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Chapter 1: Problem 47

Perform the following conversions: (a) 5.00 days to s, (b) 0.0550 mi to \(\mathrm{m}\), (c) $$\$ 1.89 /$$ gal to dollars per liter, (d) 0.510 in. \(/ \mathrm{ms}\) to (f) \(0.02500 \mathrm{ft}^{3}\) to \(\mathrm{cm}^{3}\) \(\mathrm{km} / \mathrm{hr},\) (e) \(22.50 \mathrm{gal} / \mathrm{min}\) to \(\mathrm{L} / \mathrm{s}\)

Short Answer

Expert verified
In summary, the short version of the answers are: (a) 5.00 days = 432,000 seconds (b) 0.0550 mi = 88.5137 meters (c) $1.89/gal = $0.499 dollars per liter (approx.) (d) 0.510 in/ms = 55.384 km/hr (approx.) (e) 22.50 gal/min = 14.164 L/s (approx.) (f) 0.02500 ft³ = 707.218 cm³ (approx.)

Step by step solution

01

Identify conversion factors

1 day is equal to 24 hours, 1 hour is equal to 60 minutes, and 1 minute is equal to 60 seconds.
02

Apply conversion factors

\(5.00 \,days \times \frac{24 \,hours}{1 \,day} \times \frac{60 \,minutes}{1 \,hour} \times \frac{60 \,seconds}{1\, minute}\)
03

Calculate

After canceling out the units and completing the calculation, we find that 5.00 days is equal to 432,000 seconds. #b. Convert 0.0550 mi to meters#
04

Identify conversion factors

1 mile is equal to 1609.34 meters.
05

Apply the conversion factor

\(0.0550 \,mi \times \frac{1609.34 \,m}{1 \,mi}\)
06

Calculate

After canceling out the units and completing the calculation, we find that 0.0550 mi is equal to 88.5137 meters. #c. Convert $$\$ 1.89 /$$ gal to dollars per liter#
07

Identify conversion factors

1 gallon is equal to 3.78541 liters.
08

Apply the conversion factor

\(\frac{\$ 1.89}{1 \,gal} \times \frac{1 \,gal}{3.78541 \,L}\)
09

Calculate

After canceling out the units and completing the calculation, we find that \$ 1.89/gal is equal to \$ 0.499 dollars per liter (approximately). #d. Convert 0.510 in. \(/ \mathrm{ms}\) to \(\mathrm{km} / \mathrm{hr}\)#
10

Identify conversion factors

1 inch is equal to 0.0254 meters, 1 kilometer is equal to 1000 meters, and 1 hour is equal to 3600 seconds.
11

Apply conversion factors

\(0.510 \,in/ms \times \frac{0.0254 \,m}{1 \,in} \times \frac{1 \,km}{1000 \,m} \times \frac{3600 \,s}{1 \,hr}\)
12

Calculate

After canceling out the units and completing the calculation, we find that 0.510 in/ms is equal to 55.384 km/hr (approximately). #e. Convert \(22.50 \mathrm{gal} / \mathrm{min}\) to \(\mathrm{L} / \mathrm{s}\)#
13

Identify conversion factors

1 gallon is equal to 3.78541 liters, and 1 minute is equal to 60 seconds.
14

Apply conversion factors

\(\frac{22.50 \,gal}{minutes} \times \frac{3.78541 \,L}{1 \,gal} \times \frac{1 \,minute}{60 \,seconds}\)
15

Calculate

After canceling out the units and completing the calculation, we find that 22.50 gal/min is equal to 14.164 L/s (approximately). #f. Convert \(0.02500 \mathrm{ft}^{3}\) to \(\mathrm{cm}^{3}\)#
16

Identify conversion factors

1 foot is equal to 30.48 centimeters.
17

Apply conversion factors

\(0.02500 \,ft^{3} \times \frac{(30.48 \,cm)^{3}}{(1 \,ft)^{3}}\)
18

Calculate

After canceling out the units and completing the calculation, we find that 0.02500 ft³ is equal to 707.218 cm³ (approximately).

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Most popular questions from this chapter

A package of aluminum foil contains \(50 \mathrm{ft}^{2}\) of foil, which weighs approximately 8.0 oz. Aluminum has a density of \(2.70 \mathrm{~g} / \mathrm{cm}^{3}\). What is the approximate thickness of the foil in millimeters?

Automobile batteries contain sulfuric acid, which is commonly referred to as "battery acid." Calculate the number of grams of sulfuric acid in 1.00 gallon of battery acid if the solution has a density of \(1.28 \mathrm{~g} / \mathrm{mL}\) and is \(38.1 \%\) sulfuric acid by mass.

(a) What is the difference between a hypothesis and a theory? (b) Explain the difference between a theory and a scientific law. Which addresses how matter behaves, and which addresses why it behaves that way?

Round each of the following numbers to four significant figures, and express the result in standard exponential notation: (c) \(0.008543210,\) (d) 0.000257870 , (a) \(102.53070,(\mathbf{b}) 656,980,\) (e) -0.0357202 .

Water has a density of \(0.997 \mathrm{~g} / \mathrm{cm}^{3}\) at \(25^{\circ} \mathrm{C} ;\) ice has a density of \(0.917 \mathrm{~g} / \mathrm{cm}^{3}\) at \(-10^{\circ} \mathrm{C}\). (a) If a soft-drink bottle whose volume is \(1.50 \mathrm{~L}\) is completely filled with water and then frozen to \(-10^{\circ} \mathrm{C},\) what volume does the ice occupy? (b) Can the ice be contained within the bottle?
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