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**Unformatted text preview: **1. What is the total mass of the Earth's atmosphere? (The radius of the Earth is 6.37 × 10 6 m, and atmospheric pressure at the surface is 1.013 × 10 5 N/m 2 ). Solution: The Earths surface area is 2 4 R π . The force pushing inward over this area amounts to ( ) 2 4 F P A P R π = = . This force is the weight of the air: ( ) 2 4 g F m g P R π = = so the mass of the air is ( ) ( ) ( ) 2 5 2 6 2 18 2 1.013 10 N m 4 6.37 10 m 4 5.27 10 kg 9.80 m s P R m g π π × × = = = × . 2. (a) Calculate the absolute pressure at an ocean depth of 1 000 m. Assume the density of seawater is 1 024 kg/m 3 and that the air above exerts a pressure of 101.3 kPa. (b) At this depth, what force must the frame around a circular submarine porthole having a diameter of 30.0 cm exert to counterbalance the force exerted by the water? Solution: (a) ( ) ( ) ( ) 5 3 2 1.013 10 Pa 1024 kg m 9.80 m s 1000 m P P gh ρ = + = × + 7 1.01 10 Pa P = × (b) The gauge pressure is the difference in pressure between the water outside and the air inside the submarine, which we suppose is at 1.00 atmosphere. 7 gauge 1.00 10 Pa P P P gh ρ =- = = × The resultant inward force on the porthole is then ( ) 2 7 5 gauge 1.00 10 Pa 0.150 m 7.09 10 N F P A π = = × = × . 3. The small piston of a hydraulic lift has a cross-sectional area of 3.00 cm 2 and its large piston has a cross-sectional area of 200 cm 2 (Figure 14.4). What force must be applied to the small piston for the lift to raise a load of 15.0 kN? (In service stations, this force is usually exerted by compressed air.) Solution: Since the pressure is the same on both sides, 1 2 1 2 F F A A = In this case, 2 15000 200 3.00 F = or 2 225 N F = 4. Figure on the right shows Superman attempting to drink water through a very long straw. With his great strength he achieves maximum possible suction. The walls of the tubular straw do not collapse. (a) Find the maximum height through which he can lift the water. (b) What If? Still thirsty, the Man of Steel repeats his attempt on the Moon, which has no atmosphere. Find the difference between the water levels inside and outside the straw. ...

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