Essential University Physics: Volume 1 (4th Edition)

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

Chapter 1 - Exercises and Problems - Page 12: 32

Answer

$131.58 Nm$

Work Step by Step

$$(5.1\times10^{-2} cm + 6.8\times10^{3} \mu m)\times(1.8\times10^{4}) N$$ To solve this problem first we need to add the first two physical quantities. When we are adding two physical quantities both of them should have the same units. Therefore we can convert centimeters and micrometers into meters and solve the problem as follows. $5.1\times10^{-2} cm = 5.1\times10^{-2}\times10^{-2} m = 5.1\times10^{-4} m$ $6.8\times10^{3} \mu m = 6.8\times10^{3}\times10^{-6} m = 6.8\times10^{-3} m$ Now we can simply add these values. $5.1\times10^{-4} m + 6.8\times10^{-3} m = (5.1\times10^{-4} + 68\times10^{-4} )m$ $\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space=(5.1+68)\times10^{-4} m$ $\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space= 73.1\times10^{-4} m$ Now we can multiply the above value by $1.8\times10^{4} N$ to get the final answer. $$(73.1\times10^{-4} m) \times(1.8\times10^{4} N) = 131.58 Nm$$ $$ The\space answer\space is= 131.58Nm$$
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