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A body of mass 100g is at rest on a smooth surface .A force of 0.2N acts on it 5seconds.Calculate the distance travelled by the body |
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Answer» Answer: The distance travelled 25m Given that mass = 100g = 0.1kg Weight = m \TIMES g = 100 \times 10m×g=100×10 = 1000 N Force APPLIED F = 0.2 N Time (t) = 5 sec Initial velocity (u) = 0 We now that Force = mass \times accelerationmass×acceleration 0.2 = 0.1 \times a0.1×a => a = 2 m/sec^2m/sec 2 S\QUAD =\quad ut+\frac { 1 }{ 2 } a{ t }^{ 2 }S=ut+ 2 1 at 2 S = 0 + 0.5 \times 2 \times (5)^2S=0+0.5×2×(5) 2 S = 0.5 \times 2 \times 25 = 25 mS=0.5×2×25=25m Therefore, distance traveled by the body = 25 m Step-by-step explanation: please mark brainlist amswer can you follow me please |
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