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Chapter 1: Overview In-Class Exercises 1.1.

a 1.2. a) 4 b) 3 c) 5 d) 6 e) 2 1.3. e 1.4. a) 4th b) 2nd c) 3rd d) 1st

Multiple-Choice 1.1. c 1.2. c 1.3. d 1.4. b 1.5. a 1.6. b 1.7. b 1.8. c 1.9. c 1.10. b

Questions 1.11.

(a) In Europe, gas consumption is in L/100 km. In the US, fuel efficiency is in miles/gallon. Let’s relate these two: 1 mile = 1.609 km, 1 gal = 3.785 L. 1 mile 1.609 km 1.609 æ 1 ö km 1 1 = = ( 100) L = ( 0.00425) æç L/100 km ö÷ = 235.24L/100 km gal 3.785 L 3.785çè 100 ÷ø è ø Therefore, 1 mile/gal is the reciprocal of 235.2 L/100 km. 1L 1 12.2 L = (b) Gas consumption is . Using from part (a), 100 km 235.24 miles/gal 100 km 1 1 æ ö 12.2 L æ 1L ö = 12.2ç ÷ = 12.2ç 235.24 miles/gal ÷ = 19.282 miles/gal . 100 km è 100 km ø è ø Therefore, a car that consumes 12.2 L/100 km of gasoline has a fuel efficiency of 19.3 miles/gal. (c) If the fuel efficiency of the car is 27.4 miles per gallon, then 27.4 miles 27.4 1 = . = gal 235.24 L/100 km 8.59 L/100 km Therefore, 27.4 miles/gal is equivalent to 8.59 L/100 km. (d)

1.12.

A vector is described by a set of components in a given coordinate system, where the components are the projections of the vector onto each coordinate axis. Therefore, on a two-dimensional sheet of paper there are two coordinates and thus, the vector is described by two components. In the real three-dimensional world, there are three coordinates and a vector is described by three components. A fourdimensional world would be described by four coordinates, and a vector would be described by four components.

1.13.

A vector contains information about the distance between two points (the magnitude of the vector). In contrast to a scalar, it also contains information

1

Bauer/Westfall: University Physics, 1E direction. In many cases knowing a direction can be as important as knowing a magnitude. 1.14.

In order to add vectors in magnitude-direction form, each vector is expressed in terms of component vectors which lie along the coordinate axes. The corresponding components of each vector are added to obtain the components of the resultant vector. The resultant vector can then be expressed in magnitude-direction form by computing its magnitude and direction.

1.15.

The advantage to using scientific notation is two-fold: Scientific notation is more compact (thus saving space and writing), and it also gives a more intuitive way of dealing with significant figures since you can only write the necessary significant figures and extraneous zeroes are kept in the exponent of the base.

1.16.

The SI system of units is the preferred system of measurement due to its ease of use and clarity. The SI system is a metric system generally based on multiples of 10, and consisting of a set of standard measurement units to describe the physical world. In science, it is paramount to communicate results in the clearest and most widely understood manner. Since the SI system is internationally recognized, and its definitions are unambiguous, it is used by scientists around the world, including those in the United States.

1.17.

It is possible to add three equal-length vectors and obtain a vector sum of zero. The vector components of the three vectors must all add to zero. Consider the following arrangement with T1 = T2 = T3 :

The horizontal components of T1 and T2 cancel out, so the sum T1 + T2 is a vertical vector whose magnitude is T cosq + T cosq = 2T cosq . The vector sum T1 + T2 + T3 is zero if 2T cosq - T = 0 1 cosq = 2 q = 60Â° Therefore it is possible for three equal-length vectors to sum to zero. 1.18.

Mass is not a vector quantity. It is a scalar quantity since it does not make sense to associate a direction with mass.

1.19.

The volume of a sphere is given by V = ( 4/ 3) p r 3. Doubling the volume gives 2V = 2( 4/ 3) p r3 = ( 4/ 3) p (23/3)r3 = ( 4/ 3) p (21/3r)3. Now, since the distance between the flies is the diameter of the sphere, d = 2r , and doubling the volume increases the radius by a factor of 21/3 , the distance between the flies is then increased to 2(21/3r) = 21/3(2r) = 21/3 d. Therefore, the distance is increased by a factor of 21/3.

2

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The volume of a cube of side r is Vc = r 3, and the volume of a sphere of radius r is Vsp = ( 4/ 3) p r 3 . The ratio of the volumes is:

Vc r3 3 = = . Vsp 4 3 4p pr 3 The ratio of the volumes is independent of the value of r. 1.21.

The surface area of a sphere is given by 4p r 2 . A cube of side length s has a surface area of 6s2. To determine s set the two surface areas equal: 6s2 = 4p r2 Þ s =

1.22.

4p r2 2p =r . 6 3

The mass of Sun is 2× 1030 kg, the number of stars in the Milky Way is about 100× 109 = 1011, the number of galaxies in the Universe is about 100× 109 = 1011, and the mass of an H-atom is 2× 10-27 kg. (a) The total mass of the Universe is roughly equal to the number of galaxies in the Universe multiplied by the number of stars in a galaxy and the mass of the average star: M universe = (1011)(1011)(2× 1030) = 2× 10(11+11+ 30) kg = 2× 1052 kg. (b) nhydrogen »

1.23.

M universe 2×1052 kg = = 1079 atoms M hydrogen 2× 10-27 kg

The volume of 1 teaspoon is about 4.93× 10-3 L , and the volume of water in the oceans is about 1.35× 1021 L. 1.35×1021 L = 2.74× 1023 tsp 4.93× 10-3 L/tsp There are about 2.74× 1023 teaspoons of water in the Earth’s oceans.

1.24.

The average arm-span of an adult human is d = 2 m. Therefore, with arms fully extended, a person takes up a circular area of p r2 = p ( d / 2) = p (1 m)2 = p m2. Since 2

there are approximately 6.5× 109 humans, the amount of land area required for all humans to stand without being able to touch each other is 9 2 9 2 10 2 6.5× 10 m (p ) = 6.5× 10 m (3.14) = 2.0× 10 m . The area of the United States is about 3.5×106 square miles or 9.1× 1012 m2 . In the United States there is almost five hundred times the amount of land necessary for all of the population of Earth to stand without touching each other. 1.25.

The diameter of a gold atom is about 2.6× 10-10 m. The circumference of the neck of an adult is roughly 0.40 m. The number of gold atoms necessary to link to make a necklace is given by: circumference of neck 4.0× 10-1 m n= = = 1.5× 109 atoms. diameter of atom 2.6× 10-10 m/atom The Earth has a circumference at the equator of about 4.008× 107 m . The number of gold atoms necessary to link to make a chain that encircles the Earth is given by: circumference of Earth 4.008×107 m N= = = 1.5×1017 atoms. -10 diameter of a gold atom 2.6× 10 m

3

Bauer/Westfall: University Physics, 1E

Since one mole of substance is equivalent to about 6.022× 1023 atoms , the necklace of

(

) ( atoms) / ( 6.022× 10

) atoms/mol ) = 2.5× 10

9 23 -15 gold atoms has 1.5× 10 atoms / 6.022× 10 atoms/mol = 2.5× 10 moles of gold.

(

17 gold chain has 1.5× 10

1.26.

23

-7

The

moles of gold.

The average dairy cow has a mass of about 1.0× 103 kg. Estimate the cow’s average density to be that of water, r = 1000. kg/m3. mass 1.0× 103 kg = = 1.0 m3 r 1000. kg/m3 Relate this to the volume of a sphere to obtain the radius. volume=

(

)

1/3

é 3 1.0 m3 ù ú » 0.62 m =ê ê 4p ú ë û A cow can be roughly approximated by a sphere with a radius of 0.62 m. 4 é 3V ù volume= p r3 Þ r = ê ú 3 ë 4p û

1.27.

1/3

The mass of a head can be estimated first approximating its volume. A rough approximation to the shape of a head is a cylinder. To obtain the volume from the circumference, recall that the circumference is C = 2p r , which gives a radius of r = C / 2p . The volume is then:

(

2

)

C2h æ C ö 2 V = pr h = p ç h = . ÷ 4p è 2p ø The circumference of a head is about 55 cm = 0.55 m, and its height is about 20 cm = 0.20 m. These values can be used in the volume equation:

( 0.55 m)

2

( 0.20 m) = 4.8×10-3 m3. 4p Assuming that the density of the head is about the same as the density of water, the mass of a head can then be estimated as follows: V=

(

)(

)

mass =density × volume = 1.0× 103 kg/m3 4.8× 10-3 m3 = 4.8 kg. 1.28.

The average adult human head is roughly a cylinder 15 cm in diameter and 20. cm in height. Assume about 1/3 of the surface area of the head is covered by hair. 1 1 2p 2 2p é 2 Ahair = Acylinder = 2p r 2 + 2p rh = r + rh = 7.5 cm) + ( 7.5 cm) ( 20. cm) ùú ( ê ë û 3 3 3 3 2 2 » 4.32× 10 cm

(

)

(

)

(

)

On average, the density of hair on the scalp is rhair = 2.3× 102 hairs/cm2. Therefore, you have Ahair ´ dhair hairs on your head.

(

)(

)

Ahair dhair = 4.32× 102 cm2 2.3× 102 hairs/cm2 = 9.9×104 hairs.

Problems 1.29.

(a) Three (b) Four (c) One (d) Six (e) One (f) Two (g) Three

1.30.

THINK: The known quantities are: F1 = 2.0031 N and F2 = 3.12 N. Both F1 and F2 are in the same direction, and act on the same object. The total force acting on the object is Ftotal . SKETCH:

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RESEARCH:

Forces that act in the same direction are summed, Ftotal = å Fi .

SIMPLIFY: Ftotal = å Fi = F1 + F2 CALCULATE: Ftotal = 2.0031 N + 3.12 N = 5.1231 N ROUND: When adding (or subtracting), the precision of the result is limited by the least precise value used in the calculation. F1 is precise to four places after the decimal and F2 is precise to only two places after the decimal, so the result should be precise to two places after the decimal: Ftotal = 5.12 N. DOUBLE-CHECK: This result is reasonable as it is greater than each of the individual forces acting on the object. 1.31.

The result should have the same number of decimal places as the number with the fewest of them. Therefore, the result is 2.0600 + 3.163 + 1.12 = 6.34.

1.32.

In a product of values, the result should have as many significant figures as the value with the smallest number of significant figures. The value for x only has two significant figures, so w = (1.1× 103)(2.48× 10-2)(6.000) = 1.6× 102.

1.33.

Write “one ten-millionth of a centimeter” in scientific notation. One millionth is 6 7 -7 1/106 = 1× 10-6 . Therefore, one ten-millionth is 1/ éë10× 10 ùû = 1/10 = 1× 10 cm.

1.34.

153,000,000 = 1.53× 108

1.35.

There are 12 inches in a foot and 5280 feet in a mile. Therefore there are 63,360 inch/mile. 30.7484 miles · 63,360 inch/ mile = 1948218.624 inches. Rounding to six significant figures and expressing the answer in scientific notation gives 1.94822·10 6 inches.

1.36.

(a) kilo (b) centi (c) milli

1.37.

æ 1000 m öæ 1000 mm ö 6 1 km = 1 kmç ÷ç ÷ = 1,000,000 mm = 1× 10 mm è 1 km øè 1 m ø

1.38.

1 hectare = 100 ares, and 1 are = 100 m2, so: æ ( 1000) 2 m2 ö æ 1 are öæ 1 hectareö ÷ 1 km ç ÷ = 100 hectares. ç 1 km2 ÷ çè 100 m2 ÷ç øè 100 ares ø è ø 2

1.39.

1 milliPascal

1.40.

THINK: The known quantities are the masses of the four sugar cubes. Crushing the sugar cubes doesn’t change the mass. Their masses, written in standard SI units, using scientific notation are m1 = 2.53× 10-2 kg , m2 = 2.47× 10-2 kg , m3 = 2.60× 10-2 kg and m4 = 2.58× 10-2 kg. SKETCH: A sketch is not needed to solve this problem. RESEARCH: 4

(a) The total mass equals the sum of the individual masses: M total = å mj . j =1

(b) The average mass is the sum of the individual masses, divided by the total number of masses:

5

Bauer/Westfall: University Physics, 1E

M average = SIMPLIFY: (a) M total = m1 + m2 + m3 + m4

m1 + m2 + m3 + m4 . 4

M total 4 CALCULATE: M total = 2.53× 10-2 kg + 2.47× 10-2 kg+ 2.60× 10-2 kg + 2.58× 10-2 kg (b) M average =

(a)

= 10.18× 10-2 kg =1.018× 10-1 kg

(b) M average =

10.18× 10-2 kg = 2.545× 10-2 kg 4

ROUND: (a) Rounding to three significant figures, M total = 1.02× 10-1 kg. -2 (b) Rounding to three significant figures, M average = 2.55× 10 kg.

DOUBLE-CHECK: There are four sugar cubes weighing between 2.53× 10-2 kg and 2.60× 10-2 kg, so it is reasonable that their total mass is M total = 1.02× 10-1 kg and their average mass is 2.55× 10-2 kg. 1.41.

THINK: The cylinder has height h = 20.5 cm and radius r = 11.9 cm. SKETCH:

RESEARCH: The surface area of a cylinder is A = 2p rh + 2p r2. SIMPLIFY: A = 2p r(h + r) CALCULATE: A = 2p (11.9 cm)(20.5 cm + 11.9 cm) = 2422.545 cm2 ROUND: Three significant figures: A = 2.42× 103 cm2. DOUBLE-CHECK: The units of area are a measure of distance squared so the answer is reasonable. 1.42.

THINK: When you step on the bathroom scale, your mass and gravity exert a force on the scale and the scale displays your weight. The given quantity is your mass m1 = 125.4 lbs. Pounds can be converted to SI units using the conversion 1 lb = 0.4536 kg. Let your mass in kilograms be m2. SKETCH: A sketch is not needed to solve this problem. æ 0.4536 kg ö RESEARCH: m2 = m1 ç ÷ è 1 lb ø SIMPLIFY: It is not necessary to simplify. æ 0.4536 kg ö CALCULATE: m2 = 125.4 lbsç ÷ = 56.88144 kg lb è ø ROUND: The given quantity and conversion factor contain four significant figures, so the result must be rounded to 56.88 kg. DOUBLE-CHECK: The SI units of mass are kg, so the units are correct.

6

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THINK: The orbital distance from the center of the Moon to the center of the Earth ranges from 356,000 km to 407,000 km. Recall the conversion factor 1 mile = 1.609344 kilometer. SKETCH:

RESEARCH: Let d1 be a distance in kilometers, and d2 the equivalent distance in miles. The formula to convert from kilometers to miles is d2 = d1 /1.609344 . SIMPLIFY: It is not necessary to simplify. 1 mile ö æ 356,000 kmç ÷ = 221208.144 miles è 1.609344 km ø CALCULATE: 1 mile ö æ 407,000 kmç ÷ = 252898.0752 miles è 1.609344 km ø ROUND: The given quantities have three significant figures, so the calculated values must be rounded to 221,000 miles and 253,000 miles respectively. DOUBLE-CHECK: A kilometer is roughly 2/3 of a mile, and the answers are roughly 2/3 of the given values, so the conversions appear correct. 1.44.

THINK: It is a distance d = 60 feet, 6 inches from the pitcher’s mound to home plate. Recall the conversion factors: 1 foot = 12 inches, 1 inch = 2.54 cm, 100 cm = 1 m. SKETCH:

RESEARCH: If the distance is x in meters and y in feet, then using the conversion æ 12 inches öæ 2.54 cm öæ 1 m ö factor c, x = cy. c = 1 footç ÷ç ÷ç ÷ / foot è 1 foot øè 1 inch øè 100 cm ø SIMPLIFY: c = 0.3048 meters/foot CALCULATE: 60 feet plus 6 inches = 60.5 feet. Then, converting 60.5 feet to æ 0.3048 m ö meters: d = 60.5 ftç ÷ = 18.440 m. è 1 ft ø

7

Bauer/Westfall: University Physics, 1E ROUND: Rounding to three significant figures, the distance is 18.4 m. DOUBLE-CHECK: The answer is a reasonable distance for a pitcher to throw the ball. 1.45.

THINK: The given quantities, written in scientific notation and in units of meters, are: the starting position, xo = 7× 10-3 m and the lengths of the flea’s successive hops, x1 = 3.2× 10-2 m ,

x2 = 6.5× 10-2 m ,

x3 = 8.3×10-2 m ,

x4 = 10.0× 10-2 m ,

x5 = 11.5× 10-2 m

and

-2

x6 = 15.5×10 m. The flea makes six jumps in total. SKETCH:

6

RESEARCH: The total distance jumped is xtotal = å xn . The average distance covered n=1

in a single hop is: 6

xavg =

1 xn. 6n = 1

å

SIMPLIFY: xtotal = x1 + x2 + x3 + x4 + x5 + x6 , xavg =

xtotal 6

CALCULATE: xtotal = (3.2 m+ 6.5 m + 8.3 m + 10.0 m + 11.5 m + 15.5 m) ×10-2 = 55.0× 10-2 m 55.0× 10-2 m = 9.16666× 10-2 m 6 ROUND: Each of the hopping distances is measured to 1 mm precision. Therefore the total distance should also only be quoted to 1 mm precision: xtotal = 55.5× 10-2 m. Rounding the average distance to the right number of significant digits, however, requires a few more words. As a general rule of thumb the average distance should be quoted to the same precision as the least precision of the individual distances, if there are only a few measurements contributing to the average. This is the case -2 here, and so we state xavg = 9.2×10 m. However, suppose we had 10,000 xavg =

measurements contributing to an average. Surely we could then specify the average to a higher precision. The rule of thumb is that we can add one additional significant digit for every order of magnitude of the number of independent measurements contributing to an average. You see that the answer to this problem is yet another indication that specifying the correct number of significant figures can be complicated and sometimes outright tricky! DOUBLE-CHECK: The flea made 6 hops, ranging from 3.2× 10-2 m to 15.5× 10-2 m , so the total distance covered is reasonable. The average distance per hop falls in the range between 3.2× 10-2 m and 1.55× 10-1 m, which is what is expected. 1.46.

THINK: The question says that 1 cm3 of water has a mass of 1 g, that 1 mL = 1 cm3 , and that 1 metric ton is 1000 kg. SKETCH:

8

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RESEARCH: For the first part of the question, use the conversion equation: æ 1000 mL ö æ 1 cm3 öæ 1 g ö æ 1 kg ö 1L = 1Lç ÷. ÷ç ÷ç ÷ç è 1 L ø è 1 mL ø è 1 cm3 ø è 1000 g ø For the second part of the question, use: æ 1000 kg ö æ 1000 g ö æ 1 cm3 ö 1 metric ton = 1 metric tonç ÷ç ÷. ÷ç è 1 metric ton ø è 1 kg øè 1 g ø For the last part, recall that the volume of a cube is V = l 3. SIMPLIFY: Re-arranging the formula for the volume of the cubical tank to solve for the length gives l = 3 Vc . CALCULATE:

æ 1000 mL ö æ 1 cm3 öæ 1 g ö æ 1 kg ö 1L = 1Lç ÷ = 1 kg ÷ç ÷ç ÷ç è 1 L ø è 1 mL ø è 1 cm3 ø è 1000 g ø

æ 1000 kg ö æ 1000 g ö æ 1 cm3 ö 3 1 metric ton = 1 metric tonç ÷ç ÷ = 1000000 cm ÷ç è 1 metric ton ø è 1 kg øè 1 g ø l = 3 1,000,000 = 100 cm = 1 m ROUND: No rounding is necessary. DOUBLE-CHECK: In each calculation the units work out correctly, so the answers are reasonable. 1.47.

THINK: The given quantity is the speed limit, which is 45 miles per hour. The question asks for the speed limit in millifurlongs per microfortnight. The conversions 1 furlong = 1/8 mile, and 1 fortnight = 2 weeks are given in the question. SKETCH: A sketch is not needed. RESEARCH: ö 1 fortnight 1 mile 1 mile æ 8 furlongs ö æ 103 millifurlongs ö æ 24 hours öæ 14 days ö æ = ÷ç ÷ ÷ç ÷ç 6 ç ÷ç 1 hour 1 hour è 1 mile ø è 1 furlong ø è 1 day øè 1 fortnight ø è 10 microfortnights ø SIMPLIFY:

millifurlongs 1 mile = 2.688 1 hour microfortnight

CALCULATE:

æ millifurlongs ö millifurlongs 45 miles = 45ç 2.688 ÷ = 120.96 hour microfortnight ø microfortnight è

ROUND: Because the given quantity contains two significant figures, the result must be rounded to remain consistent. A speed of 45 miles per hour is equivalent to a speed of 120 millifurlongs/microfortnight. DOUBLE-CHECK: The conversion factor works out to be roughly 3 millifurlongs per microfortnight to each mile per hour, so the answer is reasonable. 1.48.

THINK: The density of water is r = 1000. kg/m3. Determine if a pint of water weighs a pound. Remember that 1.00 kg = 2.21 lbs and 1.00 fluid ounce = 29.6 mL. SKETCH: A sketch is not needed. RESEARCH: 1 pint = 16 fluid ounces, mass = density × volume æ 16 fl. oz ö æ 29.6 mL ö æ 1 cm3 öæ 1 m3 ö = 4.736× 10-4 m3 SIMPLIFY: 1 pintç ÷ç ÷ ç 1 mL ÷ç 3 3÷ 1 pint 1.00 fl. oz (100) cm øè è øè øè ø

9

Bauer/Westfall: University Physics, 1E

CALCULATE:

(

)

m= 1000. kg/m3 (4.736× 10-4 m3) = 0.4736 kg

æ 2.21 lbs ö In pounds m is equal to 0.4736 kgç ÷ = 1.046656 lbs . è 1.00 kg ø ROUND: Rounding to three significant figures, the weight is 1.05 lbs. DOUBLE-CHECK: A pint is still a common measure for beverages, such as beer. A beer is relatively light and mainly comprised of water, so the answer is reasonable. 1.49.

THINK: The radius of a planet, rp , is 8.7 times greater than the Earth’s radius, rE . Determine how many times bigger the surface area of the planet is compared to the Earth’s. Assume the planets are perfect spheres. SKETCH:

2 RESEARCH: The surface area of a sphere is A = 4p r 2 , so AE = 4p rE2, and Ap = 4p rp ,

and rp = 8.7rE . SIMPLIFY: Ap = 4p ( 8.7rE )

2

2 CALCULATE: Ap = (75.69)4p rE , and AE = 4p rE2 . By comparison, Ap = 75.69AE .

ROUND: Rounding to two significant figures, the surface area of the planet is 76 times the surface area of Earth. DOUBLE-CHECK: Since the area is proportional to the radius squared, it is expected that the surface area of the planet will be much larger than the surface area of the Earth, since its radius is 8.7 times Earth’s radius. 1.50.

THINK: The radius of the planet rp is 5.8 times larger than the Earth’s radius rE . Assume the planets are perfect spheres. SKETCH:

RESEARCH: The volume of a sphere is given by V = ( 4/ 3) p r3 . The volume of the 3 planet is Vp = ( 4/ 3) p rp . The volume of the Earth is VE = ( 4/ 3) p rE3 . rP = 5.8rE .

SIMPLIFY:

Vp = ( 4/ 3) p (5.8rE )3

3 3 CALCULATE: Vp = ( 4/ 3) p (5.8rE ) = 195.112rE ( 4/ 3) p . Recall, VE = ( 4/ 3) p rE3. Comparing the

expressions, Vp = 195.112VE . ROUND: To two significant figures, so Vp = 2.0× 10 VE . 2

DOUBLE-CHECK: The volume of the planet is about 200 times the volume of the Earth. The area of a sphere is proportional to the radius cubed, it is reasonable to get a much larger volume for the planet compared to the Earth’s volume.

10

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THINK: The given quantity is 1.56 barrels. Calculate how many cubic inches are in 1.56 barrels. 1 barrel = 7056 cubic inches. SKETCH: A sketch is not needed. RESEARCH: If a volume V1 is given in barrels then the equivalent volume V2 in 7056 cu. in. cubic inches is given by the formula V2 = V1 1 barrel SIMPLIFY: Not applicable. æ 7056 cu. in ö CALCULATE: 1.56 barrelsç ÷ = 11007.36 cu. in. è 1 barrel ø ROUND: The value given in the question has three significant figures, so the final answer is that 1.56 barrels is equivalent to 1.10× 104 cubic inches. DOUBLE-CHECK: Barrels are not commonly used units. However, since the proper conversion factor of 7056 cubic inches per barrel was used, the answer is accurate.

1.52.

THINK: The car’s gas tank has the shape of a rectangular box with a square base whose sides measure l = 62 cm. The tank has only 1.5 L remaining. The question asks for the depth, d of the gas remaining in the tank. The car is on level ground, so that d is constant. SKETCH:

RESEARCH:

Atank = l2 . The volume of gas remaining is Vgas = Atank ´ d. Convert the

volume 1.5 L to 1500 cm3 by using 1 mL = 1 cm3. SIMPLIFY: d = Vgas / Atank , but Atank = l2 , so substitute this into the expression for d: d = Vgas / l2 . 1500 cm3 = 0.390218 cm (62 cm)2 ROUND: To two significant figures d = 0.39 cm. DOUBLE-CHECK: The car’s gas tank will hold 52 L but only has 1.5 L remaining. The sides of the gas tank are 62 cm and because the gas tank is almost empty, there should be a small depth of gas in the bottom of the tank, so the answer is reasonable. CALCULATE: d =

1.53.

3 The formula for the volume of a sphere is given by Vsphere = ( 4/ 3) p r . The formula for density is given by r = m/ V . Refer to Appendix B in the text book and express the answers in SI units using scientific notation. SKETCH:

THINK:

11

Bauer/Westfall: University Physics, 1E

RESEARCH:

The radius of the Sun is rS = 6.96× 108 m, the mass of the Sun is

mS = 1.99× 1030 kg, the radius of the Earth is rE = 6.37× 106 m, and the mass of the Earth is mE = 5.98× 1024 kg. SIMPLIFY: Not applicable. CALCULATE: 4 4 (a) VS = p rS3 = p (6.96× 108 )3 = 1.412265× 1027 m3 3 3 4 4 (b) VE = p rE3 = p (6.37× 106)3 = 1.082696× 1021 m3 3 3 mS 1.99× 1030 = = 1.40908× 103 kg/m3 (c) rS = 27 VS 1.412265× 10 mE 5.98× 1024 = = 5.523249× 103 kg/m3 VE 1.082696× 1021 ROUND: The given values have three significant figures, so the calculated values should be rounded as: (a) VS = 1.41× 1027 m3 (d) rE =

(b) VE = 1.08× 1021 m3 (c) rS = 1.41× 103 kg/m3 (d) rE = 5.52× 103 kg/m3 DOUBLE-CHECK: The radius of the Sun is two orders of magnitude larger than the radius of the Earth. Because the volume of a sphere is proportional to the radius cubed, the volume of the Sun should be (102)3 or 106 larger than the volume of the Earth, so the calculated volumes are reasonable. Because density depends on mass and volume, and the Sun is roughly 106 times larger and more massive than the Earth, it is not surprising that the density of the Sun is on the same order of magnitude as the density of the Earth (e.g. 106 / 106 = 1). Earth is primarily solid, but the Sun is gaseous, therefore it is reasonable that the Earth is denser than the Sun. 1.54.

THINK: The tank is in the shape of an inverted cone with height h= 2.5 m and radius r = 0.75 m. Water is poured into the tank at a rate of w = 15 L/s. Calculate how long it will take to fill the tank. Recall the conversion 1 L = 1000 cm3. SKETCH:

1 RESEARCH: The volume of a cone is Vcone = p r 2h . The rate water enters the cone is 3 V 1 w = water where t is time. When the cone is full, Vcone = Vwater , therefore p r 2h = wt. t 3 2 pr h SIMPLIFY: t = 3w p (75 cm)2(250 cm) t= = 98.1748 s æ 15000 cm3 ö CALCULATE: 3ç ÷ s è ø 12

Full file at http://testbank360.eu/solution-manual-university-physics-with-modernphysics-1st-edition-bauer ROUND: To two significant figures, t = 98 s. DOUBLE-CHECK: The calculation resulted in the correct units, so the answer is reasonable. 1.55.

THINK: The rate of water flow is 15 L/s, the tank is cubical, and the top surface of the water rises by 1.5 cm/s. Let h be the height of the water in the tank. SKETCH:

RESEARCH: The change in the volume of the water, DVwater , is 15 L/s = 15000 cm3 /s. The change in the height of the water is Dh = 1.5 cm/s. An equation to find the side length of the tank is DVwater = l2Dh. SIMPLIFY: l =

DVwater Dh

æ 15000 cm3 / s ö CALCULATE: l = ç ÷ = 100. cm è 1.5 cm/s ø ROUND: l = 1.0× 102 cm DOUBLE-CHECK: The flow rate of 15 L/s is quite fast, but the level of the water is rising by only 1.5 cm/s, so it is reasonable that the tank is relatively large. 1.56.

THINK: The atmosphere has an effective weight of 15 pounds per square inch. By computing the surface area of the Earth, it will be easy to compute the mass of the atmosphere. Then, since the atmosphere is assumed to have a uniform density of 1.275 kg/m3, the mass can be converted to a volume. The volume of the atmosphere is the difference of two spheres, whose radii are the radius of the Earth, rE , and the radius of the Earth plus the thickness of the atmosphere, Dr. The result will be a cubic equation with one real root which can be approximated to give the thickness of the atmosphere. SKETCH:

RESEARCH: Recall the conversions 1 inch = 0.0254 m and 1 kg = 2.205 lbs. The radius of the Earth is about 6378 km. The surface area of the Earth is AE = 4p rE2 . The mass of the atmosphere is mA = AE ( 15 lb/sq in) . The volume of the atmosphere can be

computed using the ratio VA = mA / r A , where r A is the density of the atmosphere. This volume is the difference of the two spheres, as shown in the sketch. The volume of the Earth (without its atmosphere) is VE = ( 4/ 3) p rE3, and the volume of the Earth

13

Bauer/Westfall: University Physics, 1E and atmosphere is VEA = ( 4/ 3) p ( rE + Dr ) . A second method of computing the volume 3

of the atmosphere is VA = VEA - VE . Set the two values of VA equal and solve for r. SIMPLIFY: The first expression for the volume of the atmosphere is ö m 4p rE2 æ 15 lb VA = A = ç ÷. rA r A è 1 squareinch ø

(

)

3 3 The second expression is VA = ( 4p / 3) (rE + Dr) - rE . Setting these expressions equal to

each other gives an equation to solve for Dr. æ 1000 m öæ 1 in ö 8 CALCULATE: rE = 6378 kmç ÷ç 0.0254 m ÷ = 2.511×10 in 1 km è øè ø 3

kg æ 0.0254 m ö -5 kg ç ÷ = 2.089× 10 m3 è 1 in ø in3 Substituting into the first equation for VA gives:

rA = 1.275

(

)

2

4p 2.511×108 in æ ö æ 1 kg ö 15 lb 23 3 VA = ç ÷ç ÷ = 2.580× 10 inch . 2.089× 10-5 kg/in3 è 1 square inch ø è 2.205 lb ø The second equation for VA becomes: 3 4p VA = (2.511× 108 in + Dr)3 - 2.511×108 in = 4.1888Dr3 + 3.155×109 Dr2 + 7.923× 1017 Dr. 3 Setting the two equations for VA equal results in the equation:

(

(

)

(

)

)

(

(

)

(

)

)

4.1888Dr3 + 3.155×109 Dr2 + 7.923× 1017 Dr = 2.580× 1023 , a cubic equation in Dr . This equation can be solved by a number of methods. A graphical estimate is sufficient. It has one real root, and that is at approximately 0.0254 m Dr = 325300 in = 325300 in = 8263 m. 1 in ROUND: The least precise value given in the question had two significant figures, so the answer should be rounded to 8300 m. DOUBLE-CHECK: The result has units of distance, which is what is expected. What may not be expected is that our result is not as big as the height of the tallest mountain on Earth, Mt. Everest, which has a height of 8.8 km. We can obtain a simple approximation of our result by realizing that our calculated value for Dr is small compared to the radius of Earth. This means that the surface of a sphere with radius RE + Dr and one with radius RE are not very different, allowing us to write an approximation to our result as

(

) (

)

(

(

)

Dr » VA / 4p rE2 = 2.580× 1023 inch3 / 4p 2.511×108 inch 1.57.

2

) = 3.256×10 inch = 8.3 km. 5

r r THINK: Let L be the position vector. Then L = 40.0 m and q = 57.0° (above x-axis). SKETCH:

14

Full file at http://testbank360.eu/solution-manual-university-physics-with-modernphysics-1st-edition-bauer r r RESEARCH: From trigonometry, sinq = Dy / L and cosq = Dx / L . The length of the r r vector L is given by the formula L = Dx2 + Dy2 . r r SIMPLIFY: Dx = L cosq , Dy = L sinq CALCULATE: Dx = ( 40.0 m) cos(57.0°) = 21.786 m , Dy = ( 40.0 m) sin(57.0°) = 33.547 m ROUND: Dx = 21.8 m and Dy = 33.5 m. r L = Dx2 + Dy2 = (21.8 m)2 + (33.5 m)2 » 40.0 m, to three significant DOUBLE-CHECK: figures. 1.58.

THINK: a = 6.6 cm, b= 13.7 cm, and c= 9.2 cm are the given quantities. SKETCH:

RESEARCH: Law of cosines: c2 = a2 + b2 - 2abcosg c2 = a2 + b2 - 2abcosg 2abcosg = a2 + b2 - c2 SIMPLIFY:

a2 + b2 - c2 2ab æ a2 + b2 - c2 ö g = cos-1 ç ÷ 2ab è ø

cosg =

2 2 2 -1 é (6.6 cm) + (13.7 cm) - (9.2 cm) ù CALCULATE: g = cos ê ú = 35.83399° 2(6.6 cm)(13.7 cm) ë û ROUND: g = 36° DOUBLE-CHECK: The angle g in the figure is less than 45° , so the answer is reasonable.

1.59.

THINK: The lengths of the x and y components of the vectors can be read from the provided figure. Remember to decompose the vectors in terms of their x and y components. SKETCH:

RESEARCH:

r ˆ + Vy y ˆ , where Vx = xfi - x A vector can be written as V = Vx x

Vy = yfi - y . SIMPLIFY:

Not applicable.

15

and

Bauer/Westfall: University Physics, 1E r r ˆ + (3.5- 2) y ˆ = 2.5x ˆ + 1.5y ˆ, B = (4- (-1.5))x ˆ + (1- 2.5) y ˆ = 5.5x ˆ - 1.5y ˆ CALCULATE: A = (-1.5- (-4))x r ˆ - (4 - (-1)) y ˆ = -6x ˆ - 3y ˆ C = (-3- 3)x ROUND: Not applicable. DOUBLE-CHECK: Comparing the signs of the x- and y-components of the vectors r r r A , B and C, to the provided figure, the calculated components all point in the correct directions. The answer is therefore reasonable. 1.60.

THINK: The question asks for the length and direction of the three vectors. The x and y components of the vectors can be read from the provided figure. Remember when dealing with vectors, the components must be treated separately. SKETCH:

r RESEARCH: The length of a vector is given by the formula L = x2 + y2 . The direction of a vector (with respect to the x-axis) is given by tanq = y / x . -1 æ y ö SIMPLIFY: q = tan ç ÷ è xø r æ 1.5 ö A = (2.5)2 + (1.5)2 = 2.9, q A = tan-1 ç ÷ = 30.9638° è 3ø r æ -1.5ö B = (5.5)2 + (-1.5)2 = 5.700877, q B = tan-1 ç ÷ = -15.2551° è 5.5 ø CALCULATE: r C = (-6)2 + (-3)2 = 6.7082,

æ -3 ö qC = tan-1 ç ÷ = 26.565° = 180° + 26.565° = 206.565° è -6 ø ROUND: The figure can reasonably be read to two significant digits, so the rounded r r r values are A = 3.4, q A = 31°, B = 5.7, q B = -15°, C = 6.7, and qC = 210°.

1.61.

DOUBLE-CHECK: Comparing the graphical values to the calculated values, the calculated values are reasonable. r r r r Vectors add tip to tail, A + B + C = D .

16

r By inspecting the image, it is clear that D = (2,-3). 1.62.

THINK: To subtract two vectors, reverse the direction of the vector being r r r subtracted, and treat the operation as a sum. Denote the difference as E = B - A. SKETCH:

r r r r r RESEARCH: E = B - A = B + - A

( )

SIMPLIFY: No simplification is necessary. r CALCULATE: By inspection, E = (3,-3). ROUND: No rounding is necessary. r DOUBLE-CHECK: The resultant vector E points from the origin to the fourth quadrant, so its x-component should be positive and its y-component should be negative. This gives some support to the reasonableness of the answer. 1.63.

THINK: When adding vectors, you must add the components separately. SKETCH:

17

Bauer/Westfall: University Physics, 1E

r r r r RESEARCH: D = A + B + C r ˆ + (Ay + By + Cy )y ˆ SIMPLIFY: D = (Ax + Bx + Cx )x r ˆ + (1.5- 1.5- 3) y ˆ = 2x ˆ - 3y ˆ CALCULATE: D = (2.5+ 5.5- 6)x ROUND: The answers are precise, so no rounding is necessary. DOUBLE-CHECK: The calculation seems consistent with the provided figure. 1.64.

THINK: When subtracting vectors, you must subtract the x and y components separately. SKETCH:

Fx = Cx - Ax - Bx and Fy = Cy - Ay - By . The length is computed using RESEARCH: ur ˆ + Fy y ˆ. F = Fx2 + Fy2 with F = Fx x ur SIMPLIFY: F = (Cx - Ax - Bx )2 + (Cy - Ay - By )2 ur F = CALCULATE:

( ( -6.0) - 2.5- 5.5) + ( ( -3.0) - (1.5) - (-1.5)) 2

2

= 205 = 14.318 ur ROUND: To two significant figures, the length of F is 14. ur DOUBLE-CHECK: The size of F is reasonable. 1.65.

r r r THINK: The lengths of the vectors are given as A = 75.0, B = 60.0, C = 25.0 r and D = 90.0 . The question asks for the vectors to be written in terms of unit vectors. Remember, when dealing with vectors, the x- and y-components must be treated separately. SKETCH:

18

r ˆ + Vy y ˆ . Since RESEARCH: The formula for a vector in terms of unit vectors is V = Vx x sinq =

Ay Ax adjacent opposite and cosq = , sinq = r and cosq = r . A hypotenuse hypotenuse A

q A = 30.0°, q B = 19.0° = 161.0° (with respect to the positive x-axis), qC = 52.0° = 232.0° (with respect to the positive x-axis), q D = 27.0° = 333.0° (with respect to the positive x-axis). r r r r r r SIMPLIFY: Ax = A cosq A , Ay = A sinq A , Bx =B cosq B , By =B sinq B , Cx =C cosqC , Cy =C sinqC , r r Dx =D cosq D , and Dy =D sinq D .

CALCULATE:

1.66.

ˆ, Ay = 75.0sin30.0° = 37.5y ˆ Ax = 75.0cos30.0° = 64.9519x ˆ ˆ Bx = 60.0cos161.0° = -56.73x, By = 60.0sin161.0° = 19.534y ˆ, Cy = 25.0sin232.0° = -19.70027y ˆ Cx = 25.0cos232.0° = -15.3915x ˆ, Dy = 90.0sin333.0° = -40.859144y ˆ Dx = 90.0cos333.0° = 80.19058x

ROUND: The given values had three significant figures so the answers must be rounded to: r r r ˆ = -56.7x ˆ + 37.5y ˆ, B ˆ + 19.5y ˆ, C = -15.4x ˆ - 19.7y ˆ, D = 80.2x ˆ - 40.9y ˆ. A = 65.0x DOUBLE-CHECK: Comparing the calculated components to the figure provided shows that this answer is reasonable. r r r THINK: Use the components in Question 1.65 to find the sum of the vectors A , B , C r r r r and D . Also, calculate the magnitude and direction of A - B + D . Remember, when dealing with vectors the x and y components must be treated separately. Treat the values given in the question as accurate to the nearest decimal, and hence as having two significant figures. SKETCH: Not applicable. RESEARCH: r r r r r (a) The resultant vector is V = A + B + C + D . r r (b) The magnitude of a vector is V = (Vx )2 + (Vy )2 . The direction of the vector V is

(

)

qV = tan-1 Vy / Vx . SIMPLIFY: r r r r ˆ + (Ay + By + Cy + Dy )y ˆ (a) A + B + C + D = (Ax + Bx + Cx + Dx )x r r r r V = A - B + D = (Ax - Bx + Dx )2 + (Ay - By + Dy )2 (b)

æ Ay - By + Dy ö qV = tan-1 çç ÷÷ è Ax - Bx + Dx ø

19

Bauer/Westfall: University Physics, 1E CALCULATE: r r r r ˆ + (37.5+ 19.5- 19.7- 40.9) y ˆ A + B + C + D = (65.0- 56.7 - 15.4+ 80.2)x (a) ˆ - 3.6y ˆ = 73.1x r r r A - B + D = (65.0- (-56.7) + 80.2)2 + (37.5- 19.7- 40.9)2 = 203.217 (b) é 37.5- 19.7. - 40.9 ù qV = tan-1 ê ú = -6.5270° ë 65.0- (-56.7) + 80.2û ROUND: (a) Not necessary.

r r r (b) The given magnitudes have three significant figures, so A - B + D = 203 , at -6.53°

1.67.

(below the x-axis). DOUBLE-CHECK: The length of the resulting vector is less than the sum of the lengths of the component vectors. Since the vector points into the fourth quadrant, the angle made with the x-axis should be negative, as it is. THINK: The problem involves adding vectors, therefore break the vectors up into r their components and add the components. SW is exactly 45° south of W. d1 = 4.47 r r km N, d2 = 2.49 km SW, d3 = 3.59 km E. SKETCH:

r r r r ˆ + Dy y ˆ , and recall the formula for the length of RESEARCH: Use D = d1 + d2 + d3 = Dx x r r r 2 2 ˆ D : D = Dx + Dy . Decompose each summand vector into components di = di x x r r ˆ , with summand vectors: ˆ + d2y y ˆ = -d2 sin(45o )x ˆ - d2 cos(45o ) y ˆ, ˆ, d2 = d2x x + di y y d1 = d1y r ˆ. d3 = d3x r r r r SIMPLIFY: Therefore, D = d1 + d2 + d3 = (d3 - d2 sin(45°)) + (d1 - d2 cos(45°)) and r D = (d3 - d2 sin(45°))2 + (d1 - d2 cos(45°))2 . r CALCULATE: D = (3.59- 2.49cos(45°))2 + (4.47- 2.49sin(45°))2 = 3.269 km r ROUND: D = 3.27 km DOUBLE-CHECK: Given that all vectors are of the same order of magnitude, the distance from origin to final position is less than d1 , as is evident from the sketch. This means that the calculated answer is reasonable. 1.68.

THINK: The problem involves adding vectors, therefore break the vectors up into r their components and add the components. NW is exactly 45° north of west. d1 = 20 r r paces N, d2 = 30 paces NW, d3 = 10 paces S. Paces are counted to the nearest integer, so treat the number of paces as being precise. SKETCH:

20

r r r r ˆ + Dy y ˆ , and recall the formula for the length of RESEARCH:, Use D = d1 + d2 + d3 = Dx x r r r 2 2 ˆ + D : D = Dx + Dy . Decompose each summand vector into components di = di x x r r r ˆ , with summand vectors: d1 = d1y ˆ + d2y y ˆ = -d2 sin(45o )x ˆ - d2 cos(45o )y ˆ, d3 = d3x di y y ˆ, d2 = d2x x ˆ. r r r r ˆ + (d1 - d3 + d2 cos(45°))y ˆ and SIMPLIFY: D = d1 + d2 + d3 = -d2 sin(45°)x r D = (-d2 sin(45°))2 + (d1 - d3 + d2 cos(45°))2 . r CALCULATE: D = (-30sin(45°))2 + (20- 10+ 30cos(45°))2 = 37.739 paces ROUND: 38 paces DOUBLE-CHECK: Given that d1 > d3 , the calculated answer makes sense since the distance D should be greater than d2 . 1.69.

THINK: The problem involves adding vectors, therefore break the vectors up into r their components and add the components. NW is 45° north of west. d1 = 20 paces r r r r N, d2 = 30 paces NW, d3 = 12 paces N, d4 = 3 paces into ground ( d4 implies 3 dimensions). Paces are counted to the nearest integer, so treat the number of paces as being precise. SKETCH:

r r r r r r r r ˆ + diy y ˆ + diz zˆ , D = Dx2 + Dy2 + Dz2 , d1 = d1y ˆ, RESEARCH: D = d1 + d2 + d3 + d4 , di = dix x r r r ˆ + d2y y ˆ = -d2 cos(45°)x ˆ + d2 sin(45°)y ˆ, d3 = d3 y ˆ, and d4 = -d4zˆ. d2 = -d2x x r r r r r ˆ - d4zˆ and SIMPLIFY: D = d1 + d2 + d3 + d4 = -d2 cos(45°)xˆ + (d1 + d3 + d2 sin(45°))y r D = (-d2 cos(45°))2 + (d1 + d3 + d2 sin(45°))2 + (-d4 )2 . r æ 2 2ö ˆ + ç 20+ 12+ 30 ÷ y ˆ - 3zˆ CALCULATE: D = -30 x ç 2 2 ÷ø è

21

Bauer/Westfall: University Physics, 1E r D = (-21.213)2 + (53.213)2 + (-3)2 = 57.36 paces r ˆ + 32+ 15 2 y ˆ - 3zˆ and round the number of paces to the nearest ROUND: D = -15 2x r integer: D = 57 paces.

(

)

DOUBLE-CHECK: Distance should be less than the sum of the magnitudes of each vector, which is 65. Therefore, the calculated answer is reasonable. 1.70.

THINK: Consider the Sun to be the centre of a circle with the distance from the Sun to Venus, as the radius. Earth is located a distance rE = 1.5× 1011 m from the Sun, so that the three bodies make a triangle and the vector from Earth to the Sun is at 0˚. The vector pointing from Earth to Venus intersects Venus’ orbit one or two times, depending on the angle Venus makes with the Earth. This angle is at a maximum when the vector intersects the orbit only once, while all other angles cause the vector to intersect twice. If the vector only intersects the circle once, then that vector is tangential to the circle and therefore is perpendicular to the radius vector of the orbit. This means the three bodies make a right triangle with rE as the hypotenuse. Simple trigonometry can then be used to solve for the angle and distance. SKETCH:

RESEARCH: rE2 = rV 2 + R2, rE sinq = rV SIMPLIFY: R = rE2 - rV 2 , q = sin-1 ( rV / rE ) 11 ö æ 11 2 11 2 11 -1 1.1× 10 ÷ = 47.17° CALCULATE: R = (1.5× 10 ) - (1.1× 10 ) = 1.0198× 10 m , q = sin çç 11 ÷ è 1.5× 10 ø

ROUND: R = 1.0× 1011 m, q = 47° DOUBLE-CHECK: If it had been assumed that q = tan-1 ( rV / rE ) when the E-to-S-to-V angle was 90°, then tanq would be about 36°. Therefore the maximum angle should be greater than this, so the answer is reasonable. 1.71.

THINK: All angles and directions of vectors are unknown. All that is known are the distances walked, d1 = 550 m and d2 = 178 m, and the distance d3 = 432 m that the friend is now away from you. Since the distances are the sides of a triangle, use the cosine law to determine the internal (and then external) angles. Also, since d3 < d1 , he must have turned back towards you partially, i.e. he turned more than 90°, but less than 180°. SKETCH:

22

RESEARCH: d22 = d12 + d32 - 2d1d3 cos f , d32 = d12 + d22 - 2d1d2 cosa , q + a = 180° 2 2 æ 2 ö 2 2 2 -1 d1 + d3 - d2 SIMPLIFY: 2d1d3 cos f = d1 + d3 - d2 Þ f = cos çç ÷÷ 2 d d 1 3 è ø 2 2 2 æ ö -1 d1 + d2 - d3 Likewise, a = cos çç ÷÷ . 2d1d2 è ø 2 æ (550m) + (432m)2 - (178m)2 ö -1 ÷ = 15.714° CALCULATE: f = cos çç ÷ 2(550m)(432m) è ø 2 2 2 æ (550m) + (178m) - (432m) ö a = cos-1 ç ÷ = 41.095°, q = 180° - 41.095° = 138.905° ç ÷ 2(550m)(178m) è ø d = 550m ROUND: Since 1 has two significant figures (which is the fewest) the answers should be rounded to two significant figures. This means: f = 16° , a = 41° and q = 140° . The two possibilities are that the friend turned to the right or the left (a right turn is shown in the diagram). The direction the friend turned doesn’t matter, he turns by the same amount regardless of which direction it was. DOUBLE-CHECK: The friend turned through an angle of 140 degrees. The angle between the initial departure and the final location is 16 degrees. These are both reasonable angles.

Additional Problems 1.72.

THINK: Assume that the Earth is a perfect sphere with radius, rE = 6378 km, and treat the circumference of Earth as the circumference of a circle. SKETCH:

RESEARCH: The circumference of a circle is given by C = 2p r. SIMPLIFY: C = 2p rE

CALCULATE: C = 2p ( 6378 km) = 40074 km ROUND: The instructions from the question say to round to three significant figures: C = 4.01× 104 km. DOUBLE-CHECK: Assuming a hot air balloon has an average velocity of 20 km/h, then it would take about 80 days to travel, hence the phrase around the world in 80 days.

(

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)

1.73.

4,308,229 » 4× 106 ; 44 » 4× 101, 4× 106 4× 101 = 16× 107 = 2× 108

1.74.

r ˆ + 6y ˆ - 10zˆ + C = -7x ˆ + 14y ˆ, 3x

r ˆ - 3x ˆ) + (14y ˆ - 6y ˆ) + 10zˆ = -10x ˆ + 8y ˆ + 10zˆ C = (-7x

23

Bauer/Westfall: University Physics, 1E

1.75.

r THINK: The two vectors are A = (Ax ,Ay ) = (30.0 m,-50.0 m) and r B = (Bx , By ) = (-30.0 m, 50.0 m) . Sketch and find the magnitudes. SKETCH:

r r ˆ + Cy y ˆ is C = Cx2 + Cy2 RESEARCH: The length of a vector C = Cx x r r SIMPLIFY: A = Ax2 + Ay2 , B = Bx2 + By2 r r CALCULATE: A = (30)2 + (-50)2 = 58.3095 m, B = (-30)2 + (50)2 = 58.3095 m r r ROUND: A = 58.3 m, B = 58.3 m DOUBLE-CHECK: The calculated magnitudes are larger than the lengths of the component vectors, and are less than the sum of the lengths of the component vectors. Also, the vectors are opposites, so they should have the same length. 1.76.

THINK: Use trigonometry to find the angles as indicated in the sketch below. r A = (Ax , Ay ) = (30.0 m,-50.0 m) . SKETCH:

RESEARCH: tanq =

opposite adjacent

-1 SIMPLIFY: tanq1 = (Ay / Ax ) Þ q1 = tan (Ay / Ax ),

CALCULATE: q1 = tan-1(-50/ 30) = -59.036°,

1.77.

tanq 2 = (Ax / Ay ) Þ q 2 = tan-1(Ax / Ay )

q2 = tan-1(30/ -50) = -30.963° ROUND: Drop the signs of the angles and just use their size: q1 = 59.0°, q2 = 31.0°. DOUBLE-CHECK: The two angles add up to 90˚, which they should. The answers are reasonable. r A = (Ax , Ay ) = (-30.0 m, - 50.0 m) THINK: The two vectors are and r B = (Bx , By ) = (30.0 m, 50.0 m) . Sketch and find the magnitudes.

24

Full file at http://testbank360.eu/solution-manual-university-physics-with-modernphysics-1st-edition-bauer SKETCH:

r RESEARCH: C = Cx2 + Cy2 r r SIMPLIFY: A = Ax2 + Ay2 , B = Bx2 + By2 r r CALCULATE: A = (-30.0 m)2 + (-50.0 m)2 = 58.3095 m, B = (30.0 m)2 + (50.0 m)2 = 58.3095 m r r ROUND: A = 58.3 m, B = 58.3 m DOUBLE-CHECK: The magnitudes are bigger than individual components, but not bigger than the sum of the components. Therefore, the answers are reasonable. 1.78.

THINK: Using trigonometry find the angles indicated in the diagram below. The r vector B = (Bx , By ) = (30.0 m,50.0 m) . SKETCH:

RESEARCH: tanq = SIMPLIFY:

opposite adjacent

æ ö æ By ö æ opposite ö -1 -1 Bx q = tan-1 ç ÷ , q1 = tan çç ÷÷ , q 2 = tan çç ÷÷ è adjacent ø è Bx ø è By ø

-1 æ 50.0 m ö -1 æ 30.0 m ö CALCULATE: q1 = tan ç ÷ = 59.036°, q 2 = tan ç 50.0 m ÷ = 30.963° 30.0 m è ø è ø ROUND: q1 = 59.0° , q 2 = 31.0° DOUBLE-CHECK: The angles sum to 90°, which is expected from the sketch. Therefore, the answers are reasonable.

1.79.

THINK: An angle is measured counter-clockwise from the positive x-axis (0˚). r C = (34.6 m, - 53.5 m). It is also possible to measure clockwise from the positive x-axis and consider the measure to be negative.

SKETCH: 25

Bauer/Westfall: University Physics, 1E

r æ Cy ö RESEARCH: C = Cx2 + Cy2 , tanq = çç ÷÷ è Cx ø æ Cy ö -1 SIMPLIFY: q = tan çç ÷÷ è Cx ø r 2 2 -1 æ -53.5 m ö CALCULATE: C = (34.6 m) + (-53.5 m) = 63.713 m, q = tan ç ÷ = -57.108° è 34.6 m ø r ROUND: C = 63.7 m , q = -57.1° or 303˚ (equivalent angles). DOUBLE-CHECK: The magnitude is greater than each component but less than the sum of the components and the angle is also in the correct quadrant. The answer is reasonable. 1.80.

THINK: Assume Mars is a sphere whose radius is rM = 3.39× 106 m. SKETCH:

4 C = 2p r, A = 4p r 2, V = p r 3 3 4 SIMPLIFY: C = 2p rM , A = 4p rM 2, V = p rM 3 3 RESEARCH:

C = 2p (3.39× 106 m) = 2.12999× 107 m A = 4p (3.39× 106 m)2 = 1.44419× 1014 m2 4 V = p (3.39× 106 m)3 = 1.63188× 1020 m3 3 ROUND: C = 2.13× 107 m, A = 1.44× 1014 m2 , V = 1.63× 1020 m3 DOUBLE-CHECK: The units are correct and the orders of magnitude are reasonable. r r THINK: The two vectors are A = (23.0,59.0) and B = (90.0,-150.0) . Sketch and find the magnitude and angle with respect to the positive x-axis. CALCULATE:

1.81.

SKETCH:

26

RESEARCH:

r ˆ + Cy y ˆ , the magnitude is given by the formula For any vector C = Cx x

r Cy C = Cx2 + Cy2 , and the angle made with the x-axis is tanqC = . Cx r r æ Ay ö æ By ö -1 A = Ax2 + Ay2 , B = Bx2 + By2 , q A = tan-1 çç ÷÷ , q B = tan çç ÷÷ è Ax ø è Bx ø uv u v A = (23.0)2 + (59.0)2 = 63.3246, B = (90.0)2 + (-150.0)2 = 174.9286 CALCULATE: æ 59.0 ö æ -150.0 ö q A = tan-1 ç = 68.7026°, q B = tan-1 ç ÷ ÷ = -59.0362° è 23.0 ø è 90.0 ø r r ROUND: Three significant figures: A = 63.3 at 68.7°, B = 175 at - 59.0° or 301.0°. DOUBLE-CHECK: Each magnitude is greater than the components but less than the sum of the components and the angles place the vectors in the proper quadrants. SIMPLIFY:

1.82.

THINK: Add the components of the vectors. Find the magnitude and the angle from r r the positive x-axis of the resultant vector. A = (23.0,59.0) and B = (90.0,-150.0). SKETCH:

r r RESEARCH: C = (Cx ,Cy ), Ci = nAi + mBi with n= -1 and m= +1 , C = Cx2 + Cy2 , tanqC =

Cy Cx

.

æ Cy ö -1 SIMPLIFY: Cx = - Ax + Bx , Cy = - Ay + By , qC = tan çç ÷÷ è Cx ø

27

Bauer/Westfall: University Physics, 1E CALCULATE: Cx = -23.0+ 90.0 = 67.0, Cy = -59.0+ (-150) = -209.0, r æ -209.0 ö C = (67.0)2 + (-209.0)2 = 219.477, and qC = tan-1 ç ÷ = -72.225° . è 67.0 ø r ROUND: C = 219 at - 72.2° or 288° DOUBLE-CHECK: The magnitude is greater than each component but less than the sum of the components and the angle is also in the correct quadrant. The answer is reasonable. 1.83.

THINK: Add the components of the vectors (with applicable multiplication of each vector). Find the magnitude and the angle from the positive x-axis of the resultant r r vector. A = (23.0,59.0) and B = (90.0,-150.0). SKETCH:

r r Cy RESEARCH: C = (Cx ,Cy ), Ci = nAi + mBi with n= -5 and m= +1, C = Cx2 + Cy2 , tanqC = Cx . r æ Cy ö -1 SIMPLIFY: Cx = -5Ax + Bx , Cy = -5Ay + By , C = Cx2 + Cy2 , qC = tan çç ÷÷ è Cx ø Cx = -5(23.0) + 90.0 = -25.0, Cy = -5(59.0) + (-150) = -445.0 uv C = (-25.0)2 + (-445.0)2 = 445.702 CALCULATE: æ -445.0 ö qC = tan-1 ç ÷ = 86.785° Þ 180° + 86.785° = 266.785° è -25.0 ø uv ROUND: C = 446 at 267° or -93.2° DOUBLE-CHECK: The magnitude is greater than each component but less than the sum of the components and the angle is also in the correct quadrant. The answer is reasonable. 1.84.

THINK: Add the components of the vectors (with applicable multiplication of each vector). Find the magnitude and the angle from the positive x-axis of the resultant r r vector. A = (23.0,59.0) and B = (90.0,-150.0) .

SKETCH:

28

r r Cy RESEARCH: C = (Cx ,Cy ), Ci = nAi + mBi with n= 3 and m= -7, C = Cx2 + Cy2 , tanqC = . Cx æ Cy ö -1 SIMPLIFY: Cx = 3Ax - 7Bx , Cy = 3Ay - 7By , qC = tan çç ÷÷ è Cx ø Cx = 3(23.0) - 7(90.0) = -561.0, Cy = 3(59.0) - 7(-150) = 1227.0 r C = (-561.0)2 + (1227.0)2 = 1349.17 CALCULATE: æ 1227.0 ö qC = tan-1 ç ÷ = -65.43° Þ 180° - 65.43° = 114.57° è -561.0 ø r 3 ROUND: C = 1.35×10 at 115° DOUBLE-CHECK: The magnitude is greater than each component but less than the sum of the components and the angle is also in the correct quadrant. 1.85.

THINK: Sum the components of both vectors and find the magnitude and the angle r r from the positive x-axis of the resultant vector. A = (23.0,59.0) and B = (90.0,-150.0) . SKETCH: (a)

(b)

29

Bauer/Westfall: University Physics, 1E

r r Cy RESEARCH: C = (Cx ,Cy ), Ci = nAi + mBi , C = Cx2 + Cy2 , tanqC = Cx SIMPLIFY: -1 (a) Since n= -3 and m= 9, Cx = -3Ax + 9Bx and Cy = -3Ay + 9By . Also, qC = tan Cy / Cx . -1 (b) Since n= -5 and m= 8, Cx = -5Ax + 8Bx and Cy = -5Ay + 8By . Also, qC = tan

( (C

y

) /C ) . x

CALCULATE: (a) Cx = -3(23.0) + 9(90.0) = 741.0, Cy = -3(59.0) + 9(-150) = -1527.0 r C = (741.0)2 + (-1527.0)2 = 1697.29 æ -1527.0 ö qC = tan-1 ç ÷ = -64.11° è 741.0 ø (b) Cx = -5(23.0) + 8(90.0) = 605.0, Cy = -5(59.0) + 8(-150) = -1495.0 r C = (605.0)2 + (-1495.0)2 = 1612.78

1.86.

æ -1495.0 ö qC = tan-1 ç ÷ = -67.97° è 605.0 ø ROUND: r (a) C = 1.70×103 at -64.1° or 296° r (b) C = 1.61× 103 at -68.0° or 292° DOUBLE-CHECK: Each magnitude is greater than the components but less than the sum of the components and the angles place the vectors in the proper quadrants. The calculated answers are reasonable. r r THINK: The vectors are A = (Ax , Ay ) = (-30.0 m,-50.0 m) and B = (Bx ,By ) = (30.0 m,50.0 m). Sketch and find the magnitude and angle with respect to the positive x-axis for each.

SKETCH:

30

r Cy RESEARCH: C = Cx2 + Cy2 , tanqC = Cx uv uv æ Ay ö æ By ö -1 A = Ax2 + Ay2 , B = Bx2 + By2 , q A = tan-1 çç ÷÷ , q B = tan çç ÷÷ è Ax ø è Bx ø CALCULATE: uv uv A = (-30.0 m)2 + (-50.0 m)2 = 58.3095 m, B = (30.0 m)2 + (50.0 m)2 = 58.3095 m SIMPLIFY:

æ -50.0 m ö q A = tan-1 ç ÷ = 59.036° Þ 180° + 59.036° = 239.036° è -30.0 m ø æ 50.0 m ö q B = tan-1 ç ÷ = 59.036° è 30.0 m ø r r ROUND: A = 58.3 m at 239° or -121° , and B = 58.3 m at 59.0°. DOUBLE-CHECK: Each magnitude is greater than the components of the vector but less than the sum of the components and the angles place the vectors in the proper quadrants. 1.87.

THINK: A variable is proportional to some other variable by a constant. This means the ratio of one variable to another is a constant. Therefore, both ratios are equal. F1 = 200. N , x1 = 8.00 cm and x2 = 40.0 cm. SKETCH:

RESEARCH:

F1 F2 = x1 x2

SIMPLIFY: F2 =

F1x2 x1

31

Bauer/Westfall: University Physics, 1E

CALCULATE: F2 =

(200. N)(40.0 cm) = 1000.0 N 8.00 cm

ROUND: F2 = 1.00× 103 N DOUBLE-CHECK: The ratio of force to distance remains 1:25 for the two distances. The answers are reasonable. 1.88.

THINK: When a variable is proportional to another, it is equal to the other variable multiplied by a constant. Call the constant “a”. SKETCH: A sketch is not needed to solve this problem. RESEARCH: d = at2 SIMPLIFY: d0 = at02 , d0 ' = a(3t0)2 CALCULATE: d0 ' = 9at02 = 9d0 ROUND: The distance increases by a factor of 9. DOUBLE-CHECK: Acceleration is a quadratic relationship between distance and time. It makes sense for the amount of time to increase by a factor larger than 3.

1.89.

THINK: Consider the 90° turns to be precise turns at right angles. (a) The pilot initially flies N, then heads E, then finally heads S. Determine the vector r D that points from the origin to the final point and find its magnitude. The vectors r r r are d1 = 155.3 miles N , d2 = 62.5 miles E and d3 = 47.5 miles S. (b) Now that the vector pointing to the final destination has been computed, r ˆ + (d1 - d3) y ˆ = ( 62.5 miles) x ˆ + ( 107.8 miles) y ˆ, determine the angle the vector makes D = d2x with the origin. The angle the pilot needs to travel is then 180° from this angle. (c) Before the pilot turns S, he is farthest from the origin. This is because when he starts heading S, he is negating the distance travelled N. The only vectors of interest r r are d1 = 155.3 miles N and d2 = 62.5 miles E . SKETCH: (a)

(b)

(c)

32

RESEARCH: r r r r r r ˆ + Dy y ˆ, di = di x x ˆ + di y y ˆ, D = Dx2 + Dy2 (a) D = d1 + d2 + d3 = Dx x (b) tanq =

Dy

, q ' = q ± 180° Dx r r r r r ˆ + di y y ˆ, D = Dx2 + Dy2 (c) Dmax = d1 + d2, di = di x x SIMPLIFY: r r r ˆ, d2 = d2x ˆ, d3 = -d3 y ˆ (a) d1 = d1y r r ˆ + (d1 - d3) y ˆ and D = d22 + (d1 - d3 )2 . Therefore, D = d2x æ Dy ö æ Dy ö -1 -1 (b) q = tan çç ÷÷ and q ' = tan çç ÷÷ ± 180° è Dx ø è Dx ø r r uu v uu v 2 2 $ $ $, D = d x (c) d1 = d1 $ y, d2 = d2 x 2 + d1 y Þ D = d2 + d1 CALCULATE: r D = (62.5 miles)2 + (155.3 miles- 47.5 miles)2 (a) = 124.608 miles æ 107.8 miles ö q ' = tan-1 ç ÷ ± 180° (b) è 62.5 miles ø = 59.896° ± 180° = 239.896° or - 120.104° r (c) D = (62.5 miles)2 + (155.3 miles)2 = 167.405 miles ROUND: r (a) D = 125 miles (b) q ' = 240.° or - 120.° (from positive x-axis or E) r (c) D = 167 miles DOUBLE-CHECK: (a) The total distance is less than the distance travelled north, which is expected since the pilot eventually turns around and heads south. (b) The pilot is clearly NE of the origin and the angle to return must be SW. (c) This distance is greater than the distance which included the pilot travelling S, as it should be. 1.90.

THINK:

33

Bauer/Westfall: University Physics, 1E (a) If an observer sees the Moon fully cover the Sun, then light rays from the outer edge of the Sun are blocked by the outer edge of the Moon. This means a line pointing to the outer edge of the Moon also points to the outer edge of the Sun. This in turn means that the lines share a common angle. The radii of the Moon and Sun are, respectively, rM = 1.74× 106 m and rS = 6.96× 108 m. The distance from the Moon to the Earth is dEM = 3.84× 108 m. (b) In part (a), the origin of the light ray is assumed to be the centre of the Earth. In fact, the observer is on the surface of the Earth, rE = 6378 km. This difference in observer position should then be related to the actual distance to the Moon. The observed Earth to Moon distance remains the same, dEM = 3.84× 108 m, while the actual distance is the observed distance minus the radius of the Earth. (c) Given the relative error of 1.69%between the actual and observed distance to the Moon, there should be the same relative error in the difference between the observed and actual distance to the Sun. dES(observed) = 1.54× 1011 m. SKETCH: (a)

(b)

(c) Not applicable. RESEARCH: æ oppositeö (a) tanq = ç ÷ è adjacent ø dEM (observed) - dEM (actual) (b) relative error = dEM (actual) (c) dES(actual) = (1- relativeerror)dES(observed) SIMPLIFY: æ r ö æ r ö rd (a) tanq = ç M ÷ = ç S ÷ Þ dES = S EM rM è dEM ø è dES ø dEM (observed) - (dEM (observed) - rE ) dEM (observed) - rE rE = dEM (observed) - rE

relative error = (b)

dES(actual) = (1- 0.0169)dES(observed) = 0.9831dES(observed) CALCULATE: (c)

34

Full file at http://testbank360.eu/solution-manual-university-physics-with-modernphysics-1st-edition-bauer (6.96× 108 m)(3.84×108 m) = 1.536× 1011 m (1.74× 106 m) 6378000 m = 0.01689 (b) relative error = 3.84× 108 m - 6378000 (c) dES(actual) = 0.9831(1.54× 1011 m) = 1.513× 1011 m ROUND: (a) dES = 1.54×1011 m (b) relative error = 1.69% (c) dES(actual) = 1.51× 1011 m DOUBLE-CHECK: (a) The distance from the Earth to the Sun is about 300 times the distance from the Earth to the Moon, so the answer is reasonable. (b) The radius of Earth is small compared to the distance from the Earth to the Moon, so the error calculated is small. (c) The relative error is small so there should be a small difference between the actual and the observed distance from the Earth to the Sun. (a) dES =

1.91.

THINK: The problem involves adding vectors. Break the vectors into components r r and sum the components. The vectors are: d1 = 1.50 km due N , d2 = 1.50 km 20.0° N of W r and d3 = 1.50 km due N . Find the length of the resultant, and the angle it makes with the vertical. Let a = 20.0°. SKETCH:

æD ö r r r r r r ˆ + diy y ˆ, D = Dx2 + Dy2 , tanq = ç x ÷ RESEARCH: D = d1 + d2 + d3, di = dix x ç Dy ÷ è ø r r r ˆ, d2 = -d2x x ˆ + d2y y ˆ = -d2 ( cosa ) x ˆ + d2 ( sina ) y ˆ, d3 = d3 y ˆ d1 = d1y r D = (-d2 cosa )2 + (d1 + d3 + d2 sina )2 SIMPLIFY: æD ö q = tan-1 ç x ÷ ç Dy ÷ è ø

35

Bauer/Westfall: University Physics, 1E r D = (-1.50cos(20.0°) km)2 + (3.00 km + 1.50sin(20.0°) km)2 CALCULATE:

= 7.9868 km2 + 12.3414 km2 = 3.7852 km æ -1.4095 km ö q = tan-1 ç ÷ = -21.862° è 3.5130 km ø

r ROUND: D = 3.79 km at 21.9° W of N

DOUBLE-CHECK: The only directions travelled were N or NW, so the final direction should be in the NW region. 1.92.

THINK: If the number of molecules is proportional to the volume, then the ratio of volumes should be the same as the ratio of the molecules. 1 mol = 6.02×1023 molecules, volume of mol = 22.4 L and the volume of one breath is 0.500 L. Only 80.0% of the volume of one breath is nitrogen. SKETCH: Not applicable. VNitrogen #molecules in one breath Nbreath = = RESEARCH: VNitrogen = 0.800Vbreath , Vmol #molecules in a mol Nmol SIMPLIFY: N breath =

VNitrogen (Nmol )

CALCULATE: N breath =

Vmol

=

0.800VBreath (Nmol ) Vmol

0.800(0.500 L)(6.02× 1023 molecules) = 1.07500×1022 molecules (22.4 L)

ROUND: N breath = 1.08× 1022 molecules DOUBLE-CHECK: Since the volume of one breath is about 50 times smaller than the volume of one mole of gas, the number of nitrogen molecules in one breath should be about 50 times smaller than the number of molecules in a mole. 1.93.

THINK: 24.9 seconds of arc represents the angle between the observer and the point where the radius subtends. This value must be converted to radians. The radius of Mars is rM = 6784 km. SKETCH:

1° æ öæ 2p radians ö -4 RESEARCH: s = rq , 24.9 arc secondsç ÷ç ÷ = 1.207× 10 radians è 3600 arc seconds øè 360° ø ær ö SIMPLIFY: rM = dEMq Þ dEM = ç M ÷ èq ø CALCULATE:

rM 6784 km = = 5.6197× 107 km q (1.207× 10-4 radians)

ROUND: dEM = 5.62× 107 km DOUBLE-CHECK: The difference between the mean orbital radii of Earth and Mars is about 7× 107 km. 1.94.

THINK: If the quarterback is in the exact centre of a rectangular field, then each corner should be the same distance from the centre. Only the angle changes for

36

Full file at http://testbank360.eu/solution-manual-university-physics-with-modernphysics-1st-edition-bauer each corner. The width of the field is 53 1/3 yards and the length is 100. yards. Since the question states that the length is exactly 100 yards, the precision of the final answer will be limited by the width. SKETCH:

di y r r ˆ + diy y ˆ, di = di x2 + di y2 , tanqi = RESEARCH: di = dix x di x éæ wö ù 2 2 êç ÷ ú r r r r r 2 æ wö æ l ö æ wö d1 = d2 = d3 = d4 = ç ÷ + ç ÷ = d , q1 = tan-1 ê è ø ú = tan-1 ç ÷ êæ l öú è 2 ø è 2ø è l ø ê ç 2÷ ú ëè øû

SIMPLIFY: CALCULATE:

2 2 r æ 53 1/3 yards ö æ 100 yards ö -1 æ 53 1/3 ö (a) d = ç + ÷ ç ÷ = 56.667 yards, q1 = tan ç ÷ = 28.072° 2 2 è 100 ø è ø è ø q2 = 180° - q1 = 180° - 28.072° = 151.928° (b) q3 = 180° + q1 = 180° + 28.072° = 208.072° q 4 = 360° - q1 = 360° - 28.072° = 331.928° ROUND: r (a) d1 = 56.7 yards at 28.1° r r r (b) d2 = 56.7 yards at 152°, d3 = 56.7 yards at 208°, d4 = 56.7 yards at 332° r r r r DOUBLE-CHECK: d1 & d3 and d2 & d4 are 180 ° apart. This is expected when throwing at opposite corners of the field. The answers are reasonable.

1.95.

THINK: Assume the Cornell Electron Storage Ring is a perfect circle with a circumference of C = 768.4 m. Recall the exact conversion1 m = (100/ 2.54) inches. SKETCH:

= 2d , r= 2 RESEARCH: Cπr SIMPLIFY:

d=

C æ 100 in ö π çè 2.54 m ÷ø

CALCULATE: d =

( 768.4 m) æ 100 in ö = 9629.5007 inches π

ç 2.54 m ÷ è ø

ROUND: d = 9630. inches DOUBLE-CHECK: There are 12 inches in a foot and 5280 feet in a mile. Therefore there are 63,360 inch/mile. Our answer for the Cornell ring is thus about 1/6 th of a mile, which seems the right order of magnitude. 37

Bauer/Westfall: University Physics, 1E 1.96.

THINK: 4.00% of the 0.500 L for each exhalation is composed of carbon dioxide. Assume 1 mole ( 6.02×1023 molecules) has a volume of 22.4 L. SKETCH: Not applicable. RESEARCH: Assume on average a person exhales 2.50×104 times a day. V N VCO2 = 0.0400Vbreath , CO2 = #molecules in one breath = breath Vmol #molecules in a mol N mol SIMPLIFY: N breath =

VCO2 Vmol

(N mol ) =

0.04Vbreath (Nmol ) Vmol

0.0400(0.500 L) (6.02× 1023 molecules) = 5.375× 1020 molecules 22.4 L N day = #molecules exhaled in a day

CALCULATE: N breath =

(a)

( ) = ( 2.5× 10 ) ( 5.375× 10 = 2.5× 104 N breath 4

20

)

molecules

= 1.34375× 10 molecules 25

(b) mCO2 =

1.34375×1025 molecules æ 365 days ö æ 1 mole ö æ 44 kg ö 5 ç ÷ç ÷ ç 1 mole÷ = 3.58482× 10 kg/year 23 1 day 1 year 6.02 × 10 molecule s øè ø è øè

ROUND: 25 (a) N day = 1.34× 10 molecules 5 (b) mCO2 = 3.58×10 kg/year

DOUBLE-CHECK: Since the volume of carbon dioxide in one breath is about 1000 times less than the volume of one mole of an ideal gas, the number of molecules should also be about 1000 times less, which it is. It then follows that the number of carbon dioxide molecules exhaled in one day should be on the order of 105 magnitude greater. The conversion to mass of carbon dioxide exhaled in a year should result in an answer with an order of magnitude of about 105 . 1.97.

THINK: Consider the Sun to be at the centre of a circle with Mercury on its circumference. This gives rM = 4.6× 1010 m as the radius of the circle. Earth is located a distance rE = 1.5× 1011 m from the Sun so that the three bodies form a triangle. The vector from Earth to the Sun is at 0° . The vector from Earth to Mercury intersects Mercury’s orbit once when Mercury is at a maximum angular separation from the Sun in the sky. This tangential vector is perpendicular to the radius vector of Mercury’s orbit. The three bodies form a right angle triangle with rE as the hypotenuse. Trigonometry can be used to solve for the angle and distance.

SKETCH:

38

RESEARCH: rE2 = rM 2 + R2, rE sinq = rM -1 æ r ö SIMPLIFY: R = rE2 - rM 2 , q = sin ç M ÷ è rE ø 10 11 2 10 2 11 -1 æ 4.6× 10 ö = 17.858° CALCULATE: R = (1.5×10 ) - (4.6×10 ) = 1.4277×10 m, q = sin ç 11 ÷ è 1.5× 10 ø ROUND: R = 1.4× 1011 m, q = 18° DOUBLE-CHECK: If it had been assumed that the maximum angular separation -1 occurred when the Earth to Sun to Mercury angle was 90 ° , q = tan ( rM / rE ) would be about 17°. The maximum angle should be greater than this and it is, so the answer is reasonable.

39

Solution manual university physics with modern physics 1st edition bauer

Published on Jan 13, 2017

solution manual university physics with modern physics 1st edition bauer. Full file at http://testbank360.eu/solution-manual-university...

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