Explanation Given info: The radius of the piston is 1.20 cm . The force constant of the spring is 1250 N ⋅ m -1 . When the gauge is in water, the pressure exerted on the gauge is, P = ρ g h (1) P is the pressure ρ is the density of water g is the free fall acceleration due to gravity h is the depth of water The area of the piston is, A = π r 2 (2) A is the area r is the radius of the piston According to the Hooke’s law, F = − k x (3) F is the force k is the force constant of the spring x is the change in length of the spring The pressure is defined as, P = F A (4) Substitute (1), (2) and (3) in (4), ρ g h = k x π r 2 Re-arranging the above equation for an expression for the depth, h = k x ρ g π r 2 Substitute 1.20 cm for r , 1250 N ⋅ m -1 for k , 0.750 cm for x , 9.8 m ⋅ s -2 for g , 1000 kg ⋅ m -3 for ρ to determine the depth of water, h = ( 1250 N ⋅ m -1 ) ( 0

10th Edition

ISBN: 9781337520379

Author: Vuille

Publisher: YUZU

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Variation of pressure

From the equation of pressure, pressure is directly proportional to the applied force.

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Chapter 9, Problem 20P

To determine

What depth in water causes the piston to move 0.750 cm .

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What depth in water causes the piston to move 0.750 cm .

Solution Summary: The author explains how the piston compresses at 2.11m depth in water. The pressure exerted on the gauge is P=rho gh.

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What depth in water causes the piston to move 0.750 cm .

Chapter 9 Solutions