PHYSICS

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Continuation of Vocabulary Averaged angu- Cepennc xyrose Otolith OTOJIiT lar acceleration npHCKopeHWI Cordiac muscle Cepueemi M'~3 Proprioception Flponpiopeuenuis Contraction CKOpO'-leHH~ Resultant force Pe3YJIbTYIO'-Ia CMJIa Cross-bridges Ilonepexni Rigid body Tsepne TiJIO MicTKM Cross-striated Ilonepexao-no- Saccule CaKyJIa, muscle cvyroaaaait M'~3 MllIIO'-lOK Curvilinear KpMBOJIiHiMHHM Sarcomere CapKoMep movement pyx Density I'ycrana Semicircular HaniBKpyrJIMM canal KaHaJI Displacement 3MimeHH~ Sliding-filament MOL\eJIb KOBmodel 3aHH~ HMTOK Elasticity n pyxnicrs Smooth fJIaL\eHbKMH: muscle M'~3 Fiber BOJIOKHO Strain Binaocna nediopuauis Filaments HMTKa Stress Hanpyacenicrs Fulcrum TO'-lKa onopa Tactile TaKTHJIbHa sensitivity '-IYTJIMBicTb Golgi tendom CYXO)/(MJIbHMH: Tensile .n:iarpaMa opraa fOJIbL\)/(i diagram P03THY Instantaneous MMTT€Be Tension P03T~r acceleration npacxopenna Instantaneous MMTT€Be xyrose The stiffness Koeoinicar angular npMCKopeHH~ constant »copcrxocri acceleration Instantaneous MMTT€Ba xyrona The vestibular BecTH6YJI~pangular velocity lIIBML\KICTb system HHH: annapar Instantaneous MMTT€Ba To compress CTMcKyBaTH velocity lIIBML\KicTb Isometry I3oMeTPi~ To stretch P03T~rYBaTH La teral line CMcTeMa 60KO Torque MOMeHT system BOl ninii 06epTaJIbHHH: Lengthening BHL\OB)/(eHH~ Touch .n:OTMK Lever Baacins Utricle Ilepennsep's Lever Tlnexe Weber Be6epoB arm Ba)/(eJI~ ossicles annapaT Mammal Ccaseus Weight Bara Control Works Modulus 2 Mechanics, Biomechanics, Mechanobiology Problem. The velocity of a particle moving along the x axis varies in time according to the expression V = (20 - 31 2 ) m/s, where 1 is in s. I). Find the average acceleration in the time interval 1 = 0 to 1 = 2 s. 2). Determine the instantaneous acceleration at 1 = 2 s. Answers: -8 m/s'; -12 m/s-. Problem. A load of 98 kg is supported by a wire of length 1.8 m and cross-sectional area 0.12 cm-. The wire is stretched by 0.2 cm. Find the stress, strain, and Young's modulus for the wire. Answer: 8 . 10 7 N/m 2 ; 0.11 . 10- 2 ; 7.2 . 10 10 N/m 2 • Problem. The bone of the dog has the length / = 20 ern, surface of cross-sectional area S = 2.8 cm-, and Young's modulus Y = 1.9 . 10 10 N . m". Find the force of elasticity which appears during the compression of 0.4 mm of the bone. Answer: 10.6 KN. Example. Consider the diagram of biceps muscles which lift a mass 0.6 kg in the hand (fig. 3.4). The biceps are at an angle e = 15° to the elbow. Find the force exerted by the biceps. Answer: 30.4 N. Home Work. Write an allometric equation, draw an allometric diagram, and determine constants a and b for the following forestry data which have been measured for calculating relationshi p between trunk dry weight and diameter at breast height. l Trunk dry weight, kg T Diameter at chest height, em 1 4.737 5.0 2 137.329 23.4 3 27.484 11.8 4 71.174 16.7 5 2.755 4.2 6 4.655 5.6 7 2.085 3.8 8 8.557 10.0 9 1.593 4.3 10 6.132 6.5 11 159.522 21.9 12 86.525 17.7 13 131.157 25.5 40 41 Sample
-. .,.,'. Chapter 6. FLUID AND GAS MECHANICS I~JIllIIIlI!";"f!II'~, 6.1. PRESSURE 6.1.1. Definition of Pressure The pressure is a physical quantity which characterizes the intensity of normal (perpendicular to the surface) forces with which one body acts on the surface of another. If the forces exhibit a uniform distribution along the surface, the pressure is determined as the ratio of force to area: F p=- (6.1) S where F is the magnitude of the normal force on the surface and S is the area of this surface. If the pressure is not uniform across the surface, the following expression defines the pressure at a specific point: p = L t1F = dF M--.O t1S dS (6.2) The SI unit for pressure is the Pascal (l N/m 2 = 1 Pa). The following units are also used, especially so in older literature: 1 atm = 1.01325 . 10 5 Pa = 1013.25 mbar = 760 mm Hg; 1 mbar = 100 Pa = 1 hPa = 0.75006 mm Hg; 1 mm Hg = 1 torr = 1.3332 mbar = 133.32 Pa; 1Pa = 9.87 . 10 -6 atm = 7,5 . 10 -3 mm Hg; 1 pound-force per square inch (psi) = 1 lbf/in- = 6894.74 Pa. 6.1.2. Variations in Pressure with Depth The pressure at depth (d) in opened vessel is determined by: P = P A + pgd (6.3) where PA ('" 1.01 . 10 5 Pa) is the atmospheric pressure, p is the density of the fluid, and g is the acceleration due to gravity. The absolute pressure (p) at a depth (d) below the surface of a liquid open to the atmosphere is greater than atmospheric pressure by pgd. Example. Calculate the pressure at the bottom of the Marianas Trench (depth 11043 m). Assume the density of water is 1 . 10 3 kg/rn' and Po = 1.01 . 105Pa. Solution. Using formula (6.3), the pressure is equal to: p = PA + pgd = 1.01 . 105 Pa + + (1.0 . 10 3 kg/m 3)(9.8 m/s 2)(1l043 m) = 109 MPa. 6.1.3. Physiological Effects of Increased Pressure on Humans Pressure Problems. Some professions (e.g., Arabian sponge divers, Australian pearl divers, Japanese and Korean Ama) make their living by diving in the sea without any special equi pment. When divers descend, they are sensitive to the surrounding water pressure which increases at a rate of one atmosphere for each 10 m in descent (i.e., IJp = 1 atm for IJd = 10 m). The pressure of the surrounding water is transmitted to all internal parts of the body and a state of pressure equilibrium is established where the internal pressure of the body is equal to the surrounding pressure. During ascent and descent the diver must support the pressure of air in his lungs the same as that of the surrounding water. For example, if the diver goes to 30 m underwater by holding a full breath of air, he would have 4 atm of pressure in his lungs according to Equation (6.3). When ascending, the pressure at the water surface is 1 atm, while the pressure in his lungs would be 3 atm. This pressure difference can lead to the rupture of his lungs, a phenomenon termed pneumotorax. Influence of Nitrogen. The air we breathe contains 79% of nitrogen; as a consequence, our blood is full of dissolved nitrogen. According to Henry s law, the amount of dissolved gas in a liquid at constant temperature is directly proportional to the partial pressure of the gas. Liquids which are under high pressure can dissolve more gas than liquids which are under low pressure. At substantial depths, the greatly elevated pressure of the air in the lungs results in the formation of the gas bubbles in the blood vessels. The spherical shape of the bubbles is distorted due to the bloodstream (fig. 6.1) and the left and right hemispheres have different radii of curvature. According to the Laplace equation [see Equation (6.22)], additional pressure produced by the surface tension is inversely proportional to the radius of curvature R of 42 43
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layer of glass called the cladding,
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its bathing medium), there will be
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emission of radiation. It is import
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tri phosphate (ATP). A detailed des
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