Scientific Notation Word Problems | PDF | Speed | Physical Sciences - Free Printable

Let's solve each problem step by step and explain the reasoning.

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1. The circumference of the Earth at the equator is about 24,900 miles. Express this number in scientific notation.

Step-by-step:
- Move the decimal point so that there is only one non-zero digit to the left.
- 24,900 → 2.49 × 10⁴ (because we moved the decimal 4 places to the left)

✔ Answer: $ 2.49 \times 10^4 $ miles

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2. The speed of light is approximately $ 6.7 \times 10^8 $ miles per hour.

#### a. Express this number in standard form.

- $ 6.7 \times 10^8 = 6.7 \times 100,000,000 = 670,000,000 $

✔ Answer: 670,000,000 miles per hour

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#### b. If light travels $ 6.7 \times 10^8 $ miles in one hour, how many miles will it travel in 1 minute?

- There are 60 minutes in an hour.
- So, divide the hourly distance by 60:

$$
\frac{6.7 \times 10^8}{60} = \frac{6.7}{60} \times 10^8 = 0.111666... \times 10^8
$$

Now convert to proper scientific notation:

$$
0.111666... \times 10^8 = 1.11666... \times 10^7
$$

Rounded to two significant figures (since 6.7 has two):

✔ Answer: $ 1.1 \times 10^7 $ miles (approximately)

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#### c. If it takes light 8.3 minutes to reach the Sun from Earth, what is the distance of the Sun from Earth?

We already found that light travels $ 1.1 \times 10^7 $ miles per minute (from part b).

So, multiply:

$$
\text{Distance} = \text{Speed} \times \text{Time} = (1.1 \times 10^7) \times 8.3
$$

First, multiply the numbers:

$$
1.1 \times 8.3 = 9.13
$$

Then combine with the power of 10:

$$
9.13 \times 10^7
$$

✔ Answer: $ 9.13 \times 10^7 $ miles

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3. A speck of dust in an electron microscope is $ 1.2 \times 10^2 $ millimeters wide. The image is $ 5 \times 10^2 $ times larger than the actual size. How many millimeters wide is the actual speck of dust?

- The image size is $ 1.2 \times 10^2 $ mm
- This is $ 5 \times 10^2 $ times larger than the actual size

So, to find the actual size, divide:

$$
\text{Actual size} = \frac{1.2 \times 10^2}{5 \times 10^2}
$$

Break it down:

$$
= \frac{1.2}{5} \times \frac{10^2}{10^2} = 0.24 \times 10^0 = 0.24
$$

✔ Answer: 0.24 millimeters

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4. The SR-71 "Blackbird" travels at Mach 3, or three times the speed of sound. The speed of sound is $ 7.6 \times 10^2 $ miles per hour. What is Mach 3 in miles per hour? Write your answer in scientific notation.

- Speed of sound: $ 7.6 \times 10^2 $ mph
- Mach 3 = $ 3 \times 7.6 \times 10^2 = 22.8 \times 10^2 $

Now write in proper scientific notation:

$$
22.8 \times 10^2 = 2.28 \times 10^3
$$

✔ Answer: $ 2.28 \times 10^3 $ miles per hour

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5. The half-life of uranium-238 is $ 4.5 \times 10^9 $ years. The half-life of uranium-234 is $ 2.5 \times 10^5 $ years. How many times greater is the half-life of uranium-238 than that of uranium-234?

Divide:

$$
\frac{4.5 \times 10^9}{2.5 \times 10^5} = \frac{4.5}{2.5} \times 10^{9-5} = 1.8 \times 10^4
$$

✔ Answer: $ 1.8 \times 10^4 $ times greater

That’s 18,000 times greater.

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6. Colorado covers about $ 1.04 \times 10^5 $ square miles. The Indian Ocean covers about $ 2.808 \times 10^7 $ square miles. How many times bigger than Colorado is the Indian Ocean?

Divide:

$$
\frac{2.808 \times 10^7}{1.04 \times 10^5} = \frac{2.808}{1.04} \times 10^{7-5} = 2.7 \times 10^2
$$

Check:
$ 2.808 ÷ 1.04 = 2.7 $ exactly?

Let’s compute:
- $ 1.04 \times 2.7 = 1.04 \times 2 + 1.04 \times 0.7 = 2.08 + 0.728 = 2.808 $ ✔

So yes.

✔ Answer: $ 2.7 \times 10^2 $ times bigger → 270 times bigger

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✔ Final Answers Summary:

1. $ 2.49 \times 10^4 $
2. a. 670,000,000
b. $ 1.1 \times 10^7 $ miles
c. $ 9.13 \times 10^7 $ miles
3. 0.24 mm
4. $ 2.28 \times 10^3 $ mph
5. $ 1.8 \times 10^4 $ times greater
6. $ 2.7 \times 10^2 $ times bigger (270 times)

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