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Answer:
a) initial position = 10m
b) average velocity = 30m/s
Explanation:
The general equation of a motion with constant velocity is given by:
(1)
So: initial position
v: average velocity
t: time
You have the following equation:
(2)
If you compare the equations (1) and (2) you obtain that:
![S_o=10m\\\\v=30\frac{m}{s}](https://tex.z-dn.net/?f=S_o%3D10m%5C%5C%5C%5Cv%3D30%5Cfrac%7Bm%7D%7Bs%7D)
hence:
a) initial position = 10m
b) average velocity = 30m/s
3 hours and 30 minutes.Tell Juan to get a life.
Answer:
Work done, W = 255.21 J
Explanation:
It is given that,
Rotational inertia of the wheel, ![I=16\ kg-m^2](https://tex.z-dn.net/?f=I%3D16%5C%20kg-m%5E2)
Angular displacement, ![\theta=2\ rev=12.56\ rad](https://tex.z-dn.net/?f=%5Ctheta%3D2%5C%20rev%3D12.56%5C%20rad)
Initial angular velocity, ![\omega_i=7\ rad/s](https://tex.z-dn.net/?f=%5Comega_i%3D7%5C%20rad%2Fs)
Final angular velocity, ![\omega_f=9\ rad/s](https://tex.z-dn.net/?f=%5Comega_f%3D9%5C%20rad%2Fs)
Firstly finding the angular acceleration of the wheel using the equation of rotational kinematics as :
![\alpha =\dfrac{\omega_f^2-\omega_i^2}{2\theta}](https://tex.z-dn.net/?f=%5Calpha%20%3D%5Cdfrac%7B%5Comega_f%5E2-%5Comega_i%5E2%7D%7B2%5Ctheta%7D)
![\alpha =\dfrac{9^2-7^2}{2\times 12.56}](https://tex.z-dn.net/?f=%5Calpha%20%3D%5Cdfrac%7B9%5E2-7%5E2%7D%7B2%5Ctimes%2012.56%7D)
![\alpha =1.27\ rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D1.27%5C%20rad%2Fs%5E2)
Work done by the torque is given by :
![W=\tau\times \theta](https://tex.z-dn.net/?f=W%3D%5Ctau%5Ctimes%20%5Ctheta)
![W=I\times \alpha \times \theta](https://tex.z-dn.net/?f=W%3DI%5Ctimes%20%5Calpha%20%5Ctimes%20%5Ctheta)
![W=16\times 1.27 \times 12.56](https://tex.z-dn.net/?f=W%3D16%5Ctimes%201.27%20%5Ctimes%2012.56)
W = 255.21 J
Out of given options, the correct option for the work done by the torque is 255.21 J. Hence, this is the required solution.