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The coefficient of friction between a body and the surface of an inclined plane at `45^(@)` is 0.5. if `g=9.8m//s^(2)`, the acceleration of the body downwards I `m//s^(2)` is
A. `(4.9)/sqrt(2)`
B. `4.9sqrt(2)`
C. `19.6sqrt(2)`
D. `4.9`

1 Answer

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Best answer
Correct Answer - A
`a=g sin theta-mug cos theta`
`g((1)/sqrt(2)-0.5xx(1)/sqrt(2))=(9.8xx0.5)/sqrt(2)=(4.9)/sqrt(2)`

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