Questions: Part 3 of 4 (c) 4.07 cm × 3.0015 cm = □ cm^2 □ × 10 Part 4 of 4 (d) (2 × 10^3 g + 7.345 g) / (0.091 cm^3 - 0.039 cm^3) = □ g/cm^3 □ × 10

Part 3 of 4 (c) 4.07 cm × 3.0015 cm = □ cm^2 □ × 10 Part 4 of 4 (d) (2 × 10^3 g + 7.345 g) / (0.091 cm^3 - 0.039 cm^3) = □ g/cm^3 □ × 10
Transcript text: Part 3 of 4 (c) $4.07 \mathrm{~cm} \times 3.0015 \mathrm{~cm}=$ $\square$ $\mathrm{cm}^{2}$ $\square$ $\times 10$ Part 4 of 4 (d) $\frac{\left(2 \times 10^{3} \mathrm{~g}+7.345 \mathrm{~g}\right)}{\left(0.091 \mathrm{~cm}^{3}-0.039 \mathrm{~cm}^{3}\right)}=$ $\square$ $\frac{\mathrm{g}}{\mathrm{cm}^{3}}$ $\square$ $\times 10$
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Solution

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Solution Steps

Step 1: Calculate the product in part (c)

To find the product of \(4.07 \, \text{cm} \times 3.0015 \, \text{cm}\), we multiply the two numbers:

\[ 4.07 \times 3.0015 = 12.214105 \, \text{cm}^2 \]

Step 2: Round the result to four significant digits

The result \(12.214105 \, \text{cm}^2\) should be rounded to four significant digits:

\[ 12.214105 \approx 12.21 \, \text{cm}^2 \]

Final Answer for Part (c)

\(\boxed{12.21 \, \text{cm}^2}\)

Step 3: Calculate the numerator in part (d)

First, add the two masses:

\[ 2 \times 10^3 \, \text{g} + 7.345 \, \text{g} = 2000 \, \text{g} + 7.345 \, \text{g} = 2007.345 \, \text{g} \]

Step 4: Calculate the denominator in part (d)

Next, subtract the two volumes:

\[ 0.091 \, \text{cm}^3 - 0.039 \, \text{cm}^3 = 0.052 \, \text{cm}^3 \]

Step 5: Divide the numerator by the denominator

Now, divide the total mass by the total volume:

\[ \frac{2007.345 \, \text{g}}{0.052 \, \text{cm}^3} = 38602.7885 \, \frac{\text{g}}{\text{cm}^3} \]

Step 6: Round the result to four significant digits

The result \(38602.7885 \, \frac{\text{g}}{\text{cm}^3}\) should be rounded to four significant digits:

\[ 38602.7885 \approx 38600 \, \frac{\text{g}}{\text{cm}^3} \]

Final Answer for Part (d)

\(\boxed{38600 \, \frac{\text{g}}{\text{cm}^3}}\)

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