hw01 - PROBLEM(1.23 GIVEN The diameter of the moon is dmoon...

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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 A s of the moon and ( b ) the ratio of the Earth’s and moon’s surface areas. GOVERNING EQUATIONS: A s = 4 πr 2 = πd 2 (1) SOLUTION: ( a ) Apply Eqn. (1) to find A s, moon : A s, moon = πd 2 moon = π (3480 × 10 3 m) 2 A s, moon = 3 . 80 × 10 13 m 2 ( b ) Using Eqn. (1) , A s, Earth A s, moon = πd 2 Earth πd 2 moon = d Earth d moon 2 = 12760 × 10 3 m 3480 × 10 3 m 2 A s, Earth A s, moon = 13 . 4
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PROBLEM (1.34) GIVEN: The sun sets, fully disappearing over the horizon as you lie on the beach, your eyes h 1 = 30 cm above the sand. You immediately jump up, your eyes now h 2 = 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: ω = dt (1) SOLUTION: Referring to the diagrams and using trig, the change in rotation Δ θ of the Earth is Δ θ = θ 2 - θ 1 = cos - 1 R h 2 + R - cos - 1 R h 1 + R Δ θ = cos - 1 6380 × 10 3 m 1 . 3 m + 6380 × 10 3 m - cos - 1 6380 × 10 3 m 0 . 3 m + 6380 × 10 3 m Δ θ = 3 . 317 × 10 - 4 rad From Eqn. (1) , t = Δ θ ω Earth = 3 . 317 × 10 - 4 rad 1 rev day = 3 . 317 × 10 - 4 rad 2 π rad 8 . 640 × 10 4 s t = 4 . 6 s
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PROBLEM (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/m 3 , and 1 metric ton is equivalent to 10 3 kg. GOAL: Find ( a ) the mass m of rain water in metric tons and ( b ) the volume V in gallons. GOVERNING EQUATIONS: ρ = m V (1) SOLUTION: ( a ) From Eqn. (1) , m = ρV = ρ ( Lwh ) = (1000 kg/m 3 )(9 × 10 3 m)(8 × 10 3 m)(0 . 01 m) 1metric ton 10 3 kg m = 7 × 10 5 metric tons ( b ) Using the unit conversions in the front of the class text, V = Lwh = (9 × 10 3 m)(8 × 10 3 m)(0 . 01 m) 1 L 10 - 3 m 3 1 gal 3 . 785 L V = 2 × 10 8 gal
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PROBLEM (1.57) GIVEN: Jean camps beside a wide river and wonders how wide it is. She spots a large rock on the bank directly across from her. She then walks upstream until she judges that the angle between her and the rock, which she can still see clearly, is now at an angle of θ = 30 downstream. Jean measures her stride to be about 1 yard long. The
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