Take Home Exam 2-2

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1. The Marianas trench is beated in the Pacific Ocean at a depth of about 11,000m below the surface of the water. The density of seawater is 1025 kg/m3. a. If an underwater vehicle were to explore such a depth, what force would the water exert in the vehicles observation window (diameter= 200cm)? For comparison, determine the weight of a jetliner whose mass is 1.2x105 kg. Given: h= 11,000 m ρ= 1025 kg/m3 m= 1.2x105 kg d= 200cm Find: F= ? W= ? SOLUTIONS: F= ρgh W= mg *Convert cm to m: d= 200cm x 1m100cm d= 2m *Solve for the area: A= πd24 A= 3.1416(2m)24 A= 3.1416 m2 *Solve for the force: F= ρgh FA= ρgh F= ρghA F= (1025 kg/m3)(9.8m/s2)(11,000m)( 3.1416 m2) F= 347,131,092 kg m/s2 or N *Solve for the weight: W= mg W= 1.2x105 kg(9.8m/s2) W= 1,176,000 kg m/s2 or N 2. How high must the reservoir of a water system be situated above a faucet to produce a pressure of 30N per square cm? Given: P= 30N Find: h= ? SOLUTIONS: *Convert N/cm2 to N/m2: P= 30 N/cm2 x 10,000cm21 m2 P= 300,000N/m2 *Solve for h: P= ρgh h= Pρg h= 300,000 N/m21000 kg/m3(9.8ms2) h= 30.61 m 3. Find the total force due to the water on the slide of the pool which is 20 meters long and 10 meters wide if the water is at a uniform depth of 1.2 meters. Given: H = 12m W = 10m L= 20m Find: FTota=? Solve for Area: A= LxW A=20mx10m A=200m2 Solve for Pgauge: Pgauge= ᵨgh Pgauge= 1000kg/m2 (9.8m/s2)(1.2m) Pgauge = 1170 Pa Solve for Fgauge: P= F/A Fgauge= PAgauge Fgauge= 11760 Pa (200m2) Fgauge= 2,352,000 N Convert bar to Pa: Patm = 1.013 bar x 〖10〗^5/(1 bar) Patm = 101,300 Pa

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