Question

US Help System Announcements PRINTER VERSION s BACK NEXT Chapter 14, Problem 017 Crew members attempt to escape from a damaged submarine 52.1 m below the surface. What force must be applied to a pop-out hatch, which is 0.866 m by 0.433 m, to push it out at that depth? Assume that the density of the ocean water is 1024 kg/m and the internal air pressure is at 1.00 atm. Number Units the tolerance is +/-2% Click If you would like to Show Work for this question: Open Show Work LINK TO TEXT LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM Question Attempts: 0 of 5 used SAVE FOR LATER SUBMIT ANSWER

Answer 1

atm depth = 52.1m gravity g = 9.8 ms? dimension = 0.866 m X 0.433m A = 0.375 m² water density = 1024 kg/m3 internal pressure a = 10 N/m² Now pressure at depth P= Pot g gh = 105 + 1094x 9.8*52-1 = 6.228 x 105 Pa - 6.23x105 Pa force = (P-Po A -C6-4380-703)x03 = S-23X10 N/m?x 0.375m? = 1.96810 N 1.96x105 N

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