# the distance travelled by light in one yearin scientific notation.

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

#### Solutions

Chapter 1.2, Problem 81E

a)

To determine

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

Expert Solution

5,900,000,000=5.9×1012 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

5,900,000,000,000=5.9×1000000000000=5.9×1012

Hence, 5,900,000,000=5.9×1012 mi.

b)

To determine

### To write: the diameter of electron in scientific notation.

Expert Solution

0.0000000000004=4×1013 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

0.0000000000004=4÷100000000000000=4÷1013=4×1013

Hence, 0.0000000000004=4×1013 cm.

c)

To determine

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

Expert Solution

33billionbillion=3.3×1019

### Explanation of Solution

Given information:

The number of molecules in a drop of water is

33billionbillion

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.

33billionbillion

So, the number can be expressed in scientific notation as

33billionbillion=3.3×1000000000000000000=3.3×1019

Hence, 33billionbillion=3.3×1019 .

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