Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 17 - Exercises and Problems - Page 326: 65

Answer

$3.97^{\circ}C$

Work Step by Step

We simplify the equation to find T using the quadratic formula and obtain: $ T = \frac{-b \pm \sqrt{b^2-4ca}}{2c}$ Using the values for a, b, and c, we find: $ T = \frac{-1.7 \times 10^{-5} \pm \sqrt{(-1.7 \times 10^{-5})^2-4(-6.43 \times 10^{-5})(-2.02\times 10^{-7})}}{2(-2.02\times 10^{-7})}=$$3.97^{\circ}C$
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