P = f / S = 6N / 4 × 10 negative third power M & # 178; = 1.5 × 10 third power PA how can we get 1.5 × 10 third power pa by this formula for pressure

P = f / S = 6N / 4 × 10 negative third power M & # 178; = 1.5 × 10 third power PA how can we get 1.5 × 10 third power pa by this formula for pressure

P=F/S=6N/(4×10-³m²)=(6/4)×(1/10-³)pa=1.5×10³pa

The fifth power PA of 1.013 * 10, why didn't the roof collapse under such atmospheric pressure

There's so much pressure in the house

The area of the roof is 45m2. How much pressure does the atmosphere have on the roof? Why didn't the roof collapse under such pressure?

∵ P = 1.01 × 105Pa, s = 45m2, ∵ f = PS = 1.01 × 105Pa × 45m2 = 4.545 × 106n, because there is atmospheric pressure inside and outside the house, and its effect counteracts each other, so the roof will not collapse. A: the atmospheric pressure on the roof is 4.545 × 106n, because there is atmospheric pressure inside and outside the house, and its effect counteracts each other, so the roof will not collapse