PhysI_09082009 - 3000 m 2000 km 15 cm (Vol. Candy aft er 1...

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Unformatted text preview: 3000 m 2000 km 15 cm (Vol. Candy aft er 1 minute) 17.0cm 14.0cm +7-6-5 +10-5 +16 +3 +15 +15-10-10-20-2-55-70-67-67-67-67-67-67 +58 +58 +58 +58 +58 +58 +17 +15 +16 09-08-2009 171.101, General Physics for Physical Science Majors I Clicker Problem: Chapter 1, #7 Problem: Volume of Antarctica in cm 3 : Convert to like units 2000km x 1000m /1km x 100cm/1m = 2 x 10 8 cm . 3000m x 100cm/1m = 3 x 10 5 cm Base Area = r 2 / 2 (2 x 10 8 cm) 2 /2 = 2 x 10 16 cm 2 Volume = Base Area x Thickness 2 x 10 16 cm 2 x 3 x 10 5 cm = 6 x 10 21 cm 3 1.885 x 10 22 cm 3 *P. Henry did not divide by 2 for the base area until the end. He warned that when checking a problem, you need to make sure that you divided by a factor of 2when you were supposed to. Chapter 1, #31 Problem: Rate of mass of candies filling the container in kg/min: What we are given : Base of Container: Length = 17.0cm , Width = 14.0cm Height of candy increases at rate of 0.250cm/s Weight of Candy: 0.200g Volume of Candy: 50.0mm 3 *Useful Conversion Factors: 1cm = 10 mm ; 1000g = 1kg ; 60s = 1min Assume : Ignore the volume of space between candies.- Height of Candy Increase Rate (convert seconds to m inutes): 0.250cm/s x 60s/1m in x 10mm/1cm = 15cm/m in- Volume of Candy after 1 m inute in container : 15 cm x 14cm x 17cm = 3,570cm 3- Convert Volume of Candy in container after each minute to mm 3 : 3570cm 3 x (10mm/cm) 3 = 3.57 x 10 6 mm 3- Find number of candies in container after each minute : 3.57 x 10: 3....
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This note was uploaded on 10/01/2009 for the course DOGEE 171.101 taught by Professor Henry during the Fall '09 term at Johns Hopkins.

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PhysI_09082009 - 3000 m 2000 km 15 cm (Vol. Candy aft er 1...

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