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At a given instant of time the position vector of a particle moving in a circle with a velocity `3hati-4hatj+5hatk is hati+9hatj-8hatk`.Its anglular velocity at that time is:A. `((13hati-29hatj-31hatk))/(sqrt(146)`B. `((13hati-29hatj-31hatk))/(146)`C. `((13hati+29hatj-31hatk))/(sqrt(146))`D. `((13hati+29hatj+31hatk))/(146)` |
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Answer» Correct Answer - b `vecomega=(vecrxxvecv)/(|vecr|^(2))` |
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