Question

(6%) Problem 3: The Marianas Trench, near the Philippines, is the deepest part of Earth's oceans at nearly 11 km. A 33% Part (a) Calculate the pressure, in atmospheres, due to the ocean at the bottom of the Marianas Trench, assumi is a constant 1025 kg/m2 all the way down to the bottom. P= JE ( 7 8 9 E 4 5 sino cos tano cotan asino acoso atan acotano) sinh cosh) tanho cotanho • Degrees Radians 6 3 1 2 0 NO Submit Hint I give up! Hints: 0°. deduction per hint. Hints remaining: 1 Feedback: 0 deduction per feedback. 33% Part (b) Calculate the percent decrease in volume of sea water due to such a pressure, assuming its bulk modulus water, 2.20 x 10°N m², and is constant all the way to the bottom A 33 Part (c) What would be the percent increase in the density of the water?

Answer 1

Solution in the uploaded image

Lock Solution, depth, d= 11 km = 11000 m Part (a) e= 1025 kg/m² Pressure at the bottom= exgxd 1025x 9.8x11000 = 110485000 Pa = 1080.5 atm Part (6) a F/Ar Bulk modules AV V AU x 100 = P Bulk modules = 1090.5x101325 2.20xlol x100 7. die rease = 5.02% de la m part co Tol exu= Alev) = A(M) since m-constant CAUTUAL = 0

W ERI AC Dux 100- sex loo % increase in density = % decrease in volume = 5.02% # Hit like if you understood the solution |