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A uniform solid cylindrical log begins rolling without slipping down a ramp that rises 28.0° above the horizontal. After it has rolled 4.20 m along the ramp, what is the magnitude of the linear acceleration of its center of mass?

Respuesta :

Answer:

3.07 m/s

Explanation:

Cricetus

The magnitude of the linear acceleration will be "3.07 m/s²".

Given:

  • Angle, [tex]\Theta = 28^{\circ}[/tex]
  • Length = 2.40 m

The acceleration of its center of mass will be:

= [tex]\frac{2}{3}g Sin \Theta[/tex]

By putting the above given values, we get

= [tex]\frac{2}{3}\times 9.8 Sin 28^{\circ}[/tex]

= [tex]\frac{2}{3}\times 9.8\times 0.46[/tex]

= [tex]\frac{9.0167}{3}[/tex]

= [tex]3.07 \ m/s^2[/tex]

Thus the solution above is appropriate.

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