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A body of mass 0.05 kg is observed to fa...

A body of mass 0.05 kg is observed to fall with an acceleration of `9.5 ms^(-2)`. The opposing force of air on the body is `(g = 9.8 ms^(-2))`. The opposing force of air on the body is `(g = 9.8ms^(-2))`.

A

0.015 N

B

0.15 N

C

0.030 N

D

Zero

Text Solution

Verified by Experts

The correct Answer is:
A
m = 0.05kg , `a = 9.5m//s^(2)`
`mg - f_("air") = ma rArr f_("air") = m(g-a)`
` = 0.05 xx (9.8 - 9.5)`
` = 0.015N`
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