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Sagot :
Given,
The moment of inertia of the solid ball, I=(2/5)MR²
The mass of the ball, m=87.0 g=0.087 kg
The diameter of the ball, d=17.0 cm=0.17 m
The initial velocity of the ball, u=0 m/s
The angle of inclination of the slope, θ=25°
The distance covered by the ball, s=3.00 m
The acceleration of the ball is given by,
[tex]a=g\sin \theta[/tex]Where g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} a=9.8\times\sin 25^{\circ} \\ =4.14\text{ m/s}^2 \end{gathered}[/tex]From the equation of the motion,
[tex]v^2=u^2+2as[/tex]Where v is the speed of the ball after it has rolled 3.00 m
On substituting the known values.
[tex]\begin{gathered} v^2=0+2\times4.14\times3.00 \\ \Rightarrow v=\sqrt[]{24.84} \\ v=4.98\text{ m/s} \end{gathered}[/tex]Thus the speed of the ball after it rolls 3.00 m down the slope is 4.98 m/s
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