Physics - Solutions to Check for Understanding Problems

Appendix B Solutions to Check for Understanding Problems Chapter 1 Check for Understanding 1.1 1. Which of the following represents the fundamental steps of the scientific method? A. observation ! law ! hypothesis ! theory. B. observation ! hypothesis ! experiment ! theory. C. hypothesis ! theory ! experiment ! law. D. observation ! theory ! experiment ! hypothesis. Answer: B Solution You can reject A because it lacks the experimentation that is central to the scientific method. Also, a law is not formulated until numerous observations have been made. You can reject C because the steps taken between a tentative explanation (hypothesis) and a well established explanation (theory) require numerous experiments. You can reject D because a hypothesis is always developed before a theory. 2. Characterize each of the following as an example of a scientific law, scientific theory, observation, or none of these. a) The liquid in a glass of water is composed of molecules. b) Flammable materials always contain oxygen. c) When a can of soda pop is opened, a fizzing sound is heard. d) The force of gravity between two objects increases as they get closer. Answers: a) theory c) observation b) none of these d) law Solutions a) This represents an explanation for the composition and observable properties of water. Scientists cannot observe molecules directly so this is not a simple observation or a summary of repeated observations (a law). b) This cannot be a theory because it does not provide an explanation for flammability. A single example of a flammable material that does not contain oxygen will invalidate this as an observation that is always made and a law. Hydrogen is the simplest example of a flammable substance that does not contain oxygen. Others include fuels like natural gas and propane. This leaves none of these as the correct choice. c) This is a simple observation. It is not a scientific law because it is not always observed; a can of soda may be flat or not highly carbonated so no fizzing is heard. It does not provide any explanation so it cannot be a theory. d) No explanation is provided so it cannot be a theory. It is an observation that is repeatedly made (there are no known exceptions) so it represents a scientific law. Check for Understanding 1.2 1. In Figure 1.6, which quantity is the independent variable and which is the dependent variable? Answers: volume is the independent variable pressure is the dependent variable Solution In the absence of additional information you should assume that the independent variable is plotted on the x-axis and the dependent variable is plotted on the y-axis. So volume is the independent variable and pressure is the dependent variable. Check for Understanding 1.3 1. The equation for the straight-line graph in Figure 1.5 is y = 0.0038x - 0.047. Determine the Kelvin temperature (K) at which the pressure of this gas sample equals 0.94 atm. Answer: 260 K Solution The equation for the straight line in Figure 1.5 can be expressed as: pressure (atm) = 0.0038(temp (K)) - 0.047 Substituting 0.94 atm for the pressure and solving for temperature yields: temp (K)  2. 0.94  0.047  260 0.0038 Imagine doing an experiment in which you burn a candle and measure its diminishing mass at various times. a) Answer: If you plot candle mass (in grams) on the y-axis and time (in minutes) on the x-axis and fit the data with a straight line, what are the units for the slope of this line? g/min Solution y2  y1 so the units of the slope are the x2  x1 units of the y-axis variable (g) divided by the units of the x-axis variable (min). In this case the ratio of units is g/min. The slope of a straight-line graph is given by b) Answer: Do you expect the slope of this straight line to be positive or negative? Explain your answer. negative Solution Think about how this graph might look. As the candle burns its mass decreases so a plot of candle mass versus time might look like that below. Since it slopes downward from left to right, the slope is negative. You can also see this by noting that (y2 - y1) is positive but (x2 - x1) is negative so the ratio of these quantities is negative. 12 (x1, y1) Candle Mass (g) 10 x 2 - x1 (x2, y2) 8 y2 - y1 6 4 2 0 0.0 10.0 20.0 30.0 40.0 Time (min) 50.0 60.0 70.0 Chapter 2 Check for Understanding 2.1 1. Write the solution map for the conversion of pm to km. Indicate the numerical ratio that is the conversion factor in each step. Answers: The solution map is pm ö m ö km The conversion factor for the first step is 10-12 m . 1pm The conversion factor for the second step is 1km . 103 m Solutions Each step of the solution map requires a conversion factor. For pm ö m, we need a conversion factor with units of m which will cancel pm and pm introduce m into the numerator. Since 1 pm = 10-12 m we use 10-12 m . 1pm For m ö km, we need a conversion factor with units of Since 1 km = 103 m we use 1km . 103 m km to cancel m. m 2. Convert 233 kg to an equivalent number of micrograms. 2.33 x 1011 μg Answer: Solution First create the appropriate solution map. kg ö g ö μg Each step requires a conversion factor. For kg ö g, we need a conversion factor with units of g which will cancel kg and kg introduce g into the numerator. 103 g Since 1 kg = 10 g we use . 1kg 3 For g ö μg, we need a conversion factor with units of Since 1 μg = 10-6 g we use μg to cancel g. g 1μg . 10-6 g Now multiply 233 kg by both unit ratios and cancel common terms in both numerator and denominator. 103 g 1μg x -6 233 kg x = 2.33 x1011 μg 10 g 1 kg 3. Find the correct numerical value of x. 9.15 x 104 Answer: cycles ms Solution First create the appropriate solution map. cycles s ö cycles ms 9.15 x107 cycles x cycles  s ms We need a conversion factor with units of s which will cancel s and introduce ms into ms the denominator. Since 1 ms = 10-3 g we use 10-3 s . 1ms 9.15 x107 cycles Now multiply by this unit ratio and cancel common terms in both s numerator and denominator. 9.17 x107 cycles 10-3 s cycles x = 9.15 x104 s 1ms ms Note that you would expect to have fewer cycles in the shorter time interval. 4. Write each of the following in proper scientific notation. a) 0.000000203 Answers: b) 12,918 a) 2.03 x 10-7 b) 1.2918 x 104 Solutions a) First express the original quantity as a number between 1 and 10. 0.000000203 ö 2.03 Then multiply by the appropriate power of ten. This is determined by noting how many places, and in which direction, the decimal point was moved in the first step. Notice that we moved the decimal point 7 places to the right. This means that the power of ten to use is 7. However, since the decimal point was moved to the right, thereby making the number larger, the power of ten must be negative so that you are multiplying by a value smaller than 1; that is, 10-7. 0.000000203 ö 2.03 x 10-7 b) First express the original quantity as a number between 1 and 10. 12,918 ö 1.2918 Then multiply by the appropriate power of ten. This is determined by noting how many places, and in which direction, the decimal point was moved in the first step. Notice that we moved the decimal point 4 places to the left. In doing so, the original quantity has been reduced by 4 powers of ten (12,918 ö 1.2918). In order to restore the quantity to its original value you must multiply by 4 powers of ten; that is, 104. 12,918 ö 1.2918 x 104 Check for Understanding 2.2 1. How many significant figures are in each of the following measurements? a) 588.0 kg Answers: b) 12,000 miles c) 0.00700100 s a) 4 sig. fig. b) 2 sig. fig. c) 6 sig. fig. Solutions Recall that all digits in a measurement are significant except trailing zeros in numbers without a decimal point and all leading zeros. a) Because of the decimal point, the trailing zero is significant so there are 4 significant figures. b) Because there is no decimal point, the trailing zeros are not significant so there are just 2 significant figures. c) The leading zeros are not significant, but because of the decimal point the trailing zeros are significant, so there are 6 significant figures. 2. Write each of the quantities in question 1 in proper scientific notation. Answers: a) 5.880 x 102 kg b) 1.2 x 104 miles c) 7.00100 x 10-3 s Solutions Recall that when measurements are written in proper scientific notation all significant digits are shown. a) First express the original quantity as a number between 1 and 10. 588.0 ö 5.880 The trailing zero is included because it is significant. Then multiply by the appropriate power of ten. This is determined by noting how many places, and in which direction, the decimal point was moved in the first step. Notice that we moved the decimal point 2 places to the left. In doing so, the original quantity has been reduced by 2 powers of ten (588.0 ö 5.880). In order to restore the quantity to its original value you must multiply by 2 powers of ten; that is, 102. 588.0 kg ö 5.880 x 102 kg b) Note that 4 significant figures are shown. First express the original quantity as a number between 1 and 10. 12,000 ö 1.2 The trailing zeros are not included because they are not significant. Then multiply by the appropriate power of ten. This is determined by noting how many places, and in which direction, the decimal point was moved in the first step. Notice that we moved the decimal point 4 places to the left. In doing so, the original quantity has been reduced by 4 powers of ten (12,000 ö 1.2). In order to restore the quantity to its original value you must multiply by 4 powers of ten; that is, 104. 12,000 miles ö 1.2 x 104 miles Note that 2 significant figures are shown. c) First express the original quantity as a number between 1 and 10. 0.00700100 ö 7.00100 The trailing zeros are included because they are significant. Then multiply by the appropriate power of ten. This is determined by noting how many places, and in which direction, the decimal point was moved in the first step. Notice that we moved the decimal point 3 places to the right. This means that the power of ten to use is 3. However, since the decimal point was moved to the right, thereby making the number larger, the power of ten must be negative so that you are multiplying by a value smaller than 1; that is, 10-3. 0.00700100 s ö 7.00100 x 10-3 s Note that 6 significant figures are shown. Check for Understanding 2.3 1. For each of the following, calculate the result and round it off to the proper number of significant figures. Units have been omitted for clarity. a) 28.4 ÷ 0.0091 Answers: b) 63.2 - 61.04 a) 3.1 x 103 b) 2.2 Solutions a) The calculated result is: 28.4 = 3120.8... 0.0091 Note the number of significant figures in each measurement in this calculation. 28.4 (3 sig. fig.) 0.0091 (2 sig. fig.) Since 2 significant figures is the fewest number in any measurement, the result is rounded off to 2 significant figures. Since the first digit to be dropped is less than 5, the preceding digit stays the same. 3120.8... ö 3100 or 3.1 x 103 b) The calculated result is 63.2 ________ - 61.04 2.16 ö 2.2 Note that the uncertain digit (in blue) in the first measurement is in the tenths place while the uncertain digit in the second measurement is in the hundredths place. Thus, the sum is rounded off to the (larger) tenths place. Since the first digit to be dropped is greater than 5, the preceding digit is increased by 1.