a force is applied for a duration of 10 second on a body of mass 5 kg that is at rest. as a result the body acquires velocity of 2ms.find the magnitude of the force applied.
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Answered by
34
F=ma
F=m(v-u)/t
F=5(2-0)/10
F=1N
Therefore force of 1N was applied on the body
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F=m(v-u)/t
F=5(2-0)/10
F=1N
Therefore force of 1N was applied on the body
pls mark as brainliest
Answered by
14
F = mv /t
= (5 * 2) / 10
= 1 N
Magnitude of Force applied is 1 N
= (5 * 2) / 10
= 1 N
Magnitude of Force applied is 1 N
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