu
late the period of rotation of the Ferris Wheel. Enter your answer below and include units and round to 3 significant figures. Show your work on paper, scan it, and submit it in the last problem.
1 answer:
Answer:
The period of rotation is
T=8.025s
Explanation:
The person is undergoing simple harmonic motion on the wheel
Given data
mass of the person =75kg
Radius of wheel r=16m
Velocity =8.25m/s
The oscillating period of simple harmonic motion is given as
T=(2*pi)/2=2*pi √r/g
Assuming that g=9.81m/s
Substituting our data into the expression we have
T=2*3.142 √ 16/9.81
T=6.284*1.277
T=8.025s
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Answer:
Xc= 17.267 Ω, Z= 415.5 Ω, I= 0.537 A
Explanation:
Em = 223 V
f= 300 Hz, R = 222 Ω, L = 147 mH, C = 23.1 μF
a)
Capacitive reactance = Xc=?
Xc=
Xc=1/2pi *399*23.1*10^-6
Xc= 17.267 Ω
b).
Z=
Xl= 2π * f * L
Xl= 2π * 399 * 147 *
Xl= 368.5 Ω
Z= =
Z= 415.5 Ω
c).
Current:
I= V / Z= Em / Z
I= 223/415.5
I= 0.537 A