HW #1-solutions

# HW #1-solutions - guo (jg44347) HW #1 Antoniewicz (57380) 1...

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Unformatted text preview: guo (jg44347) HW #1 Antoniewicz (57380) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Assuming that 68 % of the Earths surface is covered with water at average depth of 1 . 71 mi, estimate the mass of the water on Earth. One mile is approximately 1.609 km and the radius of the earth is 6 . 37 10 6 m. Correct answer: 9 . 54003 10 20 kg. Explanation: Let : R = 0 . 68 , h = 1 . 71 mi , = 1000 kg / m 3 , and R e = 6 . 37 10 6 m . If the area of Earth covered by water is . 68(4 R 2 e ), a good approximation of the volume of water covering the earth is V = 0 . 68(4 R 2 e ) h, and the mass is m = V = 0 . 68 4 R 2 e h = 0 . 68 (1000 kg / m 3 ) 4 (6 . 37 10 6 m) 2 (1 . 71 mi) 1000 m 1 km 1 . 609 km 1 mi = 9 . 54003 10 20 kg . 002 (part 1 of 2) 5.0 points A structural I beam is made of iron. A view of its cross-section and dimensions is shown. 20 cm 27 cm 2 cm 2 cm What is the mass of a section 2 . 32 m long? The density of iron is 7560 kg / m 3 , the atomic weight of iron is 55 . 85 g / mol and Avogadros number is 6 . 02214 10 23 atoms / mol. Correct answer: 220 . 994 kg. Explanation: Let : d = 2 cm = 0 . 02 m , w = 20 cm = 0 . 2 m , h = 27 cm = 0 . 27 m , = 2 . 32 m , and = 7560 kg / m 3 . w h d d The cross-sectional area of the beam is A = 2 w d + ( h- 2 d ) d = 2(0 . 2 m) (0 . 02 m) + [0 . 27 m- 2 (0 . 02 m)] (0 . 02 m) = 0 . 0126 m 2 , so the volume of the beam is V = A and the mass is m = V = A = (7560 kg / m 3 )(0 . 0126 m 2 )(2 . 32 m) = 220 . 994 kg . 003 (part 2 of 2) 5.0 points How many atoms are there in this section? Correct answer: 2 . 38291 10 27 atoms. Explanation: Let : M = 55 . 85 g / mol and N A = 6 . 02214 10 23 atoms / mol . guo (jg44347) HW #1 Antoniewicz (57380) 2 The number of moles is n = m M , so the num- ber of atoms is N = n N A = m M N A = 220 . 994 kg 55 . 85 g / mol 1000 g 1 kg (6 . 02214 10 23 atoms / mol) = 2 . 38291 10 27 atoms ....
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## This note was uploaded on 04/02/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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HW #1-solutions - guo (jg44347) HW #1 Antoniewicz (57380) 1...

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