Calculate the force that acts on a square access hatch at the bottom of the tank

Question

Calculate the force that acts on a square access hatch at the bottom of the tank that measure 0.5m by 0.5m.
(a). gauge pressure is P=Dgh
Density of gasoline is D=680kg/m^3
Acceleration due to gravity is g = 9.8m/s^2
Height of tank is h = 30m
P = (680kg/s^3)(9.8m/s^2)(30m)
= 199,920Pa
= 2 x10^5Pa <----where does this come from? 2 x 10^5
(b). force = (Pressure)(Area)
Area = (0.5m)(0.5m)
= 0.25m^2
Force = (2 x 10^5 Pa)(0.25m^2)
= 5 x10^4 N <-------- how do you get this? can you show me how to solve this?

Explanation / Answer

that first yellow part is just the rounded answer.... like, 199,920 Pa is approximately equal to 200,000 Pa then for the second part, you have the right equations, so...(also, note that E = 10^, so 2E5 = 2 x 10^5 Pressure = Force / Area Force = (Pressure)(Area) F = PA F = (2E5 Pa)(.25 m^2) F = (200000)(.25) F = 50000 n F = 5E4 n You get that? You have all the right stuff written down, I think the notation is the confusing part. You're doing real swell though! Gluck!

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