A small block rests on a horizontal turntable at a distance r=1.0 m from its axis; the coefficient of static friction between block and table is 0.5.…

The block has tangential acceleration $a_t=\alpha r=3\ \mathrm{m\,s^{-2}}$ and centripetal acceleration $a_c=\omega^2r$, both supplied by static friction, so $f/m=\sqrt{a_t^2+a_c^2}$. Slipping begins when this equals $\mu g=5\ \mathrm{m\,s^{-2}}$: $a_c=\sqrt{25-9}=4$, hence $\omega^2=4$, $\omega=2\ \mathrm{rad\,s^{-1}}$. Then $t=\omega/\alpha=\dfrac23\ \mathrm s$: (A) is correct. The friction components are $3$ (tangential) and $4$ (towards the axis), so the angle with the radial line satisfies $\tan\phi=3/4$, $\phi\approx37^\circ$: (B) is correct. The angle turned is $\theta=\dfrac{\omega^2}{2\alpha}=\dfrac46=\dfrac23\ \mathrm{rad}$, which is only about $0.1$ revolution, so (C) is incorrect. The speed is $v=\omega r=2\ \mathrm{m\,s^{-1}}$: (D) is correct.