Chapter 1.2, Problem 81E

a)

To determine

To write: the distance travelled by light in one yearin scientific notation.

Answer to Problem 81E

  $5,900,000,000=5.9\times {10}^{12}$ mi

Explanation of Solution

Given information:

The distance travelled by light in one year is $5,900,000,000,000$ mi.

Concept used:

Scientific notation:

A notation in which a given quantity can be expressed as a number with significant digits required for a specified degree of accuracy and multiplied by 10 to the appropriate power.

Calculation:

Consider the given number.

  $5,900,000,000,000$

So, the number can be expressed in scientific notation as

  $\begin{array}{c}5,900,000,000,000=5.9\times 1000000000000\\ =5.9\times {10}^{12}\end{array}$

Hence, $5,900,000,000=5.9\times {10}^{12}$ mi.

b)

To determine

To write: the diameter of electron in scientific notation.

Answer to Problem 81E

  $0.0000000000004=4\times {10}^{-13}$ cm

Explanation of Solution

Given information:

Diameter of electron is $0.0000000000004$ cm.

Concept used:

Scientific notation:

A notation in which a given quantity can be expressed as a number with significant digits required for a specified degree of accuracy and multiplied by 10 to the appropriate power.

Calculation:

Consider the given number.

  $0.0000000000004$

So, the number can be expressed in scientific notation as

  $\begin{array}{c}0.0000000000004=4÷100000000000000\\ =4÷{10}^{13}\\ =4\times {10}^{-13}\end{array}$

Hence, $0.0000000000004=4\times {10}^{-13}$ cm.

c)

To determine

To write: the number of molecules in a drop of water in scientific notation.

Answer to Problem 81E

  $33\text{billion}\text{billion}=3.3\times {10}^{19}$

Explanation of Solution

Given information:

The number of molecules in a drop of water is

  $33\text{billion}\text{billion}$

Concept used:

Scientific notation:

A notation in which a given quantity can be expressed as a number with significant digits required for a specified degree of accuracy and multiplied by 10 to the appropriate power.

Calculation:

Consider the given number.

  $33\text{billion}\text{billion}$

So, the number can be expressed in scientific notation as

  $\begin{array}{c}33\text{billion}\text{billion}=3.3\times 1000000000000000000\\ =3.3\times {10}^{19}\end{array}$

Hence, $33\text{billion}\text{billion}=3.3\times {10}^{19}$ .