A water drop of 0.05cm3 is squeezed between two glass plates and spreads into area of 40cm2. If the surface tension of water is
70 dyne/cm then the normal force required to separate the glass plates from each other will be
Answers
Answered by
122
Given:
Surface tension (T) = 70
Water drop (d) = 0.05
Area (A) = 40
In the given content:
So:
Therefore:
Hence:
The normal force required to separate the glass plates from each other will be 45 N.
Anonymous:
Ha
Answered by
5
Final answer is 45 N.
A water drop of 0.05cm3 is squeezed between two glass plates and spreads into area of 40cm2. If the surface tension of water is
70 dyne/cm then the normal force required to separate the glass plates from each other will be
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