Question

The age of the universe is thought to be about 14 billion years. Assuming two significant figures, write this in powers of 10 in (a) years, (b) seconds.

Short answer

The universe's age in powers of 10 in years is (a) \(14 \times {10^9}\) and in seconds is (b) \(44 \times {10^{16}}\).

Step-by-step solution

Method of calculation

First, express the number in the powers of 10. Then convert the given units into the required forms.

For example, 20 m can be written in feet as the following:

\(\begin{aligned}{c}20{\rm{ m}} = \left( {20{\rm{ m}}} \right)\left( {\frac{{{\rm{100 cm}}}}{{1{\rm{ m}}}}} \right)\left( {\frac{{1{\rm{ in}}{\rm{.}}}}{{2.54{\rm{ cm}}}}} \right)\left( {\frac{{1{\rm{ ft}}}}{{12{\rm{ in}}{\rm{.}}}}} \right)\\ = 65.617{\rm{ ft}}\end{aligned}\)

Thus, 20 m is equivalent to \(65.617{\rm{ ft}}\).

(a) Conversion of the given age in years

Theage of the universe in powers of 10 in yearsis as follows:

\(\begin{aligned}{c}14{\rm{ billion years}} = 14{\rm{ billion years}}\left( {\frac{{{{10}^9}{\rm{ years}}}}{{1{\rm{ billion years}}}}} \right)\\ = 14 \times {10^9}{\rm{ years}}\end{aligned}\)

Thus, the age of the universe in years is \(14 \times {10^9}\).

(b) Conversion of the given age in seconds

Theage of the universe in powers of 10 in seconds is as follows:

\(\begin{aligned}{c}14{\rm{ billion years}} = 14{\rm{ billion years}}\left( {\frac{{{{10}^9}{\rm{ years}}}}{{1{\rm{ billion years}}}}} \right)\left( {\frac{{365{\rm{ days}}}}{{1{\rm{ year}}}}} \right)\left( {\frac{{24{\rm{ h}}}}{{1{\rm{ day}}}}} \right)\left( {\frac{{60{\rm{ min}}}}{{1\;{\rm{h}}}}} \right)\left( {\frac{{60{\rm{ seconds}}}}{{1{\rm{ min}}}}} \right)\\ = 4.415 \times {10^{17}}{\rm{ seconds}}\end{aligned}\)

This can also be expressed as

\(\begin{aligned}{c}4.415 \times {10^{17}}{\rm{ seconds}} = \left( {4.415 \times {{10}^{17}} \times \frac{{10}}{{10}}} \right){\rm{ seconds}}\\ = \left( {4.415 \times 10} \right) \times \left( {\frac{{{{10}^{17}}}}{{10}}} \right){\rm{ seconds}}\\ = 44.15 \times {10^{16}}{\rm{ seconds}}{\rm{.}}\end{aligned}\)

Thus, the required age in seconds is \(44.15 \times {10^{16}}\).

Expression of the age in seconds with two significant figures in the powers of 10

Theage of\(44.15 \times {10^{16}}{\rm{ seconds}}\)can also be expressedas

\(44.15 \times {10^{16}}{\rm{ seconds}} \approx 44 \times {10^{16}}{\rm{ seconds}}{\rm{.}}\)

Thus, the age of the universe in seconds is \(44 \times {10^{16}}\).