Express the following using the prefixes of Table 1–4: a) 1 x 10^6 volts , b) 2 x 10^{-6} meters , c) 6 x 10^3 days , d) 18 x 10^2 bucks , and e) 7 x 10^{-7} seconds .

a) 1 MV b) 2 \mu m c) 6 kdays d) 18 hbucks e) 700 ns

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Giancoli's Physics: Principles with Applications, 7th Edition

Introduction, Measurement, Estimating

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1-4: Measurement, Uncertainty, Significant Figures

1-5 and 1-6: Units, Standards, SI, Converting Units

1-7: Order-of-Magnitude Estimating

1-8: Dimensions

Problem 13

Express the following using the prefixes of Table 1–4:

$1 \times 10^6 \textrm{ volts}$ ,
$2 \times 10^{-6} \textrm{ meters}$ ,
$6 \times 10^3 \textrm{ days}$ ,
$18 \times 10^2 \textrm{ bucks}$ , and
$7 \times 10^{-7} \textrm{ seconds}$ .

1 \textrm{ MV}

2 \textrm{ }\mu\textrm{m}

6 \textrm{ kdays}

18 \textrm{ hbucks}

700 \textrm{ ns}

Giancoli 7th Edition, Chapter 1, Problem 13 solution video poster

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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Times 10 to the 6 gets the prefix mega so we have 1 megavolt in part (a). Times 10 to the minus 6 is the prefix micro so we have 2 micrometers in part (b). 10 to the positive 3 is kilo so we have 6 kilodays, 6000 days. In part (d), we have 10 to the 2 is hecto so we have 18 hectobucks. And then part (e), 7 times 10 to the minus 7 there is no prefix just for 10 to the minus 7 so we have to kind of use one of the others; you could have either used nano or micro I used nanoseconds which is times 10 to the minus 9 but we have 10 to the minus 7 here so we'll have to multiply this number by a hundred or 10 to the 2 in order for 10 to the 2 times 10 to the minus 9 to work out to 10 to the minus 7 that we started with so you can say 700 nanoseconds; you could have also said 0.7 microseconds would also work.

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