Find the force acting on the window of the submarine by the water pressure of 1.03 x 10^7 Pa
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Answered by
25
Answer:
From equation: - P=Pa+ρgh
ρgh = 1.29 kg m–3 × 9.8 m s2 × h m = 1.01 × 105 Pa
∴ h = 7989 m ≈ 8 km
In reality the density of air decreases withheight. So does the value of g. The atmospheric
cover extends with decreasing pressure over100 km. We should also note that the sea level
atmospheric pressure is not always 760 mm ofHg. A drop in the Hg level by 10 mm or more isa sign of an approaching storm.
Answered by
0
Explanation:
P total = P atm + rho×h×g
10^5+10^7=10^12 pa
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