A body of mass 0.05kg is observed to fall with an acceleration of 9.5ms-2.The opposing force of air on the body is .[take g = 9.8 ms-2]
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Answered by
101
Heya.......!!!!
Given in the question :
=> Net acceleration is = 9.5 m/s^2
=> Acceleration due to Gravity = 9.8m/s^2
Let, F be the force due to air resistance ,,
F( air ) = ma( air )
let us denote Net acceleration by "a "
Therefore :
ma = mg - ma( air )
=> F( air )= mg – ma
=> F( air ) = 0.05 (9.8 – 9.5) = 0.015 N
♦Opposing Force = 0.015 N
Hope It Helps You ^_^
Given in the question :
=> Net acceleration is = 9.5 m/s^2
=> Acceleration due to Gravity = 9.8m/s^2
Let, F be the force due to air resistance ,,
F( air ) = ma( air )
let us denote Net acceleration by "a "
Therefore :
ma = mg - ma( air )
=> F( air )= mg – ma
=> F( air ) = 0.05 (9.8 – 9.5) = 0.015 N
♦Opposing Force = 0.015 N
Hope It Helps You ^_^
Answered by
8
Answer:0.015
Explanation:Given in the question :
=> Net acceleration is = 9.5 m/s^2
=> Acceleration due to Gravity = 9.8m/s^2
Let, F be the force due to air resistance ,,
F( air ) = ma( air )
let us denote Net acceleration by "a "
Therefore :
ma = mg - ma( air )
=> F( air )= mg – ma
=> F( air ) = 0.05 (9.8 – 9.5) = 0.015 N
♦Opposing Force = 0.015 N
Hope It Helps You ^_^
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