CP During a test dive in 1939, prior to being accepted by the U.S. Navy, the submarine Squalus sank at a point where the depth of water was 73.0 m. The temperature was 27.0°C at the surface and 7.0°C at the bottom. The density of seawater is 1030 kg/m3. (a) A diving bell was used to rescue 33 trapped crewmen from the Squalus. The diving bell was in the form of a circular cylinder 2.30 m high, open at the bottom and closed at the top. When the diving bell was lowered to the bottom of the sea, to what height did water rise within the diving bell? (Hint: Ignore the relatively small variation in water pressure between the bottom of the bell and the surface of the water within the bell.) (b) At what gauge pressure must compressed air have been supplied to the bell while on the bottom to expel all the water from it?
Want to see the full answer?
Check out a sample textbook solutionChapter 18 Solutions
University Physics with Modern Physics (14th Edition)
Additional Science Textbook Solutions
Modern Physics
Physics: Principles with Applications
Lecture- Tutorials for Introductory Astronomy
University Physics Volume 1
Conceptual Integrated Science
Physics (5th Edition)
- A cylinder with a piston holds 0.50 m3 of oxygen at an absolute pressure of 4.0 atm. The piston is pulled outward, increasing the volume of the gas until the pressure drops to 1.0 atm. If the temperature stays constant, what new volume does the gas occupy? (a) 1.0 m3 (b) 1.5 m3 (c) 2.0 m3 (d) 0.12 m3 (e) 2.5 m3arrow_forwardAt 25.0 m below the surface of the sea, where the temperature is 5.00C, a diver exhales an air bubble having a volume of 1.00 cm3. If the surface temperature of the sea is 20.0C, what is the volume of the bubble just before it breaks the surface?arrow_forwardHow many cubic meters of helium are required to lift a balloon with a 400-kg payload to a height of 8 000 m? Take He = 0.179 kg/m3. Assume the balloon maintains a constant volume and the density of air decreases with the altitude z according to the expression air = 0ez/8, where z is in meters and 0 = 1.20 kg/m3 is the density of air at sea level.arrow_forward
- A sealed cubical container 20.0 cm on a side contains a gas with three times Avogadros number of neon atoms at a temperature of 20.0C. (a) Find the internal energy of the gas. (b) Find the total translational kinetic energy of the gas. (c) Calculate the average kinetic energy per atom, (d) Use Equation 10.13 to calculate the gas pressure. (e) Calculate the gas pressure using the ideal gas law (Eq. 10.8).arrow_forwardThe rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is constant. At 15°C the pressure is 101.3 kPa at sca level and 87.14 kPa at h = 1000 m. (a) What is the pressure at an altitude of 3000 m? (b) What is the pressure on top of Mount McKinley, at an altitude of 6187 m?arrow_forwardAt 28 m below the surface of the sea (density of 1476 kg/m³), where the temperature is 6°C, a diver exhales an air bubble having a volume of 0.5 cm³. If the surface temperature of the sea is 22 C, what is the volume of the bubble immediately 3 before it breaks the surface? The acceleration .2 of gravity is 9.8 m/s² and the atmospheric pressure is 1.02 × 10° Pa. 3 Answer in units of cm°.arrow_forward
- A liquid has a density r. (a) Show that the fractional change in density for a change in temperature ΔT is Δρ/ρ = -β ΔT. (b) What does the negative sign signify? (c) Fresh water has a maximum density of 1.000 0 g/cm3 at 4.0°C. At 10.0°C, its density is 0.999 7 g/cm3. What is β for water over this temperature interval? (d) At 0°C, the density of water is 0.999 9 g/cm3. What is the value for β over the temperature range 0°C to 4.00°C?arrow_forwardThe density of gasoline at 0°C is 0.68 × 10° kg/m³. (a) What is the density on a hot day, when the temperature is 33°C? (b) What is the percent change in density?arrow_forwardThe pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pressure gage reads 210 kPa. If the volume of the tire is 0.025 m³, determine the pressure rise in the tire when the air temperature in the tire rises to 44°C. Also, determine the amount of air that must be bled off to restore pressure to its original value at this temperature. kJ kPa.m³ kJ Assume the atmospheric pressure to be 100 kPa. The gas constant of air is R = 0. 287- = 0.287 kg-K The pressure rise in the tire is 210.2 kPa. The amount of air that must be bled off to restore pressure to its original value is 0.000825 kg. kPa.m³ kg.Karrow_forward
- The partial pressure of oxygen in the lungs is about 150 mm of Hg. (The partial pressure is the pressure of the oxygen alone, if all other gases were removed.) This corresponds to a concentration of 5.3 × 1024 molecules per m3. In the oxygen-depleted blood entering the pulmonary capillaries, the concentration is 1.4 × 1024 molecules per m3. The blood is separated from air in the alveoli of the lungs by a 1-μm-thick membrane. What is the rate of transfer of oxygen to the blood through the 5 × 10-9 m2 surface area of one alveolus? Give your answer in both molecules/s and μmol/s. Assume The diffusion coefficient for oxygen in tissue is 2 × 10-11 m2/s. What is the same rate of transfer in μmol/s. Give your answer to 1 significant figure.arrow_forwardAn automobile moving through the air causes the air velocity (measured with respect to the car) to decrease and fill a larger flow channel. An automobile has an effective flow channel area of 3 m2 . The car is traveling at 90 km/h on a day when the barometric pressure is 70 cm of mercury and the temperature is 20°C.arrow_forwardStep 1 (a) The ideal-gas equation describes each condition of the air in the tire. We will set up ratios in order to describe these changes. Dividing the equations PV = nRT in the initial and final states, we have PfVf Tf P¡Vi Ti Solving for the final pressure in the tire, gives Pf = = = which gives I Tf P(V)() P₁ Vf T₁ = (1.013 x 105 Pa) x 105 Pa. Note that the volume V¡ cancels out of the equation. = Vf Po = Pr (V ( ) ( ) Pd Pf Vi V₁ Step 2 (b) After the car is driven, the temperature and volume of air in the tire have changed. Let Td be the temperature and Vd be the volume of air in the tire. We have PdVd Td PfVf Tf' x 105 Pa 273 + 273 + Vf Vf Again we see that the volume Vf cancels out of the equation. °C K °℃ K = x 105 Pa.arrow_forward
- College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning