\nThe advantage of the scientific notation becomes even more pronounced when we have to compute how many times heavier a proton is than an electron. Instead of writing the value of the ratio, r, as
\n
![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-8-Module-2-Lesson-11-Example-Answer-Key-1.png\")
\nwe express it as
\nr=\\(\\frac{1.672622 \\times 10^{-27}}{9.10938291 \\times 10^{-31}}\\).
\n\u2192 Should we eliminate the power of 10 in the numerator or denominator? Explain.
\n\u2192 Using the theorem on generalized equivalent fractions, we can eliminate the negative power of 10 in the numerator and denominator to see what we are doing more clearly. Anticipating that
\n10-31<\/sup>\u00d71031<\/sup>=1, we can multiply the numerator and denominator of the (complex) fraction by 1031<\/sup>
\n
![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-8-Module-2-Lesson-11-Example-Answer-Key-2.png\")
\nUsing the first law of exponents (10) presented in Lesson 5, we get
\n
![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-8-Module-2-Lesson-11-Example-Answer-Key-3.png\")
\nNote that since we are using scientific notation, we can interpret an approximate value of r right away. For example, we see
\n
![\"Eureka](\"https:\/\/ccssmathanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-Grade-8-Module-2-Lesson-11-Example-Answer-Key-4.png\")
\nso that r is approximately \\(\\frac{1}{5}\\) \u00d710,000, which is 2,000. Thus, we expect a proton to be about two thousand times heavier than an electron.<\/p>\n
Example 2. Exercise 1. Exercise 2. Exercise 3. Exercise 4. Exercise 5. Exercise 6. Question 1. Question 2. Question 3. Question 4. Question 1. Question 2. Engage NY Eureka Math 8th Grade Module 2 Lesson 11 Answer Key Eureka Math Grade 8 Module 2 Lesson 11 Example Answer Key Example 1. The advantage of the scientific notation becomes even more pronounced when we have to compute how many times heavier a proton is than an electron. Instead of writing the value … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/82327"}],"collection":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/comments?post=82327"}],"version-history":[{"count":1,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/82327\/revisions"}],"predecessor-version":[{"id":91546,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/82327\/revisions\/91546"}],"wp:attachment":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/media?parent=82327"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/categories?post=82327"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/tags?post=82327"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nThe U.S. national debt as of March 23, 2013, rounded to the nearest dollar, is $16,755,133,009,522. According to the 2012 U.S. census, there are about 313,914,040 U.S. citizens. What is each citizen\u2019s approximate share of the debt?
\nAnswer:
\n\\(\\frac{1.6755 \\times 10^{13}}{3.14 \\times 10^{8}}\\)=\\(\\frac{1.6755}{3.14}\\)\u00d7\\(\\frac{10^{13}}{10^{8}}\\)
\n=\\(\\frac{1.6755}{3.14}\\)\u00d7105<\/sup>
\n=0.533598…\u00d7105<\/sup>
\n\u22480.5336\u00d7105<\/sup>
\n=53360
\nEach U.S. citizen\u2019s share of the national debt is about $53,360.<\/p>\nEureka Math Grade 8 Module 2 Lesson 11 Exercise Answer Key<\/h3>\n
\nThe mass of a proton is
\n0.000 000 000 000 000 000 000 000 001 672 622 kg.
\nIn scientific notation it is
\nAnswer:
\n1.672622\u00d710-27<\/sup> kg.<\/p>\n
\nThe mass of an electron is
\n0.000 000 000 000 000 000 000 000 000 000 910 938 291 kg.
\nIn scientific notation it is
\nAnswer:
\n9.10938291\u00d710-31<\/sup> kg.<\/p>\n
\nWrite the ratio that compares the mass of a proton to the mass of an electron.
\nAnswer:
\nRatio: (1.672622\u00d710-27<\/sup>) :(9.10938291\u00d710-31<\/sup>)<\/p>\n
\nCompute how many times heavier a proton is than an electron (i.e., find the value of the ratio). Round your final answer to the nearest one.
\nAnswer:
\nLet r= the value of the ratio, then:
\nr=\\(\\frac{1.672622 \\times 10^{-27}}{9.10938291 \\times 10^{-31}}\\)
\n=\\(\\frac{1.672622 \\times 10^{-27} \\times 10^{31}}{9.10938291 \\times 10^{-31} \\times 10^{31}}\\)
\n=\\(\\frac{1.672622 \\times 10^{4}}{9.10938291}\\)\u00d7104<\/sup>
\n=\\(\\frac{1.672622 \\times 10^{8}}{9.10938291 \\times 10^{8}}\\) \u00d7104<\/sup>
\n=\\(\\frac{167,262,200}{910,938,291}\\)\u00d7104<\/sup>
\n=0.183615291675\u00d7104<\/sup>
\n=1836.15291675
\n\u22481836<\/p>\n
\nThe geographic area of California is 163,696 sq.mi., and the geographic area of the U.S. is 3,794,101 sq.mi. Let\u2019s round off these figures to 1.637\u00d7105<\/sup> and 3.794\u00d7106<\/sup>. In terms of area, roughly estimate how many Californias would make up one U.S. Then compute the answer to the nearest ones.
\nAnswer:
\n\\(\\frac{3.794 \\times 10^{6}}{1.637 \\times 10^{5}}\\)=\\(\\frac{3.794}{1.637}\\)\u00d7\\(\\frac{10^{6}}{10^{5}}\\)
\n=\\(\\frac{3.794}{1.637}\\)\u00d710
\n=2.3176…\u00d710
\n\u22482.318\u00d710
\n=23.18
\nIt would take about 23 Californias to make up one U.S.<\/p>\n
\nThe average distance from Earth to the moon is about 3.84\u00d7105<\/sup> km, and the distance from Earth to Mars is approximately 9.24\u00d7107<\/sup> km in year 2014. On this simplistic level, how much farther is traveling from Earth to Mars than from Earth to the moon?
\n9.24\u00d7107<\/sup>-3.84\u00d7105<\/sup>=924\u00d7105<\/sup>-3.84\u00d7105<\/sup>
\n=(924-3.84)\u00d7105<\/sup>
\n=920.16\u00d7105<\/sup>
\n=92 016 000
\nIt is 92,016,000 km further to travel from Earth to Mars than from Earth to the moon.<\/p>\nEureka Math Grade 8 Module 2 Lesson 11 Problem Set Answer Key<\/h3>\n
\nThere are approximately 7.5\u00d71018<\/sup> grains of sand on Earth. There are approximately 7\u00d7102<\/sup>7 atoms in an average human body. Are there more grains of sand on Earth or atoms in an average human body? How do you know?
\nAnswer:
\nThere are more atoms in the average human body. When comparing the order of magnitude of each number,
\n27>18; therefore, 7\u00d71027<\/sup>>7.5\u00d71018<\/sup>.<\/p>\n
\nAbout how many times more atoms are in a human body compared to grains of sand on Earth?
\nAnswer:
\n\\(\\frac{7 \\times 10^{27}}{7.5 \\times 10^{18}}\\)=\\(\\frac{7}{7.5}\\)\u00d7\\(\\frac{10^{27}}{10^{18}}\\)
\n\u22481\u00d71027-18<\/sup>
\n\u22481\u00d7109<\/sup>
\n\u2248109<\/sup>
\nThere are about 1,000,000,000 times more atoms in the human body compared to grains of sand on Earth.<\/p>\n
\nSuppose the geographic areas of California and the U.S. are 1.637\u00d7105<\/sup> and 3.794\u00d7106<\/sup> sq.mi., respectively. California\u2019s population (as of 2012) is approximately 3.804\u00d7107<\/sup> people. If population were proportional to area, what would be the U.S. population?
\nAnswer:
\nWe already know from Exercise 5 that it would take about 23 Californias to make up one U.S. Then the population of the U.S. would be 23 times the population of California, which is
\n23\u00d73.804\u00d7107<\/sup>=87.492\u00d7107<\/sup>
\n=8.7492\u00d7108<\/sup>
\n=874,920,000.<\/p>\n
\nThe actual population of the U.S. (as of 2012) is approximately 3.14\u00d7108<\/sup>. How does the population density of California (i.e., the number of people per square mile) compare with the population density of the U.S.?
\nAnswer:
\nPopulation density of California per square mile:
\n\\(\\frac{3.804 \\times 10^{7}}{1.637 \\times 10^{5}}\\)=\\(\\frac{3.804}{1.637}\\)\u00d7\\(\\frac{10^{7}}{10^{5}}\\)
\n=2.32376…\u00d7102<\/sup>
\n\u22482.32\u00d7102<\/sup>
\n=232
\nPopulation density of the U.S. per square mile:
\n\\(\\frac{3.14 \\times 10^{8}}{3.794 \\times 10^{6}}\\)=\\(\\frac{3.14}{3.794}\\)\u00d7\\(\\frac{10^{8}}{10^{6}}\\)
\n=0.8276…\u00d7102<\/sup>
\n\u22480.83\u00d7102<\/sup>
\n=83
\nPopulation density of California compared to the population density of the U.S.:
\n\\(\\frac{232}{83}\\)=2.7951\u2026
\n\u22482.8
\nCalifornia is about 3 times as dense as the U.S. in terms of population.<\/p>\nEureka Math Grade 8 Module 2 Lesson 11 Exit Ticket Answer Key<\/h3>\n
\nTwo of the largest mammals on earth are the blue whale and the elephant. An adult male blue whale weighs about 170 tonnes or long tons. (1 tonne = 1000 kg)
\nShow that the weight of an adult blue whale is 1.7\u00d7105<\/sup> kg.
\nAnswer:
\nLet x (or any other symbol) represent the number of kilograms an adult blue whale weighs.
\n170\u00d71000=x
\n1.7\u00d7105<\/sup>=x<\/p>\n
\nAn adult male elephant weighs about 9.07\u00d7103<\/sup> kg.
\nCompute how many times heavier an adult male blue whale is than an adult male elephant (i.e., find the value of the ratio). Round your final answer to the nearest one.
\nAnswer:
\nLet r be the value of the ratio.
\nr=\\(\\frac{1.7 \\times 10^{5}}{9.07 \\times 10^{3}}\\)
\n=\\(\\frac{1.7}{9.07}\\)\u00d7102<\/sup>
\n=0.18743\u00d7102<\/sup>
\n=18.743
\n\u224819
\nThe blue whale is 19 times heavier than the elephant.<\/p>\n","protected":false},"excerpt":{"rendered":"