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The angular acceleration of a fan `alpha=(-3t^(2))/2`. At the initial moment, its angular velocity `omega=10 rad//s` and has an angular position of 1 rad.A. Its angular velocity at t=1sec. Is 9.5 rad/sB. its angular position at t=2 sec. Is 5 radC. its angular velocity at t=2 sec. is 6 rad/sD. its angualr position at t=1 sec. is `87/8` rad |
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Answer» Correct Answer - ACD `alpha =(-3t^(2))/2` `(d omega)/(dt)=(-3t^(2))/2` `rArr int d omega=|(-3t^(2))/2 dt ` `omega=(-3t^(2))/32+C" " t=0, omega=10` C=10 `rArr omega=t^(3//2)+10` `rArr (d theta)/(dt)=-t^(3//2) +10` `theta=-(t^(4))/8+10 t+1` `omega_((t=1))=9.5 rad//sec` `omega_((t=2))=6 rad//sec` `theta_((t=2))=-2+21p=19 rad` `theta_((t=1))=-166+10+1 =87/8 rad.` |
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