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Find the pressure exerted at the tip of a board pin, of area 0.1 `mm^(2)`, if it is pressed against the board with a force 15 N. |
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Answer» Normal force applied on the pin, F = 15 N Area of the tip of the board pin, A = 0.1 `mm^(2)` = `0.1 xx 10^(-6) m^(2)` Pressure, `P = (F)/(A) = (15)/(0.1 xx 10^(-6)) N m^(-2)` `= 15 xx 10^(7) N m^(2)` |
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