Chapter 1
Q. 1.5
The beds in your dorm room have extra-long matresses. These mattresses are 80 inches (2 significant figures) long and 39 inches wide. (Regular twin beds are 72 inches long.)
What is the area of the mattress top in m²? (1 inch = 2.54 cm)
ANALYSIS | |
mattress length (80 in) and width (39 in) bridge conversion (1 inch = 2.54 cm) |
Information given: |
centimeter to meter conversion | Information implied: |
area in m² | Asked for: |
STRATEGY
1. Recall equation for finding the area of a rectangle: area = length × width
2. Follow the plan: in² → cm² → m²
Step-by-Step
Verified Solution
80 in × 39 in = 3.12 × 10³ in²(We will round off to correct significant figures at the end.) | area in in² |
3.12 × 10³ in² × \frac{(2.54)² cm²}{(1)² in²} × \frac{(1)² m²}{(100)² cm²} = 2.0 m² | area in m² |
ENd POINT
There are 36 inches in one yard, so the dimensions of the mattresses are approximately 1 yd wide and 2 yd long or 2 yd² . A meter is almost equivalent to a yard (see Table 1.3) so the calculated answer is in the same ball park.
Table 1.3 Relations Between Length, Volume, and Mass Units
Metric | English | Metric-English | |||
Length | |||||
1 km | = 10³ m | 1 ft | = 12 in | 1 in | = 2.54 cm* |
1 cm | = 10^{−2} m | 1 yd | = 3 ft | 1 m | = 39.37 in |
1 mm | = 10^{−3} m | 1 mi | = 5280 ft | 1 mi | = 1.609 km |
1 nm | = 10^{−9} m = 10 Å | ||||
Volume | |||||
1 m³ | = 10^{6}cm^{3} = 10³ L | 1 gal | = 4 qt = 8 pt | 1 ft³ | = 28.32 L |
1 cm³ | = 1 mL = 10^{−3} L | 1 qt (U.S. liq) | = 57.75 in³ | 1 L | = 1.057 qt (U.S. liq) |
Mass | |||||
1 kg | = 10³ g | 1 lb | = 16 oz | 1 lb | = 453.6 g |
1 mg | = 10^{−3} G | 1 short ton | = 2000 lb | 1 g | = 0.03527 oz |
1 metric ton | = 10³ kg | 1 metric ton | = 1.102 short ton | ||
*This conversion factor is exact; the inch is defined to be exactly 2.54 cm. The other factors listed in this column are approximate, quoted to four significant figures. Additional digits are available if needed for very accurate calculations. For example, the pound is defined to be 453.59237 g. |