Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 2 - Describing Motion: Kinematics in One Dimension - General Problems - Page 48: 83

Answer

(a) $4.3 \times 10^6 ~bits/s$ (b) 67% of the bits on the CD are used for encoding and error-correction.

Work Step by Step

(a) $N = \frac{1.2 ~m/s}{0.28\times 10^{-6} ~m/bit} = 4.3 \times 10^6 ~bits/s$ The CD players reads $4.3 \times 10^6 ~bits$ every second. (b) We can calculate the percentage of bits which are used for encoding and error-correction. $\frac {N-N_0}{N} \times 100\% = \frac{4.3 \times 10^6 ~bits/s - 1.4 \times 10^6 ~bits/s}{4.3 \times 10^6 ~bits/s} \times 100\% = 67\%$ 67% of the bits on the CD are used for encoding and error-correction.
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