fluid mechanics
posted by cindy .
A brick weighs 15.0 N and is resting on the ground. Its dimensions are 0.203 m multiplied by 0.0890 m multiplied by 0.0570 m. A number of the bricks are then stacked on top of this one. What is the smallest number of whole bricks (including the one on the ground) that could be used, so that their weight creates a pressure of at least one atmosphere on the ground beneath the first brick? (Hint: First decide which face of the brick is in contact with the ground.)

1 atm= 101.3 Kilo newton/m^2
pressure= Number*15/area (decide which area, I recommend the smallest side down).
101.3*10^3*area/15 = number bricks.