## Engage NY Eureka Math 8th Grade Module 1 Lesson 10 Answer Key

### Eureka Math Grade 8 Module 1 Lesson 10 Exercise Answer Key

Exercise 1.

The speed of light is 300,000,000 meters per second. The sun is approximately 1.5Ã—10^{11} meters from Earth. How many seconds does it take for sunlight to reach Earth?

Answer:

300 000 000=3Ã—10^{8}

\(\frac{1.5 \times 10^{11}}{3 \times 10^{8}}\) = \(\frac{1.5}{3}\)Ã—\(\frac{10^{11}}{10^{8}}\)

=0.5Ã—10^{3}

=0.5Ã—10Ã—10^{2}

=5Ã—10^{2}

It takes 500 seconds for sunlight to reach Earth.

Exercise 2.

The mass of the moon is about 7.3Ã—10^{22} kg. It would take approximately 26,000,000 moons to equal the mass of the sun. Determine the mass of the sun.

Answer:

26 000 000=2.6Ã—10^{7}

(2.6Ã— 10^{7})(7.3Ã—10^{22})=(2.6Ã—7.3)(10^{7}Ã—10^{22})

=18.98Ã—10^{29}

=1.898Ã—10Ã—10^{29}

=1.898Ã—10^{30}

The mass of the sun is 1.898Ã—10^{30} kg.

Exercise 3.

The mass of Earth is 5.9Ã—10^{24} kg. The mass of Pluto is 13,000,000,000,000,000,000,000 kg. Compared to Pluto, how much greater is Earthâ€™s mass than Plutoâ€™s mass?

Answer:

13 000 000 000 000 000 000 000=1.3Ã—10^{22}

5.9Ã—10^{24}-1.3Ã—10^{22}=(5.9Ã—10^{2})Ã—10^{22}-1.3Ã—10^{22}

=(590-1.3)Ã—10^{22}

=588.7Ã—10^{22}

=5.887Ã—10^{2}Ã—10^{22}

=5.887Ã—10^{24}

The mass of Earth is 5.887Ã—10^{24} kg greater than the mass of Pluto.

Exercise 4.

Using the information in Exercises 2 and 3, find the combined mass of the moon, Earth, and Pluto.

Answer:

7.3Ã—10^{22}+1.3Ã—10^{22}+5.9Ã—10^{24}=(7.3Ã—10^{22}+1.3Ã—10^{22})+5.9Ã—10^{24}

=8.6Ã—10^{22}+5.9Ã—10^{24}

=(8.6+590)Ã—10^{22}

=598.6Ã—10^{22}

=5.986Ã—10^{2}Ã—10^{22}

=5.986Ã—10^{24}

The combined mass of the moon, Earth, and Pluto is 5.986Ã—10^{24} kg.

Exercise 5.

How many combined moon, Earth, and Pluto masses (i.e., the answer to Exercise 4) are needed to equal the mass of the sun (i.e., the answer to Exercise 2)?

Answer:

\(\frac{1.898 \times 10^{30}}{5.986 \times 10^{24}}\) = \(\frac{1.898}{5.986}\) Ã— \(\frac{10^{30}}{10^{24}}\)

=0.3170â€¦Ã—10^{6}

â‰ˆ0.32Ã—10^{6}

=0.32Ã—10Ã—10^{5}

=3.2Ã—10^{5}

It would take 3.2Ã—10^{5} combined masses of the moon, Earth, and Pluto to equal the mass of the sun.

### Eureka Math Grade 8 Module 1 Lesson 10 Problem Set Answer Key

Have students practice operations with numbers written in scientific notation and standard notation.

Question 1.

The sun produces 3.8Ã—10^{2}7 joules of energy per second. How much energy is produced in a year? (Note: a year is approximately 31,000,000 seconds).

Answer:

31 000 000=3.1Ã—10^{7}

(3.8Ã—10^{27})(3.1Ã—10^{7})=(3.8Ã—3.1)(10^{27}Ã—10^{7})

=11.78Ã—10^{34}

=1.178Ã—10Ã—10^{34}

=1.178Ã—10^{35}

The sun produces 1.178Ã—10^{35} joules of energy in a year.

Question 2.

On average, Mercury is about 57,000,000 km from the sun, whereas Neptune is about 4.5Ã—10^{9} km from the sun. What is the difference between Mercuryâ€™s and Neptuneâ€™s distances from the sun?

Answer:

57 000 000=5.7Ã—10^{7}

4.5Ã—10^{9}-5.7Ã—10^{7}=(4.5Ã—10^{2})Ã—10^{7}-5.7Ã—10^{7}

=450Ã—10^{7}-5.7Ã—10^{7}

=(450-5.7)Ã—10^{7}

=444.3Ã—10^{7}

=4.443Ã—10^{2}Ã—10^{7}

=4.443Ã—10^{9}

The difference in the distance of Mercury and Neptune from the sun is 4.443Ã—10^{9} km.

Question 3.

The mass of Earth is approximately 5.9Ã—10^{24} kg, and the mass of Venus is approximately 4.9Ã—10^{24} kg.

a. Find their combined mass.

Answer:

5.9Ã—10^{24}+4.9Ã—10^{24}=(5.9+4.9)Ã—10^{24}

=10.8Ã—10^{24}

=1.08Ã—10Ã—10^{24}

=1.08Ã—10^{25}

The combined mass of Earth and Venus is 1.08Ã—10^{25} kg.

b. Given that the mass of the sun is approximately 1.9Ã—10^{30} kg, how many Venuses and Earths would it take to equal the mass of the sun?

Answer:

\(\frac{1.9 \times 10^{30}}{1.08 \times 10^{25}}\)=\(\frac{1.9}{1.08} \times \frac{10^{30}}{10^{25}}\)

=1.75925â€¦Ã—10^{5}

â‰ˆ1.8Ã—10^{5}

It would take approximately 1.8Ã—10^{5} Venuses and Earths to equal the mass of the sun.

### Eureka Math Grade 8 Module 1 Lesson 10 Exit Ticket Answer Key

Question 1.

The speed of light is 3Ã—10^{8} meters per second. The sun is approximately 230,000,000,000 meters from Mars. How many seconds does it take for sunlight to reach Mars?

Answer:

230 000 000 000=2.3Ã—10^{11}

\(\frac{2.3 \times 10^{11}}{3 \times 10^{8}}\)=\(\frac{2.3}{3}\)Ã—\(\frac{10^{11}}{10^{8}}\)

=0.7666â€¦Ã—10^{3}

â‰ˆ0.77Ã—10Ã—10^{2}

â‰ˆ7.7Ã—10^{2}

It takes approximately 770 seconds for sunlight to reach Mars.

Question 2.

If the sun is approximately 1.5Ã—10^{11} meters from Earth, what is the approximate distance from Earth to Mars?

Answer:

(2.3Ã—10^{11})-(1.5Ã—10^{11})=(2.3-1.5)Ã—10^{11}

=0.8Ã—10^{11}

=0.8Ã—10Ã—10^{10}

=8Ã—10^{10}

The distance from Earth to Mars is 8Ã—10^{10} meters.