Question

Problem 3 (4 pts.). A bathyscaphe (a small free-diving self-propelled deep-sea submersible), is a sphere of radius R- 1.5 m located 873.5 m below the sea level. a what is the pressure exerted on its surface (here you can neglect the size of the bathyscape compared to the depth of its immersion) and the buoyant force acting on it in the water? (6) If a small leak is opened at the top of the bathyscaphe, with what speed will the water enter it? The pressure inside the bathyscaphe is p = 1.3 where Date = 1.01 - 10° Pa is the atmospheric pressure. (c) We need to make sure that our bathyscaphe will not be completely filled with the water for at least 20 minutes. What air pressure should be created inside the bathyscaphe if the area of the opens is 1.07 cm"? You may assume that the pressure inside the bathyscaphe does not change while it is being filled with water. 2R H = 873.5m R=1.5m Figure 3: Schematic graph for Problem 3.

Answer 1

3. Когч бm }н долд 8- К.». + 34Н \sv\4 = 1. 1 x165-169xiv хө73.- -36 хур” ра. с . 48 » I Buoyancy force = force everted by the displate volume Tolla mieter fe = sva - 8 4 * ) 8 Cg - (03x чт (+ 3х10 = 14).37x16°ы TB = 141.31 kn) P=2=1.3 Batm atm Редат взяторых" чам.до P, E-91-48 ( :-) ...мик Баня 4cmk } С. 4 ош уве - buttph Bovхэд эмэлт+ zero velocity ce vino Due J х өт.uө -3) «4» v, (3 х 16 хь (е-40 — 3 ) (1-оїх +5) - v. ТУ, = 13.3 ms | | |32mlѕ sv) volume water a flow in the baths Caphe in zominthe у - от vt:1.07xtwх 131, 980x64 - 16.93 Total volume of butts tatele v = 4TR 4711.533 - 1 - 13 so in rominute volume flow exceeds the actual capacity of the bathscaphe. Since Final pressure will be equal to outside Pression ное fщим А щ: 1,0 = 91- чөз» - 86 6xls rе. Ігт тогт