Unit 1 Solutions

CC2404 Applied Physics and Instrumentation in Health Care Supplementary Problems Unit 1 Measurement and unit 1. A space shuttle orbits the Earth at altitude of 280 km. What is the altitude in (a) miles and (b) mm? (a) 280 km = 280 kmx 1 = 280km x (1 mi/1.61 km) = 173.9 mi (b) 280 km = 280 x 1000 m = 280 x 1000 x 1000 mm = 2.8x108 mm 2. The Earth is approximately a sphere of radius 6.37x106 m. (a) What is its circumference in kilometers? (b) What is its surface area in square kilometers? (c) What is its volume in cubic kilometers? Note that r = 6.37x106 m = 6.37x 103 km (a) circumference = 2π R = 2π x 6.37x103 km = 4x104 km (b) surface area = 4π R2 = 4π x (6.37x103)2 km2 = 5.1x108 km2 (c) volume = (4/3)π R3 = (4/3)π x (6.37x103)3 km3 = 1.1x1012 km3 3. Enrico Fermi once pointed out that a standard lecture period is close to 1 microcentury. How long is a microcentury in minutes, and is it a long or short lecture for you? 1 century = 100 year = 100 x 365 days (neglect some years have 366 days; that would make a small difference) = 100 x 365 x 24 hours = 8.76 x 105 hours 1 microcentury = 1 x 10-6 century = 1 x 10-6 x 8.76 x 105 hours = 0.876 hours = 53 minutes 4. A person on diet might lose 1.3 kg of fat per week. Assuming to lose 100 g of fat is equal to consume 250 calories, express the energy loss rate in calories per day. (Ans. 464 cal) 100 g of fat = 250 cal, 1 = 250 cal/100 g energy loss = 1.3 kg/ week = 1300 g / 7 days = 186 g/day = (186 g/day) x (250/100) cal/g = 464 cal/day. Energy 5. (P.P Urone 2001, Problem 6.10) (a) How fast must a 3000 kg elephant move to have the same kinetic energy as a 65.0 kg sprinter running at 10.0 m/s? (b) Discuss how the larger energies needed for the movement of larger animals would relate to metabolic rates.

(a) Kinetic energy of elephant = Kinetic energy of sprinter =

1 mv 2 = 0.5 x 3000 x v2= 1500v2 2

1 mv 2 = 0.5 x 65 x 102 = 3250 J 2

For them to have equal energy, 1500 v2 = 3250; v = 1.47 m/s (b) The metabolic rate, which is he rate of energy generation in living organisms, is larger in large animals. This is because they need more energy to produce even small movements.

6. (P.P Urone 2001, Problem 6.50) (a) What is the efficiency of an out-of-condition professor who does 1.10 x 105 J of useful energy while metabolizing 500 kcal of food energy? (b) How many food calories would a well-conditioned athlete metabolize in giving the same energy with an efficiency of 20%? (giving 1 cal = 4.186 J) Efficiency =

Useful Output ×100 % Input Necessary

(a) 500 kcal = 500 x 103 cal = 500 x 103 x 4.186 J = 2.093x106 J efficiency = 100% x (1.1 x 105 J)/ (2.093 x 106) J = 5.3%

(b) 20% = 100% x (1.1 x 105 J)/ E E = 5.5 x 105 J = 5.5 x 105 J x (1 cal/ 4.186 J) = 1.31 x 105 cal = 131 kcal. Buoyant force 7. (P.P Urone 2001, Problem 10.44) A rock with a mass of 540 g in air is found to have an apparent mass of 342 g when submerged in water. (a) What mass of water is displaced? (b) What is the volume of the rock? (c) What is its average density? (density of water 1 g/ cm3). (a) By Archimede’s principle, Buoyant force = weight of water displaced by object. Now buoyant force = weight of stone in air – weight of stone in water = 0.540*10 – 0.342*10 = 1.98 N Therefore weight of water displaced = 1.98 N, so mass of water displaced = 198 g

(c) Since mass of water displaced = 1.98 /10 = 0.198 kg = 198 g volume of water displaced = mass / density = 198 g / (1 g/cm3) = 198 cm3 Therefore volume of stone = volume of water diplaced = 198 cm3 (c) Density of stone = mass of stone in air / volume of stone = 540 g / 198 cm3 = 2.73 g/cm3. 8. (P.P Urone 2001, Problem 10.47) Some fish have a density slightly less than that of water and must exert a force (swim) to stay submerged. What force must a 85.0 kg fish exert to stay submerged in salt water if its body density is 1015 kg/m3? (density of salt water 1025 kg/m3). (Ans. 8N) Volume of fish = Mass / Density = 85 kg / (1015 kg/m3) = 8.37 x 10-2 m3 Mass of sea water displaced = volume x density = (8.37 x 10-2 m3) x 1025 kg/m3 = 85.79 kg Therefore force required to push down the fish = (85.79 – 85)*g = 8 N.