Kaplunovsky - hw 01
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Kaplunovsky - hw 01

Course: PHY 303K, Spring 2007

School: University of Texas

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homework 01 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...

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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. 1 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 2 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) 4 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 5 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 6 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 7 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 8 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 9 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 9 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.

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University of Texas - PHY - 303K
homework 02 Due: Jan 31 2007, 4:00 pm Question 1, chap 2, sect 3. part 1 of 4 5 points The position versus time for a certain object moving along the x-axis is shown. The object's initial position is -3 m. 6 position (m) 4 2 0 -2 -4 0 1 2 3 4 5 tim
University of Texas - PHY - 303K
homework 03 Due: Feb 7 2007, 4:00 am Question 1, chap 2, sect 6. part 1 of 1 5 points An object is shot vertically upward into the air with a positive initial velocity. Which of the following correctly describes the velocity and acceleration of the
University of Texas - PHY - 303K
homework 04 Due: Feb 14 2007, 4:00 am Question 1, chap -1, sect -1. part 1 of 3 5 points A projectile is shot on level ground with a horizontal velocity of 12 m/s and a vertical velocity of 22 m/s . The acceleration of gravity is 9.8 m/s2 . y 22 m/
University of Texas - PHY - 303K
homework 05 Due: Feb 22 2007, 4:00 am Question 1, chap 5, sect 2. part 1 of 1 6 points An astronaut who weighs 554.9 N on the surface of the earth lifts off from planet Zuton in a space ship. The free-fall acceleration on Zuton is 5.8 m/s2 . At the
University of Texas - PHY - 303K
homework 06 Due: Mar 1 2007, 4:00 am Question 1, chap 6, sect 1. part 1 of 1 10 points A 930 kg car moves along a horizontal road at speed v0 = 22 m/s. The road is wet, so the static friction coefficient between the tires and the road is only s = 0
University of Texas - PHY - 303K
homework 07 Due: Mar 7 2007, 4:00 am Solving for the frictional force, f , Question 1, chap 7, sect 1. part 1 of 1 6 points Stan does 151 J of work lifting himself 0.1 m. The acceleration of gravity is 9.8 m/s2 . What is Stan's mass? Correct answer
University of Texas - PHY - 303K
homework 08 Due: Mar 22 2007, 4:00 am Explanation: Question 1, chap 7, sect 2. part 1 of 2 5 points The force required to stretch a Hooke's-law spring varies from 0 N to 22 N as we stretch the spring by moving one end 6.58 cm from its unstressed po
University of Texas - PHY - 303K
homework 09 Due: Mar 29 2007, 4:00 am Question 1, chap 10, sect 2. part 1 of 1 10 points Consider a 31 m high pyramide with a square base 48 m by 48 m. The pyramid is made of tightly-fitting stones of uniform density 2730 kg/m3 . Calculate the net
University of Texas - PHY - 303K
homework 10 Due: Apr 4 2007, 4:00 am 5 kg Question 1, chap 12, sect 5. part 1 of 1 6 points The rigid object shown is rotated about an axis perpendicular to the paper and through center point O. 68 kg 8m O 16 m 8m 34 kg 16 m 5 kg 8m 6m y1 5 kg
University of Texas - PHY - 303K
homework 11 Due: Apr 12 2007, 4:00 am Question 1, chap 13, sect 2. part 1 of 2 8 points A spool of wire of mass M = 5.4 kg and radius R = 0.64 m is unwound under a constant wire tension F = 4.3 N. Assume the spool is a uniform solid cylinder that r
University of Texas - PHY - 303K
homework 12 Due: Apr 19 2007, 4:00 am Question 1, chap 3, sect 4. part 1 of 3 5 points Consider the two vectors M = (a, b) = a ^+ i b ^ and N = (c, d) = c ^ + d ^ where, a and i c represent the x-displacement and b and d represent the y-displacem
University of Texas - PHY - 303K
homework 13 Due: Apr 25 2007, 4:00 am Question 1, chap 11, sect 3. part 1 of 3 5 points A clock balance wheel has a period of oscillation of 0.47 s. The wheel is constructed so that very nearly all of its 33 g of mass is concentrated around rim of
University of Texas - PHY - 303K
homework 14 Due: May 2 2007, 4:00 am 3 Question 1, chap 16, sect 3. part 1 of 1 8 points You are given two waves, a transverse wave that moves to the right f1 (x) and a transverse wave that moves to the left f2 (x), on a string. As the problem begi
University of Texas - PHY - 303K
practice 02 Due: Jan 29 2007, noon Question 1, chap 2, sect 5. part 1 of 1 10 points A sailboat is initially moving at a speed of 2 m/s. A strong wind blows up and accelerates the boat forward with a constant acceleration of 0.1 m/s2 for 10 s. What
University of Texas - PHY - 303K
practice 03 Due: Feb 5 2007, noon Question 1, chap 2, sect 7. part 1 of 3 5 points The position of a softball tossed vertically upward is described by the equation y = c1 t - c2 t2 , where y is in meters, t in seconds, c1 = 3.72 m/s, and c2 = 5.49
University of Texas - PHY - 303K
practice 04 Due: Feb 12 2007, noon Question 1, chap 4, sect 3. part 1 of 2 10 points A particle travels horizontally between two parallel walls separated by 18.4 m. It moves toward the opposing wall at a constant rate of 8.6 m/s. Also, it has an ac
University of Texas - PHY - 303K
practice 05 Due: Feb 20 2007, noon Question 1, chap 5, sect 2. part 1 of 1 5 points A 4100 kg helicopter accelerates upward at 1.8 m/s2 . The acceleration of gravity is 9.8 m/s2 . What lift force is exerted by the air on the propellers? Correct ans
University of Texas - PHY - 303K
practice 06 Due: Feb 26 2007, noon Question 1, chap 6, sect 1. part 1 of 1 5 points Consider three sheets of paper, all of the same size and weight. One sheet is un-folded, the other is folded in two, and the third is wadded into a ball. When you d
University of Texas - PHY - 303K
practice 07 Due: Mar 5 2007, noon Consequently, the friction force Question 1, chap -1, sect -1. part 1 of 5 5 points A 8.37 kg block is pushed 9.4 m up a vertical wall with constant speed by a constant force F applied at an angle of 58 with the ho
University of Texas - PHY - 303K
practice 08 Due: Mar 20 2007, noon Question 1, chap -1, sect -1. part 1 of 3 8 points Two blocks of respective masses m1 = 0.66 kg and m2 = 3.9 kg stand without motion on a frictionless horizontal table. A spring is compressed between the two block
University of Texas - PHY - 303K
practice 09 Due: Mar 27 2007, 1:00 pm Question 1, chap 11, sect 3. part 1 of 1 8 points A disc of mass m moves horizontally to the right with speed v on a table with negligible friction when it collides with a second disc of mass 7 m. The second di
University of Texas - PHY - 303K
practice 10 Due: Apr 2 2007, noon Question 1, chap 12, sect 5. part 1 of 1 10 points Calculate the moment of inertia for 4m L Axis L m L 2m L 3m L1perpendicular to the paper and through the center of mass? Correct answer: 0.624443 kg m2 (tolera
University of Texas - PHY - 303K
practice 11 Due: Apr 10 2007, noon Question 1, chap 13, sect 2. part 1 of 1 10 points A constant horizontal force of 120 N is applied to a lawn roller in the form of a uniform solid cylinder of radius 0.35 m and mass 13 kg. The acceleration of grav
University of Texas - PHY - 303K
practice 12 Due: Apr 17 2007, noon Therefore, using Question 1, chap 3, sect 4. part 1 of 2 7 points Given: Two vectors A = Ax ^ + Ay ^ i and B = Bx ^ + By ^ , i where Ax = -2, Ay = 1, Bx = 2, and By = 1. Find the z component of A B. Correct ans
University of Texas - PHY - 303K
practice 13 Due: Apr 23 2007, noon Question 1, chap 11, sect 3. part 1 of 1 5 points The equation of motion for a torsion pendulum (in small angle oscillation) is given by: : where = I = - , we find that = . I1Compare with the standard simpl
University of Texas - PHY - 303K
practice 14 Due: Apr 30 2007, noon 3 Question 1, chap 16, sect 3. part 1 of 1 8 points You are given two waves, a transverse wave that moves to the right f1 (x) and a transverse wave that moves to the left f2 (x), on a string. As the problem begins
University of Texas - PHY - 303K
practice 15 Due: May 8 2007, noon Question 1, chap 18, sect 3. part 1 of 1 5 points The pressure exerted on the ground by a man is greatest when: 1. He stands with both feet flat on the ground. 2. He stands on the toes of one foot. correct 3. The p
University of Texas - PHY - 303K
midterm 02 Due: Mar 7 2007, 10:00 pm Question 1, chap 5, sect 99. part 1 of 2 5 points A spherical mass rests upon two wedges, as seen in the figure below. The sphere and the wedges are at rest and stay at rest. There is no friction between the sph
University of Texas - PHY - 303K
midterm 03 Due: Apr 4 2007, 11:00 pm Question 1, chap 8, sect 1. part 1 of 1 10 points Consider a bungee cord of unstretched length L0 = 35 m. When the cord is stretched to L &gt; L0 it behaves like a spring and its tension follows the Hooke's law T =
University of Texas - PHY - 303K
midterm 04 Due: May 2 2007, 11:00 pm the torque Question 1, chap -1, sect -1. part 1 of 1 7 points A 3 m long rod of negligible weight is attached on one end to a ball joint which allows the rod to rotate in all directions. The free end of the rod
University of Texas - PHY - 303K
final 01 Due: May 10 2007, 11:00 pm Question 1, chap 1, sect 6. part 1 of 1 6 points Given 1 inch 2.54 cm and 1 foot 12 inches, how many square centimeters are there in 2.48 ft2 ? 1. 75.5904 cm2 2. 1.33333 cm2 3. 2304 cm2 correct 4. 0.0813648 cm2