Unit 6 Problems Solutions

CC2404 Applied Physics and Instrumentation in Health Care Problems Unit 6 1. What is the pressure on a surface when a force of 50 N acts on an area of (i) 2.0 m2, (ii) 100 m2 (iii) 0.5 m2 F 50 N = = 25 Pa A 2 .0 m 2 F 50 N = 0.5 Pa (ii) P = = A 100 m 2 F 50 N = 100 Pa (iii) P = = A 0.5m 2

(i) P =

2. A pressure of 10 Pa acts on an area of 3.0 m2. What is the force acting on the area? F = PA = 10 Pa × 3.0m 2 = 30 N

3. The aqueous humor in a person’s eye is exerting a force of 0.3 N on the 1.10 cm 2 area of the area cornea. What pressure is this in mm Hg? Notes: 1 atmosphere (atm) = 1.01× 105 N/m2 = 101 kPa = 760 mmHg P=

F 0.3N = = 2727 .3Pa A 0.00011 m 2

P=

760 × 2727 .3mmHg = 20 .52 mmHg 101000

4. The left side of the heart creates a pressure of 120 mm Hg by exerting a force directly on the blood over an effective area of 15.0 cm2. What force does it exert to accomplish this? P=

101000 ×120 Pa = 15947 .36 Pa 760

F = PA = 15947 .36 Pa × 0.0015 m 2 = 23 .92 N

5. If the pressure in the esophagus is -2.00 mm Hg while that in the stomach is +20.0 mm Hg, to what height could stomach fluid rise in the esophagus, assuming a density of 1.10 g/cm3? (This will not occur if the muscle closing the lower end of the esophagus is working properly.) Notes: 1 atmosphere (atm) = 1.01× 105 N/m2 = 101 kPa = 760 mmHg ∆P = 20 .00 mmHg − ( −2.00 mmHg ) = 22 .00 mmHg = 22 .00 ×101 ×10 3 / 760 Pa = 2923 .7 Pa ∆P = ρgh

h=

∆P 2923 .7 Pa = = 0.271 m ρg 1100 kg / m 3 × 9.8m / s 2