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Chapter 1: Problem 57

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

Short Answer

Expert verified
(a) \(432000\,seconds\) (b) \(88.5137\,m\) (c) \(\$0.49897 / L\) (d) \(465.116\frac{km}{hr}\) (e) \(14.1749\frac{L}{s}\) (f) \(707.92\,cm^3\)

Step by step solution

01

(a) Conversion from days to seconds

For this conversion, we need to know how many seconds are in a day. There are 24 hours in a day, each hour contains 60 minutes and each minute contains 60 seconds. To convert 5.00 days to seconds, we will multiply by the number of seconds in a day: \( 5.00\,days \times 24\frac{hours}{day} \times 60\frac{minutes}{hour} \times 60\frac{seconds}{minute}= 432000\,seconds \)
02

(b) Conversion from miles to meters

To perform this conversion, we need to know the conversion factor between miles and meters. 1 mile = 1609.34 meters. To convert 0.0550 miles to meters, we will multiply by the conversion factor: \( 0.0550\,mi \times 1609.34\frac{m}{mi} = 88.5137\,m \)
03

(c) Conversion from dollars per gal to dollars per liter

To perform this conversion, we need to know the conversion factor between gallons and liters. 1 gal = 3.78541 liters. To convert $1.89/gal to dollars per liter, we will divide by the conversion factor: \( \frac{\$1.89}{gal} \times \frac{1\,gal}{3.78541\,L} = \$0.49897 / L \)
04

(d) Conversion from in/ms to km/hr

We can break this conversion into two steps. Firstly, we can convert inches to kilometers, and then convert milliseconds to hours. We know that 1 inch = 2.54 cm = 0.0000254 km, and 1 ms = 0.001 seconds = 0.000000277778 hours. To convert 0.510 in/ms to km/hr, we will multiply by the conversion factors: \( 0.510\frac{in}{ms} \times 0.0000254\frac{km}{in} \times \frac{1}{0.000000277778\frac{hr}{ms}} = 465.116\frac{km}{hr} \)
05

(e) Conversion from gal/min to L/s

To perform this conversion, we need to know the conversion factor between gallons and liters, and between minutes and seconds. 1 gal = 3.78541 L, and 1 min = 60 seconds. To convert 22.50 gal/min to L/s, we will multiply by the conversion factors: \( 22.50\frac{gal}{min} \times 3.78541\frac{L}{gal} \times \frac{1}{60\frac{s}{min}} = 14.1749\frac{L}{s} \)
06

(f) Conversion from ft³ to cm³

To perform this conversion, we need to know the conversion factor between cubic feet and cubic centimeters. 1 ft³ = 28316.8 cm³. To convert 0.02500 ft³ to cm³, we will multiply by the conversion factor: \( 0.02500\,ft^3 \times 28316.8\frac{cm^3}{ft^3} = 707.92\,cm^3 \)

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Most popular questions from this chapter

A sample of ascorbic acid (vitamin C) is synthesized in the laboratory. It contains \(1.50 \mathrm{~g}\) of carbon and \(2.00 \mathrm{~g}\) of oxygen. Another sample of ascorbic acid isolated from citrus fruits contains $6.35 \mathrm{~g}$ of carbon. According to the law of constant composition, how many grams of oxygen does it contain?

The distance from Earth to the Moon is approximately \(240,000 \mathrm{mi}\). (a) What is this distance in meters? (b) The peregrine falcon has been measured as traveling up to \(350 \mathrm{~km} /\) hr in a dive. If this falcon could fly to the Moon at this speed, how many seconds would it take? (c) The speed of light is \(3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}\). How long does it take for light to travel from Earth to the Moon and back again? (d) Earth travels around the Sun at an average speed of $29.783 \mathrm{~km} / \mathrm{s}$. Convert this speed to miles per hour.

(a) The speed of light in a vacuum is $2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}$. Calculate its speed in miles per hour. (b) The Sears Tower in Chicago is \(1454 \mathrm{ft}\) tall. Calculate its height in meters. \((\mathbf{c})\) The Vehicle Assembly Building at the Kennedy Space Center in Florida has a volume of \(3,666,500 \mathrm{~m}^{3}\). Convert this volume to liters and express the result in standard exponential notation. (d) An individual suffering from a high cholesterol level in her blood has $242 \mathrm{mg}\( of cholesterol per \)100 \mathrm{~mL}$ of blood. If the total blood volume of the individual is \(5.2 \mathrm{~L}\), how many grams of total blood cholesterol does the individual's body contain?

(a) What is the mass of a silver cube whose edges measure 2.00 \(\mathrm{cm}\) each at \(25^{\circ} \mathrm{C} ?\) The density of silver is $10.49 \mathrm{~g} / \mathrm{cm}^{3}\( at \)25^{\circ} \mathrm{C}$. (b) The density of aluminum is \(2.70 \mathrm{~g} / \mathrm{cm}^{3}\) at \(25^{\circ} \mathrm{C}\). What is the weight of the aluminum foil with an area of \(0.5 \mathrm{~m}^{2}\) and a thickness of \(0.5 \mathrm{~mm} ?\) (c) The density of hexane is \(0.655 \mathrm{~g} / \mathrm{mL}\) at $25^{\circ} \mathrm{C} .\( Calculate the mass of \)1.5 \mathrm{~L}$ of hexane at this temperature.

Is the use of significant figures in each of the following statements appropriate? (a) The 2005 circulation of National Geographic was \(7,812,564 .\) (b) On July 1, 2005, the population of Cook County, Illinois, was $5,303,683 .(\mathbf{c})\( In the United States, \)0.621 \%$ of the population has the surname Brown. (d) You calculate your grade point average to be \(3.87562 .\)
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