General Chemistry: Principles and Modern Applications (10th Edition)

Published by Pearson Prentice Hal
ISBN 10: 0132064529
ISBN 13: 978-0-13206-452-1

Chapter 1 - Matter: Its Properties and Measurement - Exercises - Units of Measurement: 35

Answer

a) It would take 10 s to run 100.0 m. b) The speed is 9.8 meters per second. c) It would take 1.5 * $10^{2}$ s to run 1.45 km

Work Step by Step

a) 100 yd in 9.3 s 1 m = 1.094 yd 1.094 yd = 1 m 1 yd = 1 / 1.094 m = 0.9140767824 m 100 yd = 91.40767824 m . So the sprinter runs 91.40767824 m in 9.3 s. This is 1 m in 9.3/91.40767824 s = 0.101742 s So 100.0 m in 10.1742 s. Rounding to 2 significant figures gives 10 s. b) Speed is 100 yd per 9.3 s 100 yd = 91.40767824 m (see a), so speed is 91.40767824 m per 9.3 s. This is the same as 91.40767824 / 9.3 per second, which is 9.828782606 meters per second. Rounding to 2 significant figures gives 9.8 meters per second. c) velocity = distance / time so time = distance / velocity distance = 1.45 km = 1450 m velocity = 9.828782606 meters per second So time = 1450 / 9.828782606 = 147.5259s Rounding to 2 significant figures gives 1.5 * $10^{2}$ s
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