The maximum acceleration of the train in which a box lying on its floo

1.	The maximum acceleration of the train in which a box lying on its floor will remain stationary given that the coefficient of friction between the box and the train is 0.15(g=ms^-2? ) is(a) 15ms^-2, (b) 1.5 ms ^ -2, (C) 2.0ms^ -2, (d) 0.15ms^-2
Answer» Answer: acceleration a is GIVEN by a = coefficient of friction × G ( SINCE ma = coefficient of friction × mg) a = 0.15 × 10 = 1.5m/s^2 Hope it helps you. Mark as BRAINLIEST please

The maximum acceleration of the train in which a box lying on its floor will remain stationary given that the coefficient of friction between the box and the train is 0.15(g=ms^-2? ) is(a) 15ms^-2, (b) 1.5 ms ^ -2, (C) 2.0ms^ -2, (d) 0.15ms^-2

Answer»

Answer:

acceleration a is GIVEN by

a = coefficient of friction × G

( SINCE ma = coefficient of friction × mg)

a = 0.15 × 10 = 1.5m/s^2

Hope it helps you. Mark as BRAINLIEST please

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The maximum acceleration of the train in which a box lying on its floor will remain stationary given that the coefficient of friction between the box and the train is 0.15(g=ms^-2? ) is(a) 15ms^-2, (b) 1.5 ms ^ -2, (C) 2.0ms^ -2, (d) 0.15ms^-2​

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The maximum acceleration of the train in which a box lying on its floor will remain stationary given that the coefficient of friction between the box and the train is 0.15(g=ms^-2? ) is(a) 15ms^-2, (b) 1.5 ms ^ -2, (C) 2.0ms^ -2, (d) 0.15ms^-2