[tex]\displaystyle \sf \text{Temos} :\\\\ \text{comprimento do barbante (L)}=20\ cm\\\\ L =20\cdot 10^{-2}m \\\\ L = 2\cdot 10^{-1} \ m\\\\ \text{Achando o raio de movimento } : \\\\ cos(\theta) =\frac{R}{L}\to cos(30\º) = \frac{R}{2\cdot 10^{-1}} \\\\ \frac{\sqrt{3}}{\not 2} = \frac{10R}{\not 2} \to R = \frac{\sqrt{3}}{10} \\\\\ \text{Decompondo a tra\c c\~ao no barbante, temos } : \\\\ Ty=T\cdot sen(30\º)\to Ty=\frac{T}{2} \ ; \\\\ T_x=T\cdot cos(30\º) \to T_x=\frac{T\sqrt{3}}{2}[/tex]

[tex]\displaystyle \sf \underline{\text{Equacionando as for\c cas atuantes}} \\\\ \text{Peso} =Ty \to m\cdot g= \frac{T}{2} \to T=2mg[/tex]

[tex]\displaystyle \sf \text{Resultante centr\'ipeta} = T_x\\\\ \frac{m\cdot V^2}{R} = \frac{T\sqrt{3}}{2} \to V^2=T\cdot R\cdot \frac{\sqrt{3}}{2m} \\\\\ V^2=\not 2\not mg\cdot \frac{\sqrt{3}}{10}\cdot \frac{\sqrt{3}}{\not 2\not m}\\\\\ V^2=g\cdot \frac{3}{10}\\\\ \text{tomando g =10}m/s ^2:\\\\ V^2=10\cdot \frac{3}{10}\\\\ V^2=3\\\\ V = \sqrt{3}\ m /s[/tex]

velocidade angular :

[tex]\displaystyle\sf \omega = \frac{V}{R} \\\\ \omega = \frac{\sqrt{3}}{\displaystyle \frac{\sqrt{3}}{10}}\\\\\\ \omega = \frac{10\sqrt{3}}{\sqrt{3}} \\\\ \\\large\boxed{\sf \z \omega =10\ rad/s \ }\checkmark[/tex]

letra C