Question: Decide whether each proposed multiplication or division of measurements is possible. If it is possible, write the result in the last column of the table.
proposed
multiplication or division
Is this
possible?
result
(1.0 cm)^2 \cdot (0.16 m) = ?
yes
no
(2.0 mg)^2 \cdot (0.013 g) = ?
yes
no
\frac{27. \text{kg}^2}{9.0 \text{kg}} = ?
yes
no
3. kg
1. **Problem statement:** Decide if the following multiplications or divisions of measurements are possible and find the result if possible.
2. **Recall:** Multiplying or dividing measurements is possible if the units are compatible or can be combined meaningfully.
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### First expression: $$(1.0 \text{ cm})^2 \cdot (0.16 \text{ m})$$
3. Square the first measurement:
$$ (1.0 \text{ cm})^2 = 1.0^2 \text{ cm}^2 = 1.0 \text{ cm}^2 $$
4. Multiply by $0.16 \text{ m}$:
$$ 1.0 \text{ cm}^2 \cdot 0.16 \text{ m} = 0.16 \text{ cm}^2 \cdot \text{ m} $$
5. Units combined are $\text{cm}^2 \cdot \text{m}$ which is a valid multiplication of units resulting in cubic length units but mixed units.
6. Convert $\text{cm}^2$ to $\text{m}^2$ for consistency:
$$ 1 \text{ cm} = 0.01 \text{ m} \Rightarrow 1 \text{ cm}^2 = (0.01)^2 \text{ m}^2 = 0.0001 \text{ m}^2 $$
7. So,
$$ 1.0 \text{ cm}^2 = 0.0001 \text{ m}^2 $$
8. Multiply:
$$ 0.0001 \text{ m}^2 \cdot 0.16 \text{ m} = 0.000016 \text{ m}^3 $$
9. **Result:** Yes, possible. The result is $1.6 \times 10^{-5} \text{ m}^3$.
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### Second expression: $$(2.0 \text{ mg})^2 \cdot (0.013 \text{ g})$$
10. Square the first measurement:
$$ (2.0 \text{ mg})^2 = 4.0 \text{ mg}^2 $$
11. Multiply by $0.013 \text{ g}$:
$$ 4.0 \text{ mg}^2 \cdot 0.013 \text{ g} = 0.052 \text{ mg}^2 \cdot \text{ g} $$
12. Units are $\text{mg}^2 \cdot \text{g}$ which is not a standard unit and cannot be simplified to a single mass unit.
13. Since $\text{mg}^2$ is not a standard unit (mass squared), this multiplication is not physically meaningful.
14. **Result:** No, not possible.
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### Third expression: $$\frac{27. \text{kg}^2}{9.0 \text{kg}}$$
15. Division of units:
$$ \frac{\text{kg}^2}{\text{kg}} = \text{kg} $$
16. Divide the numbers:
$$ \frac{27.}{9.0} = 3.0 $$
17. So,
$$ \frac{27. \text{kg}^2}{9.0 \text{kg}} = 3.0 \text{ kg} $$
18. **Result:** Yes, possible. The result is $3.0 \text{ kg}$.
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**Summary:**
- First: Yes, result $1.6 \times 10^{-5} \text{ m}^3$
- Second: No
- Third: Yes, result $3.0 \text{ kg}$