Section 1.4More on the Metric System

1. (Exercises 3 and 4) A drinking glass contains 8 oz of water, which is equivalent to 236 mL. What is the mass of water in the glass (a.) in grams and (b.) in kilograms?

a. Since 1 mL of water has a mass of 1 g, the glass will contain 236 g of water.

b. The mass in kg will be 0.236 kg because 1 kg = 1000 g,
and 236 g (1 kg/ 1000 g) = 0.236 kg

Section 1.5Derived Quantities and Conversion Factors

2. (Exercises 7 and 8) The height of a flagpole is 15.0 ft. How high is this in cm and in m?

Applying the proper conversion factors, we obtain:

15.0 ft (12 in./1 ft) (2.54 cm/in.) = 457 cm
457 cm (1 m/100 cm) = 4.57 m

3. (Exercises 13 and 14) A jet airliner must reach a speed of about 300 km/h on the runway before it can lift into the air. How fast is this in mph?

300 km/h (1 mi/h / 1.61 km/h) = 186 mph

4. (Exercises 15 and 16) A metal bar is 2.50 cm high, 5.40 cm wide, and 10.3 cm long. Determine the mass of this bar if it is made of iron that has a density of 7.86 g/cm³.

First, the volume of this bar must be calculated. V = l x w x h =
(2.50 cm)(5.40 cm)(10.3 cm) = 139 cm³. Then the mass can be found using:

m = V = 7.86 g/cm³(139 cm³) = 1090 g or 1.09 x 10³ g

Section 1.6Significant Figures and

Section 1.7Powers-of-10 Notation

5. (Exercises 17 and 18) Round off the following numbers to four (4) significant figures.

a. 256,943

b. 0.025317

c. 65.9111

d. 2.63883 x 106

a. 256,943	=	256,900 or 2.569 x 105
b. 0.025317	=	0.02532
c. 65.9111	=	65.91
d. 2.63883 x 106	=	2.639 x 106

6. (Exercises 21 and 22) Convert each of the following into conventional, or standard, scientific notation (one digit to the left of the decimal point).

a. 3490 b. 0.0000612c. 822 x 10⁵d. 0.012 x 10^-6


The decimal points must be moved in these numbers as follows:

a. For 3490 move the decimal point 3 places to the left to give:
3.49 x 10³

b. For 0.0000612 move the decimal point 5 places to the right to give:
6.12 x 10^-5

c. For 822 x 10⁵ move the decimal point 2 places to the left to give:
8.22 x 10⁷

d. For 0.012 x 10^-6 move the decimal point 2 places to the right to give: 1.2 x 10^-8

Section 1.8Approach to Problem Solving

7. (Exercises 25 and 26) The books described in Exercise 26 in your textbook were sent to the bookstore in a shipping carton.

a. How tall would the carton have been if one dozen books (each is 3.50 cm thick) had come in each carton?

b. What was the height of the shipping carton in meters?

a. Assuming that the books were put into the box flat, that is parallel to the floor, when they were loaded, the height of the box would simply be the same height as 12 of the textbooks stacked on top of each other.

h = 3.50 cm x 12 = 42.0 cm

* Why is the answer 42.0 cm instead of just 42 cm? The number 3.50 has 3 significant figures and 12 is an exact whole number representing the quantity of individual textbooks involved. This means that the number of significant figures in the actual measured value, in this case three, representing the thickness of one single textbook, is the determining factor in the number of significant figures allowed in the final answer.

b. To find the height of the carton in meters, we must convert centimeters to meters using one of the conversion factors found in your textbook.

h = 42.0 cm x (1 m / 100 cm) = 0.420 m

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