Ciśnienie atmosferyczne w kołobrzegu jest równe 1000 hpa oblicz wartość siły nacisku atmosfery ziemskiej na powieszchnie o wymiarach 2m x 20 dm
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![Dane:\\p=1000[hPa]=100000[Pa]\\s=2m\cdot 20dm=2 m\cdot 2m=4[m^2]\\\\Szukane :\\F_n = ? \\\\Rozwiazanie :\\p = \frac{F_n}{s}\to F_n = ps \\\\F_n = 100000\cdot 4 = 400000 [N] = 0,4[MN]](https://tex.z-dn.net/?f=Dane%3A%5C%5Cp%3D1000%5BhPa%5D%3D100000%5BPa%5D%5C%5Cs%3D2m%5Ccdot+20dm%3D2+m%5Ccdot+2m%3D4%5Bm%5E2%5D%5C%5C%5C%5CSzukane+%3A%5C%5CF_n+%3D+%3F+%5C%5C%5C%5CRozwiazanie+%3A%5C%5Cp+%3D++%5Cfrac%7BF_n%7D%7Bs%7D%5Cto+F_n+%3D+ps+%5C%5C%5C%5CF_n+%3D+100000%5Ccdot+4+%3D+400000+%5BN%5D+%3D+0%2C4%5BMN%5D+ "Dane:\\p=1000[hPa]=100000[Pa]\\s=2m\cdot 20dm=2 m\cdot 2m=4[m^2]\\\\Szukane :\\F_n = ? \\\\Rozwiazanie :\\p = \frac{F_n}{s}\to F_n = ps \\\\F_n = 100000\cdot 4 = 400000 [N] = 0,4[MN]")
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