University of California, Berkeley PHYSICS 7Ahw01

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PROBLEM (1.23) GIVEN: The diameter of the moon is d moon = 3480 km, and Earth’s diameter is d Earth = 12760 km. GOAL: Find (a) the surface area As of the moon and (b) the ratio of the Earth’s... and moon’s surface areas. GOVERNING EQUATIONS: A s = 4πr2 = πd2 (1) SOLUTION: (a) Apply Eqn. (1) to find As; moon: A s; moon = πd2 moon = π(3480 × 103 m)2 A s; moon = 3:80 × 1013 m2 (b) Using Eqn. (1), A s; Earth A s; moon = πd2 Earth πd2 moon = ddEarth moon 2 = 12760 3480 ××10 1033mm2 A s; Earth A s; moon = 13:4PROBLEM (1.34) GIVEN: The sun sets, fully disappearing over the horizon as you lie on the beach, your eyes h1 = 30 cm above the sand. You immediately jump up, your eyes now h2 = 130 cm above the sand, and you can again see the top of the sun. The radius of the Earth is R = 6380 km. GOAL: Find the time t it takes for the sun to disappear over the horizon. DIAGRAM: GOVERNING EQUATIONS: ! = dθ dt (1) SOLUTION: Referring to the diagrams and using trig, the change in rotation ∆θ of the Earth is ∆θ = θ 2 − θ1 = cos−1 h2 R+ R − cos−1 h1 R+ R ∆θ = cos−1 1:3 m + 6380 6380 × 10×3 m103 m − cos−1 0:3 m + 6380 6380 × 10×3 m103 m ∆θ = 3:317 × 10−4 rad From Eqn. (1), t = ∆θ ! Earth = 3:317 × 10−4 rad 1 rev day = 3:317 × 10−4 rad 2π rad 8:640×104 s t = 4:6 sPROBLEM (1.53) GIVEN: A heavy rainstorm dumps h = 1:0 cm of rain on a city L = 9 km wide and w = 8 km long. The density of water is ρ = 1000 kg/m3, and 1 metric ton is equivalent to 103 kg. GOAL: Find (a) the mass m of rain water in metric tons and (b) the volume V in gallons. GOVERNING EQUATIONS: ρ = m [Show More]

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