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In the above question, the magnitude of the average acceleration over the arc of `60^(@)` isA. `10 m//s^(2)`B. `11 m//s^(2)`C. `11.5 m//s^(2)`D. `12 m//s^(2)` |
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Answer» Correct Answer - C `theta=omegat` `t=(theta)/(omega)=(pi)/(3xx(v)/(r))=(pixxr)/(3v)=(pixx300)/(3xx60)` `t=(5pi)/(3)` `a_(av)=(Deltav)/(t)=(60)/((5pi)/(3))` `a_(av)=(60xx3)/(5pi)=(12xx3)/(pi)=11.5m//s^(2)`. |
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