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A 0.11 N m torque is applied to a fan that was initially at rest. The fan has moment of inertia of 0.034 kg m2. Determine the kinetic energy of the fan after 8.0 s.

Sagot :

samuelonum1

Answer:

2.72*10-3 Joules

Explanation:

From Newton's second law of motion

F=ma

[tex]\tau=I*\alpha[/tex]

given

[tex]\tau= 0.11Nm\\\\I=0.034kgm^2\\\\t= 8s\\\\\alpha=?[/tex]

[tex]\alpha =0.11/0.034\\\\\alpha=3.23 rad/s^2[/tex]

the angular velocity is

[tex]\omega = \alpha/t\\\\\omega =3.23/8\\\\\omega =0.4 rad/s[/tex]

[tex]KE= 1/2*I* \omega^2[/tex]

[tex]KE=1/2*0.034*0.4^2\\\\KE=1/2*0.034*0.16\\\\KE=0.00272\\\\KE=2.72*10^-3J[/tex]

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