HW_1_sol

# HW_1_sol - The amount of bronze required for the statue is...

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P1112 Fall 2008 Homework #1 Solutions 1. Total distance is 20 miles, total time is 1.5 h. Average speed is 20/1.5 = 13.3 miles per hour. 2. 1 km = 1000 m and 1 hr = 3600 s, therefore 1228.0 km/h = 341.1 m/s. That’s faster than the speed of sound! 4. a. First let’s convert to km/L. 21 mi/gal = 21*1.6 km/3.8 L = 8.8 km/L. If the car goes 8.8 km with every liter of fuel, it will take 100/8.8 = 11.4 L/100km. b. 6 L/100km = 2.5 gal/100mi. That’s 40 mi/gal. c. The conversion formula : e (L/100 km) = 240/f (mi/gal). 5. The amount of varnish required is proportional to the surface area of the statue. Surface area scales as the square of the dimensions of the statue. Therefore, if the full- size model is five times larger, the sculptor will need twenty-five small cans of varnish.
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Unformatted text preview: The amount of bronze required for the statue is proportional to the volume. Volume scales as length cubed. So the full-scale model will need 125 times more bronze, or 250 kg. 6. The faster runner will overtake the slower one when he is one lap ahead. The distance between the runners thus needs to reach 200 m. The difference in speeds is 0.7 m/s, so the overtake will happen in 200/0.7 = 285.7 s. The runners will have run 1571 and 1771 meters, respectively. 7. Converting to metric units, the takeoff speed is 76.9 m/s and the distance is 92.1 meters. a) The most convenient constant-acceleration formula to use here is 2*a*x = v f 2-v i 2 . Solving for the acceleration, a = 76.9 2 /184.2 m/s 2 = 32 m/s 2 . That’s 3 g’s! b) t = v/a = (76.9/32)s = 2.4 s....
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## This note was uploaded on 08/28/2009 for the course PHYS 1112 taught by Professor Leclair,a during the Fall '07 term at Cornell.

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