Answer:
8.04 second
Explanation:
torque = 500 Nm
Length of blade, L = 2.4 m
mass of blade, m 40 kg
initial angular velocity, ωo = 0 rad/s
frequency, f = 2000 rpm = 2000 / 60 rps
final angular velocity, ωf = 2 x π x 2000 / 60 = 209.33 rad/s
Moment of inertia of the blade,
![I = \frac{1}{12}mL^{2}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B1%7D%7B12%7DmL%5E%7B2%7D)
![I = \frac{1}{12}\times 40\times 2.4^{2}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B1%7D%7B12%7D%5Ctimes%2040%5Ctimes%202.4%5E%7B2%7D)
I = 19.2 kg m^2
Torque = Moment of inertia x angular acceleration
500 = 19.2 x α
where, α be the angular acceleration
α = 26.04 rad/s^2
Use first equation of motion
ω = ωo + αt
where t is the time taken by the propeller to reach 2000 rpm.
209.33 = 0 + 26.04 x t
t = 8.04 second
Thus, the time taken by the propeller is 8.04 second.
Edges of tectonic plates where the shock waves resonate through the ground with larger amplitudes
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