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A small ball is attached to the lower end of a 0.800-m-long string, and the other end of the string is tied to a horizontal rod. The string makes a constant angle of 49.3° with the vertical as the ball moves at a constant speed in a horizontal circle. If it takes the ball 1.45 s to complete one revolution, what is the magnitude of the radial acceleration of the ball?

Respuesta :

Hania12

Answer:

a= 11.36 m/s²

Explanation:

Radius of circle=R=0.8 sin∅

                             =0.8 sin(49.3)

                             =0.606 m

 and we know that

                               ω=[tex]\frac{2\pi }{T}[/tex]

                                   =[tex]\frac{2*3.141592}{1.45}[/tex]

                                 ω=4.33 [tex]s^{-1}[/tex]

                                 a=R ω²

                                   =0.606*([tex]4.33^{2}[/tex]

                                  a=11.36 m/s²

                             

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