Bradford, Steve Homework 1 Due: Jun , 8:00 pm Inst: Randy Bybee 1

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1 Bradford, Steve Homework 1 Due: Jun , 8:00 pm Inst: Randy Bybee 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Calculate the density of a solid cube that measures 5.61 cm on each side and has a mass of 527 g. Correct answer: g/cm 3. Let : m = 527 g and s = 5.61 cm. The density of an object is defined as mass per unit value. Thus, ρ = m V = m s 3 = 527 g (5.61 cm) 3 = g/cm (part 1 of 1) 10 points A sphere of metal has a radius of 5.4 cm and a density of 9.63 g/cm 3. What is the mass of the sphere? Correct answer: g. Let : r = 5.4 cm and ρ = 9.63 g/cm 3. Density is mass per unit volume, so ρ = m V m = ρ V = ρ 4 3 π r3 = ( 9.63 g/cm 3) 4 π (5.4 cm)3 3 = g. 003 (part 1 of 2) 5 points A structural I beam is made of iron. A view of its cross-section and its dimensions is shown in the figure, where d = cm, w = 16.5 cm, and h = 22.3 cm. d d w The density of iron is 7560 kg/m 3. What is the mass of a section 2.38 m long? Correct answer: kg. Let : d = cm, w = 16.5 cm, h = 22.3 cm, l = 2.38 m. h and The cross-sectional area of the beam is A = 2 (16.5 cm) (0.969 cm) + (22.3 cm cm) (1 cm) = cm 2 = m 2. The volume of a piece l = 2.38 m long is then V = A l = m 3. The mass is then given by m = ρ V = kg. 004 (part 2 of 2) 5 points How many atoms are there in this section? Correct answer: atoms. Let : m = kg = g and N A = /mol. Presuming that the atoms are iron, we look to the periodic table and find the molar mass to

2 Bradford, Steve Homework 1 Due: Jun , 8:00 pm Inst: Randy Bybee 2 be M = g/mol. The number of moles is then n = m M = g g/mol and the number of atoms is = mol N = n N A = ( mol) ( /mol) = atoms. 005 (part 1 of 1) 10 points A cylinder, 19 cm long and 6 cm in radius, is made of two different metals bonded end-toend to make a single bar. The densities are 4.9 g/cm 3 and 6.4 g/cm 3. r l x l x Let x be the length of the lighter metal; then l x is the length of the heavier metal. Thus, m = m 1 + m 2 = ρ 1 π r 2 x + ρ 2 π r 2 (l x) = ρ 1 π r 2 x + ρ 2 π r 2 l ρ 2 π r 2 x. 19 cm Therefore m ρ 2 π r 2 l = ρ 1 π r 2 x ρ 2 π r 2 x and x π r 2 (ρ 1 ρ 2 ) = m ρ 2 π r 2 l. 6 cm What length of the lighter metal is needed if the total mass is g? Correct answer: cm. Let : l = 19 cm, r = 6 cm, ρ 1 = 4.9 g/cm 3, ρ 2 = 6.4 g/cm 3, m = g. and Volume of a bar of radius r and length l is V = π r 2 l Consequently, x = m ρ 2 π r 2 l π r 2 (ρ 1 ρ 2 ) = (11850 g) (6.4 g/cm3 ) π (6 cm) 2 (19 cm) π (6 cm) 2 (4.9 g/cm g/cm 3 ) = cm. 006 (part 1 of 4) 3 points It is given that r and s are distances with unit [L], t is a time with unit [T] and θ is an angle in radians. and its density is so that ρ = m V = m π r 2 l m = ρ π r 2 l 1. [c]=[l]/[t 2 ] 2. [c]=[l 2 ] s = c t

3 Bradford, Steve Homework 1 Due: Jun , 8:00 pm Inst: Randy Bybee 3 3. [c]=[t] 4. [c]=1/[l] 10. [c]=1/[t] 5. [c]=[l]/[t] correct 6. [c]=1/[t] 7. [c]=[l] [T 2 ] c = s t 2 = [c] = [s] [t 2 ] = [L]/[T 2 ]. 8. [c]=[l] 9. [c]=[l] [T] 10. [c]=[t]/[l] 008 (part 3 of 4) 3 points s = r cos(c t) c = s t = [c] = [s] [t] = [L]/[T ]. 007 (part 2 of 4) 2 points 1. [c]=[l] 2. [c]=[l 2 ] 3. [c]=[l] [T 2 ] 4. [c]=[l]/[t] 5. [c]=[t]/[l] s = c t 2 6. [c]=[l]/[t 2 ] correct 7. [c]=[t] 8. [c]=1/[l] 9. [c]=[l] [T] 1. [c]=[l] [T 2 ] 2. [c]=[l] 3. [c]=[l]/[t] 4. [c]=[t]/[l] 5. [c]=[t] 6. [c]=1/[t] correct 7. [c]=[l] [T] 8. [c]=1/[l] 9. [c]=[l]/[t 2 ] 10. [c]=[l 2 ] The argument of a triangular function is dimensionless (radians), and the result of the triangular function is also dimensionless. For the equation s = r cos(c t), we have [c] = 1 [t] = 1/[T ]. 009 (part 4 of 4) 2 points

4 Bradford, Steve Homework 1 Due: Jun , 8:00 pm Inst: Randy Bybee 4 1. [c]=[t]/[l] 2. [c]=[l 2 ] 3. [c]=[l] [T 2 ] 4. [c]=1/[l] 5. [c]=[t] 6. [c]=[l]/[t 2 ] 7. [c]=[l] [T] 8. [c]=1/[t] 9. [c]=[l]/[t] 10. [c]=[l] correct θ = s c c = s θ = [c] = [s] [θ] = [L]. 010 (part 1 of 2) 5 points s is a distance with unit [L], t is a time with unit [T] and θ is an angle in radians. What is the required dimension for the quantity c 1 in s = c 1 t? 1. [T]/[L] 2. 1/[T] 3. [L]/[T] 2 4. [L]/[T] correct 5. 1/[T] /[L] 7. [T] 8. [T] 2 /[L] 9. None of these 10. [L] The argument of a trig function (when in radians) is dimensionless, and the result of the trig function is also dimensionless. s = c 1 t c 1 = s t = [c 1 ] = [s]/[t] = [L]/[T ] 011 (part 2 of 2) 5 points What is the required dimension for the quantity c 4 in s = c 3 cos(c 4 t)? 1. [T] 2 /[L] 2. 1/[T] correct 3. 1/[L] 4. None of these 5. [T] 6. [L] 7. [T]/[L] 8. [L]/[T] 9. [L]/[T] /[T] 2 1 = c 4 t c 4 = 1 t = [c 4 ] = 1/[T ]

5 Bradford, Steve Homework 1 Due: Jun , 8:00 pm Inst: Randy Bybee (part 1 of 1) 10 points Rain drops fall on a tile surface at a density of 4602 drops/ft 2. There are 19 tiles/ft 2. How many drops fall on each tile? Correct answer: drops/tile. We need a drops/tile, so the unit of ft 2 must reduce. Dimensional analysis: Thus drops ft 2 tiles ft 2 = drops ft 2 ft 2 tiles = drops tiles x = ρ n 013 (part 1 of 2) 5 points Grains of fine beach sand are assumed to be spheres of radius 53 µm. These grains are made of silicon dioxide which has a density of 2600 kg/m 3. a) What is the mass of each grain of sand? Correct answer: kg. Since ρ = m, then each grain has a mass of V m = ρ V = ρ 4 3 π r3 = ( 2600 kg/m 3) 4 π (53 µm)3 3 ( ) 3 1 m 10 6 µm = kg. 014 (part 2 of 2) 5 points Consider a cube whose sides are 1.39 m long. b) How many kg of sand would it take for the total surface area of all the grains of sand to equal the surface area of the cube? Correct answer: kg. Each grain of sand has a surface area of A g = 4 π r 2, and the cube has a surface area of A c = 6 a 2, so the required number of grains of sand is n = A c A g = 6 a2 4 π r 2. Thus the total mass of the sand will be M tot = n m = 3 m a2 2 π r 2 = 3 ( kg) (1.39 m) 2 2 π (53 µm) 2 ( 10 6 ) 2 µm 1 m = kg. 015 (part 1 of 1) 10 points Given: An acre is an area equivalent to that of a rectangle 60.5 yd wide and 80 yd long. There are 36 inches in one yard. There are 2.54 cm in one inch. In May 1998, forest fires in southern Mexico and Guatemala spread smoke all the way to Austin. Those fires consumed forest land at a rate of acres/week. On the average, how many square meters of forest are burned down every minute? Correct answer: m 2 /min. A conversion factor for week can be easily min derived 1 wk 7 days 1 day 24 hr 1 hr 60 min wk min. A conversion factor for can also be derived ( 36 in 1 yd where yd2 acre 2.54 cm 1 in m2 acre ) 2 1 m yd cm acre m2 acre, is given in the problem.

6 Bradford, Steve Homework 1 Due: Jun , 8:00 pm Inst: Randy Bybee 6 Therefore the rate R in square meters per minutes is [ m 2 ] /acre R = (29400 acres/week) week/min [ ( m 2 /yd 2 )(4840 yd 2 ] /acre) = (10080 min/week) (29400 acres/week) = m 2 /min. 016 (part 1 of 1) 10 points A light year is the unit of distance that light, with a speed of m/s, travels in one year. What is the surface area of a planet whose radius is 9400 km? Correct answer: lightyears 2. Solution: The relationship between light years and meters is 1 lightyear = 1 year ( m/s) 365 day 24 h year day 3600 s h = m. Thus S = 4 π r 2 4 π ( m) 2 = ( m/lightyears) 2 = lightyears (part 1 of 1) 10 points The number x = has how many significant digits? Correct answer: 3. Figures except zeros to the left of the first nonzero digit are all of significance.