Project 3 - Ryan Fox\'s Quantitative Literacy and Reasoning PDF

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Ryan Fox's Quantitative Literacy and Reasoning...

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Project 3 Part 1: Warm Up: How long is one trillion seconds? 1. 1 Trillion seconds equals 16,666,666,666.67 minutes 2. 16,666,666,666.67 minutes equals 277,777,777.8 hours 3. 277,777,777.8 hours equals 11,574,074.07 days 4. 11,574,074.07 days equals 31,709.79 years

Packet 1: One Trillion Dollars One Dollar Bill measures 6.14 inches by 2.61 inches by 0.0043 inches and weighs about 1 gram. 1. A stack of a million bills would be 1,000,000 x 0.0043 = 4,300 inches high 2. Since there are 12 inches per foot, the stack would be 358.3 feet high 3. The AT&T building or “Batman” building downtown is 617 feet high. a. How high would a stack of one million dollars be relative to the building? 0.58 buildings 4. One trillion is a million millions. a. Thus, a trillion bills would make a stack 358,333,333.3 feet high 5. Since there are 5280 feet per mile, the stack would be 67,866.2 miles high 6. Our current national debt is almost 20 trillion dollars. a. 20 trillion bills would make a stack 135,7323.2 miles high 7. The average distance from earth to the moon is 238,857 miles. a. How high would a stack of 20 trillion bills be relative to the distance to the moon? 5.7 trips to the moon 8. A stack of a million bills would weigh 1,000,000 x 1 = 1,000,000 grams 9. Since there are 28.35 grams per ounce, the stack would weigh 35,273.4 ounces 10. Since there are 16 ounces per pound, the stack would weigh 2,204.6 pounds 11. A 2016 Mini Cooper Hatchback weighs 2,625 pounds. a. How much would a stack of one million dollars weigh relative to a MINI Cooper? 0.84 MINI Coopers

12. One trillion is a million millions. a. Thus, a trillion bills would make a stack weighing 2,204,600,000 pounds 13. Since there a 2000 pounds per ton, the stack would weigh 11,023,000 tons 14. Our current national debt is almost 20 trillion dollars. a. 20 trillion bills would make a stack weighing 220,460,000 tons 15. Scientist believe that a blue whale can weigh as much as 200 tons. a. How much would a stack of 20 trillion bills weigh relative to a blue whale? 1,102,300 blue whales 16. A string of a million bills placed end to end would be 1,000,000 x 6.14= 6,140,000 inches long 17. Since there are 12 inches per foot, the string would be 511,666.7 feet long 18. Since there are 5280 feet in a mile, the string would be 96.9 miles long 19. It is a drive of about 180 miles from Nashville to Knoxville. a. How long would a string of a million bills be relative the distance from Nashville to Knoxville? 0.54 trips from Nashville to Knoxville 20. One trillion is one million millions. a. Thus, a trillion bills would make a string 96,900,000 miles long 21. Our current national debt is almost 20 trillion dollars. a. 20 trillion bills would make a string 1,938,000,000 miles long 22. The average distance from the earth to the sun is 92,960,000 miles. a. How long would a string of 20 trillion bills be relative to the distance to the sun? 20.8 trips to the sun

Part 2: Speed Warm Up: Let’s figure out how long it takes to read a text message. 1.5 seconds to read a text 1. If I was driving “interstate speed” (65 miles per hour), how long would I have traveled reading my text message? Represent your answer in a distance and in terms of some other measurement. a.

65 𝑚𝑝ℎ 1 ℎ𝑟 x 60 𝑚𝑖𝑛𝑠 1 ℎ𝑟

x

1 𝑚𝑖𝑛 60 𝑠𝑒𝑐𝑠

x

5280 𝑓𝑡 1 𝑚𝑖𝑙𝑒

=

343200 𝑓𝑡 3600 𝑠𝑒𝑐𝑠

= 95 ft/ sec

b. 65 mph = 95 ft/sec x 1.5 sec = 142.5 ft traveled in time it took me to read a text. i.

That is almost half the distance of a football field

c. 142.5 ft ÷16 ≈9 car lengths 2. Now perform the same computations driving a “normal road” (45 miles per hour). Again, determine the distance and compare that distance to something else. a.

45 𝑚𝑝ℎ 𝑥 5280 3600

= 66 ft/ sec

b. 45 mph= 66 ft/sec x 1.5 sec = 99 ft traveled in the time it took me read that text. i.

I went 33 yards while reading that text

c. 99 ÷16 ≈6 car lengths 3. Now perform the same computations driving at “residential speeds” (25 miles per hour). Again, determine the distance and compare that distance to something else.

a.

25 𝑚𝑝ℎ 𝑥 5280 = 3600

37 ft/ sec

b. 25 mph= 37 ft/ sec x 1.5 sec = 55 ft traveled in the time it took me to read that text c. 55÷16 ≈3.5 car lengths 4. Write a brief reflection about your work. In the time it takes to look away from the road for less than 2 seconds, even at perceived slow rate of 25 mph, I travel a much farther distance than I would have ever believed. In those brief seconds a person is looking away from the road, they are not aware how far they are actually traveling and that could lead to potential danger for the driver or another. This math clearly proves that a fatal or dangerous car accident really can happen in a a few seconds, and why people should not text and drive.

Packet 2: Audio Engineering Technology: Problem 2: Problems of Marching Band Speed of sound - 1125 ft/ sec 1. A symphony orchestra is arranged in a semi circle around the conductor. The rows are about 3’ apart, with the first row about 6’ from the conductor. A midsized orchestra

typically has about 6 rows of chairs. What is the approximate delay between the instant when a player at the 6th row starts a note and when the conductor hears the note? a. 6 rows x 3 ft apart + 6 feet from conductor = 24 feet b.

1 𝑠𝑒𝑐 x 1125 𝑓𝑡

24 ≈0.021 seconds delay for the conductor to hear the noise

2. In the orchestra, when playing at 4/4 at an Allegro tempo, say, 140 beats per minute, what fraction of a 16th note is the time delay you just found? Do you think this size of delay creates a problem for the listener? (For non musicians, a “beat” in 4/4 time is a quarter note, so a 16th note is 1/4th as long as a “beat”) a. 140 bmp=

140 𝑏𝑒𝑎𝑡𝑠 1 𝑚𝑖𝑛 x 60 𝑠𝑒𝑐 1 𝑚𝑖𝑛

x

4 16𝑡ℎ 𝑛𝑜𝑡𝑒 560 = 60 1 𝑏𝑒𝑎𝑡

= 9.3 sixteenth notes/ second

b. 0.021 x 9.3 = 0.1953 sixteenth notes c. No because it is less than a full note. They are off but not enough that it would be noticable. 3. Now consider a marching band on a football field. It is quite typical for the band to stretch out, at times, so that it covers 165 ft from end to end. For some members of the audience there is 165 ft difference between the nearest and farthest band players. Suppose the band is playing 16th notes, also in 4/4 time with 140 beats per minute. How many note’s delay between a note from a nearby player and a note from a far away player. Could this be a problem for a listener? a.

1 𝑠𝑒𝑐 x 1125 𝑓𝑡

165 ft = 0.147 seconds to hear the delay from the nearest player to the

farthest. b.

140 𝑏𝑒𝑎𝑡𝑠 9.3 16𝑡ℎ 𝑛𝑜𝑡𝑒𝑠 x = 1 𝑚𝑖𝑛 1 𝑠𝑒𝑐

0.147 x 9.3 = 1.3671 16th note delay

c. This delay means they are off beat a whole note, which depending on how many people are off, could be a problem to the listener, or get drowned out by the correct notes. Problem 3: LP and Frequency 1. An LP record revolves 33.3 times per minute approximately. That means the needle which reads the record travels around a groove in the record, traveling 33.3 times the

circumference of the record each time. Suppose the LP records a sound of frequency 1000 Hertz, a typical musical frequency ( a little above the treble clef). The way the LP records this sound is that the groove has a series of bumps along it, which the needle will hit, making it vibrate. It has to hit 1000 bumps per second. How far apart, along the grooves, are the bumps? (It depends on whether the needle is near the outer edge or the middle of the LP. Assume the needle travels along the outer edge.) You will have to measure the diameter of a 33.3 rpm LP to answer this question. Do you think you would be able to see the separation of the bumps with the eye, or would you need a microscope. a. c = πd = C = (3.14)(12 in) = 37.699 in around b. In one minute, the edge of record travels = 33.3 x 37.699 = 1,256.637 inches traveled in one minute c. 1,256.6 ÷60 = 20.94 inches/ sec d.

1000 𝑏𝑢𝑚𝑝𝑠 20.94 𝑖𝑛 x 1000 𝑏𝑢𝑚𝑝𝑠 1 𝑠𝑒𝑐

i.

= 0.02 inches per second.

We would not be able to see it with our eye, we would need to use a microscope....