Chapter 2, Problem 79A

(a)

To determine

To write: the answer in correct number of significant figures.

Answer to Problem 79A

  $\text{7}\text{.4}\text{mm}$

Explanation of Solution

Introduction:

The number of digits in a certain value which hold importance in determining the level of accuracy of that value is known as significant figures.

In the given result $5.32\text{mm +2}\text{.1}\text{mm = 7}\text{.4200000}\text{mm}$

  $\text{2}\text{.1}\text{mm }$ is the least precise measure.

Therefore, the answer should be rounded off to the nearest $1/{10}^{th}$ of a mm.

  $\text{7}\text{.4200000}\text{mm = 7}\text{.4}\text{mm}$

Conclusion:

Answer in correct number of significant digits is $\text{7}\text{.4}\text{mm}$ .

(b)

To determine

To write: the answer in correct number of significant figures.

Answer to Problem 79A

  $49.6{\text{m}}^2$

Explanation of Solution

Introduction:

The number of digits in a certain value which hold importance in determining the level of accuracy of that value is known as significant figures.

In the given result $13.597\text{ m }\times \text{ 3}\text{.65 m = 49}\text{.62905}{\text{m}}^2$

  $\text{3}\text{.65}\text{m }$ is the least precise measure.

Significant digit in the product will have equal number of significant digit in least precise measurement.

Therefore, the answer should be rounded off to the nearest tenth.

  $\text{49}\text{.62905}{\text{m}}^2=49.6{\text{m}}^2$

Conclusion:

Answer in correct number of significant digits is $49.6{\text{m}}^2$ .

(c)

To determine

To write: the answer in correct number of significant figures.

Answer to Problem 79A

  $\text{70}.4\text{kg}$

Explanation of Solution

Introduction:

The number of digits in a certain value which hold importance in determining the level of accuracy of that value is known as significant figures.

In the given result $83.2\text{kg -12}\text{.804}\text{kg = 70}\text{.3960000}\text{kg}$

  $\text{83}\text{.2}\text{kg}$ is the least precise measure.

Therefore, the answer should be rounded off to the nearest $1/{10}^{th}$ of a kg.

  $\text{70}\text{.3960000}\text{kg = 70}.4\text{kg}$

Conclusion:

Answer in correct number of significant digits is $\text{70}.4\text{kg}$ .