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Sagot :
Given,
The diameter of the wheel, d=26 cm=0.26 m
Therefore the radius of the wheel, r=d/2=0.13m
Time period, t=2.0 s
Frequency of the revolution, f=1500 rpm
1 relovution=2π radian.
Therefore the angular velocity is calculated as,
[tex]\omega=\frac{1500\times2\pi}{60}=157.1\text{ rad/s}[/tex]The angular velocity and angular displacement are related as,
[tex]\omega=\frac{\theta}{t}[/tex]On rearranging the above equation and substituting the known values,
[tex]\theta=\omega t=157.1\times2.0=314.2\text{ rad}[/tex]We can calculate how far the rim has moved by using the following relationship,
[tex]s=r\theta[/tex]Where 's' is the distance of the movement of the rim,
On substituting the known values in the above equation,
[tex]s=0.13\times314.2=40.8\text{ m}\approx41\text{ m}[/tex]Therefore the correct answer is option 2, 41 m
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