01 homework Due: Jan 26 2007, 4:00 am cube. Question 1, chap 1, sect 5. part 1 of 3 5 points A solid aluminum cube has sides each of length L . A second cube of the same material has sides three times the length of the first cube, i.e., 3 L . Compared to the first cube, the density of the second cube is 1. four times as much as the first cube. 2. the same as the first cube. correct 3. twenty-four times as much as the first cube. 4. sixteen times as much as the first cube. 5. two times as much as the first cube. 6. eight times as much as the first cube. 7. nine times as much as the first cube. 8. twenty-seven times as much as the first cube. 9. None of these 10. sixty-four times as much as the first cube. Explanation: Density is defined as mass per unit volume and therefore is a property of the material, in this case aluminum, not the shape or size and so remains the same. Question 2, chap 1, sect 5. part 2 of 3 5 points Compared to the first cube, the weight of the second cube is 1. twenty-seven times as much as the first cube. correct 2. sixty-four times as much as the first 3. the same as the first cube. 4. two times as much as the first cube. 5. four times as much as the first cube. 6. eight times as much as the first cube. 7. nine times as much as the first cube. 8. None of these 9. sixteen times as much as the first cube.
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10. twenty-four times as much as the first cube. Explanation: The first cube has a volume of V1 = L L L = L3 . The second cube has a volume of V2 = 3 L 3 L 3 L = (3 L)3 = 27 L3 . The ratio of the surface areas is V2 27 L3 = = 27 . V1 L3 The second cube has 27 times the volume of the first cube and so 27 times the mass because the density is defined as mass per unit volume. The second cube has 27 times the weight because it has 27 times the mass. Question 3, chap 1, sect 5. part 3 of 3 5 points Compared to the first cube, the total surface area of the second cube is 1. twenty-four times as much as the first cube. 2. nine times as much as the first cube. correct 3. sixteen times as much as the first cube.
homework 01 Due: Jan 26 2007, 4:00 am 4. the same as the first cube. 5. twenty-seven times as much as the first cube. 6. eight times as much as the first cube. 7. sixty-four times as much as the first cube. 8. four times as much as the first cube. 9. None of these 10. ninty-six times as much as the first cube. Explanation: The surface of one side of the first cube is A1 = L L = L2 . Every cube has 6 sides so the total surface area of the first cube is A 1 = 6 L2 . The surface area of one side of the second cube is A2 = 3 L 3 L = 9 L2 . So the total surface area of the second cube is A2 = 6 9 L2 = 54 L2 . The ratio of the surface areas is A2 54 L2 = = 9. A1 6 L2 Question 4, chap 1, sect 5. part 1 of 1 10 points A newly discovered Jupiter-like planet has an average radius 10.5 times that of the Earth and a mass 311 times that of the Earth. Calculate the ratio of new planet's mass density to the mass density of the Earth. Correct answer: 0.268653 (tolerance 1 %). Explanation: Question 5, chap 1, sect 6. part 1 of 1 8 points
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the ratio of the body's mass to its volume, = M/V . Planets are spherical, so the (average) density of a planet of a given mass M and a given radius R is = M . R3
4 3
Comparing the newly discovered planet to the Earth, we have np = E = Mnp 3 Rnp ME 3 RE
3
4 3
4 3
Mnp ME 311 = (10.5)3
Rnp RE
= 0.268653 .
An acceleration of 3.9 mi/h/s is equal to: Correct answer: 1.74308 m/s2 (tolerance 1 %). Explanation:
a = 3.9 mi/h/s 3.9 mi 1.609km 1000 m 1h = hs 1 mi 1 km 3600 s = 1.74308 m/s2
Question 6, chap 1, sect 6. part 1 of 3 5 points There are roughly 1059 neutrons and protons in an average star and about 1011 stars in a typical galaxy. Galaxies tend to form in clusters of (on the average) about 103 galaxies, and there are about 109 clusters in the known part of the Universe. Approximately how many neutrons and protons "#" are there in the known Universe?
Let : Rnp = 10.5 RE and Mnp = 311 ME . The volume of a sphere of radius R is V = 4 R3 . The (average) density of a body is 3
homework 01 Due: Jan 26 2007, 4:00 am 1. # 1052 2. None of these 3. # 10 4. # 10
87 82
3
6. None of these Explanation: Let : Vp = 1.401 10-45 m3
correct
5. # 1047 Explanation: Let : Nn = 1059 , Ns = 1011 , Ng = 103 and Nc = 10 . The number of particles in the observable Universe equals the product of the numbers of particles in each astrophysical unit NnU = Nn Ns Ng Nc = 1059 1011 103 109 = 1082 neutrons and protons . Question 7, chap 1, sect 6. part 2 of 3 5 points Suppose all this matter were compressed into a sphere of nuclear matter such that each nuclear particle occupied a volume of 1.401 10-45 m3 (which is approximately the "volume" of a neutron or proton). What would be the radius of this sphere of nuclear matter? 1. R 1012 m correct 2. R 1023 m 3. R 1025 m 4. R 1014 m 5. R 10
35 9
The volume of the sphere would equal the product of the number of protons and neutrons in the observable Universe and the volume of such particles; i.e., NnU Vp =
1 3
6. # 1043
4 r3 3
r= =
3 NnU Vp 4 3 1082
1.401 10-45 4
1 3
1012 m . Question 8, chap 1, sect 6. part 3 of 3 5 points Avogadro's number is 6.023 1023 mol-1 . How many moles of nuclear particles are there in the observable Universe? 1. # 1018 mol 2. # 1058 mol correct 3. # 1054 mol 4. # 1056 mol 5. None of these 6. # 1035 mol Let : NA = 6.023 1023 .
Explanation:
There are NA particles in a mole, so the number of moles of the particles in the observable Universe amounts to n= NnU NA
m
homework 01 Due: Jan 26 2007, 4:00 am = 1082 6.023 1023 mol-1 1058 mol . 3. None of these 4. One significant figure (0.9 m/s)
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Question 9, chap 1, sect 6. part 1 of 4 5 points Use significant figures to calculate the following: a) Find the sum of the measurements 755 g, 37.3 g, 0.82 g, and 2.5 g. 1. One significant figure (800 g) 2. Two significant figures (800 g) 3. Whole number (796 g) correct 4. Hundredths (795.62 g)
5. Two significant figures (0.90 m/s) correct 6. Three significant figures (0.898 m/s) 7. Four significant figures (0.8979 m/s) 8. Thousandths (0.898 m/s) Explanation: 3.2 m = 0.897868 m/s = 0.90 m/s 3.564 s The least number of significant figures is 2.
5. None of these 6. Tens (800 g) 7. Tenths (795.6 g) 8. Three significant figures (796 g) Explanation: Basic Concept: Addition/subtraction: least precise measurement Multiplication/division: least significant figures in measured quantities Solution: 755 g + 37.3 g + 0.82 g + 2.5 g = 795.62 g = 796 g The least precise measurement is a whole number of grams. Question 10, chap 1, sect 6. part 2 of 4 5 points 3.2 m . b) Find the quotient 3.564 s 1. Tenths (0.9 m/s) 2. Hundredths (0.90 m/s) (4.61 mm)() = 14.4827 mm = 14.5 mm Question 11, chap 1, sect 6. part 3 of 4 5 points c) Find the product of 4.61 mm and . 1. Five significant figures (14.483 mm) 2. None of these 3. Three significant figures (14.5 mm) correct 4. Two significant figures (14 mm) 5. Whole number (14 mm) 6. Thousandths (14.483 mm) 7. Four significant figures (14.48 mm) 8. Hundredths (14.48 mm) 9. Tenths (14.5 mm) Explanation:
homework 01 Due: Jan 26 2007, 4:00 am The least number of significant figures is 3 ( is a constant, not a measured quantity). Question 12, chap 1, sect 6. part 4 of 4 5 points d) Find the difference of 27.56 s and 3.2 s. 1. Tens (20 s) 2. Hundredths (24.36 s) 3. Whole number (24 s) 4. One significant figure (20 s) 5. Two significant figures (24 s) 6. None of these 7. Tenths (24.4 s) correct 8. Three significant figures (24.4 s) 9. Four significant figures (24.36 s) Explanation: Therefore, A= V = 1.24263 109 m2 t
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Question 14, chap 1, sect 99. part 1 of 1 8 points A 19th century British naturalist with a penchant for archaic units of measurement described a species of snail crawling at average speed of one furlong per fortnight. (A furlong is one eighth of 1 mile or 220 yards; a yard is 3 feet or 0.9144 meter; a fortnight is a time interval of 14 days or 14 24 hours.) Recently, a biology student re-measured the snail's average speed and reported it as one centimeter per minute. Which of the following is the most likely explanation of the difference between the two measurements? Hint: Use common sense. 1. Evolution in action: even the snails got somewhat faster than they used to be. 2. The student got a much slower species of snail than the one described by the naturalist. 3. The student got a much faster species of snail than the one described by the naturalist. 4. The student's snails are crawling at exactly the same snail's pace they ever did, but he reported a slightly different value for their speed because he rounded it up in different units. correct 5. The student has smoked too much weed and lost all sense of time; his measurements are garbage. 6. Pollution in action: the snails became somewhat slower than they used to be because of some environmental toxins. Explanation: Let us convert the archaic units of the 19th century naturalist's into cm/minutes
27.56 s - 3.2 s = 24.36 s = 24.4 s The least precise measurement a is tenth of a second. Question 13, chap 1, sect 99. part 1 of 1 8 points Assume that an oil slick consists of a single layer of molecules and that each molecule occupies a cube 0.441 nm on a side. Determine the area of an oil slick formed by 0.548 m3 of oil. Correct answer: 1.24263 109 m2 (tolerance 1 %). Explanation: If we call the thickness of each molecule t, and the area of the oil slick A, then the total volume V of the oil is V = A t.
homework 01 Due: Jan 26 2007, 4:00 am and then compare the two measurements: v = 1 furlong/fortnight 220 yards = 14 days (220)(0.9144) m = (14)(24) hours (220)(0.9144)(100) cm = (14)(24)(60) minutes = 0.9979 cm/min. Since both the naturalist and the student rounded off their results (albeit in different units), it is clear that the two measurements agree with each other and the snail's pace has not changed at all. Question 15, chap 1, sect 6. part 1 of 1 10 points Water flows into a swimming pool at the rate of 12.8 gal/min. If the pool dimensions are 21.6 ft wide, 41.6 ft long and 11.2 ft deep, how long does it take to fill the pool? (1 gallon = 231 cubic inches) Correct answer: 5881.48 min (tolerance 1 %). Explanation: Basic Concepts The volume of the pool is given by V = lwh Solution The flow rate is v= Thus t= V 75283 gal = = 5881.48 min v 12.8 gal/min V t
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A water holding tank measures 120 m long, 45 m wide, and 12 m deep. Traces of mercury have been found in the tank, with a concentration of 50 mg/L. What is the total mass of mercury in the tank? Correct answer: 32400 kg (tolerance 1 %). Explanation:
Let :
l = 120 m , w = 45 m , d = 12 m , and = 50 mg/L .
The volume of the tank is V = lwd = (120 m) (45 m) (12 m) = 64800 m3 . Since the density is expressed in milligrams per liter, let's convert this volume to liters. V = 64800 m3 The mass is m = V = (50 mg/L) (6.48 107 L) = 3.24 109 mg = 32400 kg . 1000 L = 6.48 107 L . m3
Question 17, chap 1, sect 5. part 1 of 1 10 points One cubic meter (1.0 m3 ) of aluminum has a mass of 2700 kg, and a cubic meter of iron has a mass of 7860 kg. Find the radius of a solid aluminum sphere that has the same mass as a solid iron sphere of radius 3.71 cm. Correct answer: 5.29737 cm (tolerance 1 %). Explanation:
Dimensional analysis of V : in ft 12 ft
3 3
1 gal 231 in3
= gal
Question 16, chap 1, sect 6. part 1 of 1 0 points
homework 01 Due: Jan 26 2007, 4:00 am Explanation: Basic Concepts: and v= x t
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Let : mAl = 2700 kg , mFe = 7860 kg , rFe = 3.71 cm .
m Density is = . Since the masses are the V same, Al VAl = Fe VFe 4 4 3 rAl = Fe r3 3 3 Fe rAl rFe
3
Al
Solution: The average speed of the second car is given by the distance d divided by the time it takes the second car to travel that distance d v2 = . t2 t2 is given by t1 , the time it takes the first car to travel the distance d, minus the time difference in which the two cars started t2 = t1 - tdiff .
=
Fe Al Fe Al
1 3
rAl = rFe
= (3.71 cm)
7860 kg 2700 kg
1 3
We know that t1 is given by t1 = therefore d , v1 d d - tdiff v1 .
= 5.29737 cm . Question 18, chap 2, sect 2. part 1 of 1 8 points A glacier moves with a speed of 33 nm/s. How many years would it take for the glacier to move 0.65 km? Correct answer: 622.424 yr (tolerance 1 %). Explanation: The time can be calculated as d 650 m t= = v 3.3 10-8 m/s = 1.9697 1010 s = 622.424 yr. Question 19, chap 2, sect 1. part 1 of 1 10 points A car makes a 204 km trip at an average speed of 32.5 km/h. A second car starting 1 h later arrives at their mutual destination at the same time. What was the average speed of the second car for the period that it was in motion? Correct answer: 38.6589 km/h (tolerance 1 %). v2 =
Question 20, chap 1, sect 6. part 1 of 1 8 points A graph of a straight line going through two points is shown below. y 5 4 3 2 1 0 -1 -2 -3 -4 -5 -5
x
-3
-1 0 1 2 3 4 5
What is the equation of this line? 2 7 1. y = - x + 3 3
homework 01 Due: Jan 26 2007, 4:00 am 7 2 2. y = + x + 3 7 3 2 3. y = - x + 7 3 7 2 4. y = + x - 3 3 3 2 5. y = + x + 7 7 2 7 6. y = - x - 3 7 7 2 7. y = + x - 3 7 3 2 8. y = + x + 7 3 7 2 9. y = + x + correct 3 3 2 3 10. y = - x - 7 3 Explanation: Solution: Let (x1 , y1 ) = (-2, -4) (x2 , y2 ) = (1, 3) . The slope is m= (3) - (-4) (1) - (-2) 7 =+ . 3
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Note: This problem has a different line for each student. keywords: Question 21, chap 2, sect 1. part 1 of 2 7 points A person travels by car from one city to another. She drives for 37.1 min at 81.4 km/h, 15.3 min at 118 km/h, 51.3 min at 48.3 km/h, and spends 16.6 min along the way eating lunch and buying gas. Determine the distance between the cities along this route. Correct answer: 121.719 km (tolerance 1 %). Explanation: Distances traveled are x1 = v1 t1 x2 = v2 t2 x3 = v3 t3 Then the total distance traveled is x = x1 + x2 + x3 = (50.3323 km) + (30.09 km) + (41.2965 km) = 121.719 km . Question 22, chap 2, sect 1. part 2 of 2 7 points Determine the average speed for the trip. Correct answer: 60.7076 km/h (tolerance 1 %). Explanation: And, time spent is t = t1 + t2 + t3 + tother = (37.1 min) + (15.3 min) + (51.3 min) + (16.6 min) = 2.005 h . (3) Hence vav = x t 121.719 km = 2.005 h = 60.7076 km/h .
(2)
Using 2 points (y - y1 ) =m (x - x1 ) (y - y1 ) = m(x - x1 ) y = m(x - x1 ) + y1 7 T hus y = + (x + 2) - 4 3 7 2 =+ x+ , 3 3
7 where the slope of the line is + and the 3 2 y-intercept is + (as can be seen from the 3 graph).
homework 01 Due: Jan 26 2007, 4:00 am Basic Concept: Question 23, chap 2, sect 2. part 1 of 1 10 points The velocity of the transverse waves produced by an earthquake is 5.6 km/s, while that of the longitudinal waves is 9.128 km/s. A seismograph records the arrival of the transverse waves 65.7 s after that of the longitudinal waves. How far away was the earthquake? Correct answer: 951.92 km (tolerance 1 %). Explanation: If the distance to the earthquake is d, then the time lag is t = d d (vl - vt ) d - = vt vl vl vt (average speed) = (total distance traveled) time
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The total distance traveled by the student is 100 miles from Austin to Waco, plus another 100 miles from Waco back to Austin, plus 80 miles from Austin to San Antonio, or 280 miles altogether. His average speed therefore is |v| = L 280 miles = = 46.1285 MPH t 6.07 hours
Question 25, chap 2, sect 1. part 2 of 2 8 points What was the student's average velocity during his trip? Take your positive direction to be southbound on I35. Correct answer: 13.1796 MPH (tolerance 1 %). Explanation: Basic Concept: (net displacement) time
t vl vt d= vl - vt (65.7 s) (9.128 km/s) (5.6 km/s) = (5.6 km/s) - (9.128 km/s) = 951.92 km . Note: The longitudinal wave travels faster than the transverse waves. In fact, the ratio of the longitudinal velocity to transverse the wave velocity should be about 3 . Question 24, chap 2, sect 1. part 1 of 2 8 points A student wanted to drive from Austin to San Antonio, 80 miles south of Austin on highway I35. Unfortunately, he entered the highway in the wrong direction and drove all the way to Waco -- 100 miles north of Austin -- before he noticed his error. In Waco, he turned around, drove back to Austin and continued to San Antonio. The whole trip took 6.07 hours. What was the student's average speed during this trip? Correct answer: 46.1285 MPH (tolerance 1 %). Explanation:
(average velocity) =
The student's net displacement is from Austin to San Antonio -- the fact that he went their by way of Waco is irrelevant for this question -- or 80 miles in the positive southbound direction. His average velocity is therefore X 80 miles = = 13.1796 MPH t 6.07 hours
v=
Question 26, chap 2, sect 1. part 1 of 1 6 points The position-versus-time graph below describes the motion of three different bodies (labelled 1, 2, 3).
homework 01 Due: Jan 26 2007, 4:00 am x position ( 5 m) xA A 3 6 5 4 3 2 1 0 tA tB t 0 1 2 3 4 5 6 time ( 5 s) 7 8
10
1 xB
2
B
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Consider the average velocities of the three bodies. Which of the following statements is correct? 1. v1 > v2 and v3 > v2 2. v1 < v2 < v3 3. v1 > v2 > v3 4. v1 = v2 = v3 correct Explanation: The definition of average velocity is
What is the average velocity of motion depicted on this graph between times tinit = 0 s and tfin = 45 s? Correct answer: 0.333333 m/s (tolerance 1 %). Explanation: Looking at the graph -- and carefully paying attention to the units -- we see that at time tinit = 0 s the body in question was at xinit = 0 m, while at time tfin = 45 s it was at xfin = 15 m. Overall, we have net displacement x = xfin - xinit = 15 m during time t = tfin - tinit = 45 s. Consequently, the average velocity is v= x 15 m = = 0.333333 m/s. t 45 s
v=
displacement time xB - xA . = tB - tA
Note that the average velocity does not depend on the details of the motion, but only on the net displacement and the net time it took.
All three bodies have exactly same displacement in exactly same time, hence all three average velocities are exactly equal, v1 = v2 = v3 . Question 27, chap -1, sect -1. part 1 of 1 10 points The scale on the horizontal axis is 5 s per division and on the vertical axis 5 m per division.