Here are five math problems involving dimensional analysis, along with their solutions:

1. Speed Conversion

A car is traveling at a speed of 90 kilometers per hour. Convert this speed to meters per second using dimensional analysis.

Solution: $\text{Given: } 90 \ \text{km/h}$ Using conversion factors: $1 \ \text{km} = 1000 \ \text{m}, \quad 1 \ \text{hour} = 3600 \ \text{seconds}$ $90 \ \text{km/h} \times \frac{1000 \ \text{m}}{1 \ \text{km}} \times \frac{1 \ \text{hour}}{3600 \ \text{seconds}} = 90 \times \frac{1000}{3600} = 25 \ \text{m/s}$ Answer: 25 meters per second (m/s)

---

2. Time Conversion

Convert 5 years into seconds using dimensional analysis.

Solution: $\text{Given: } 5 \ \text{years}$ Using conversion factors: $1 \ \text{year} = 365.25 \ \text{days}, \quad 1 \ \text{day} = 24 \ \text{hours}, \quad 1 \ \text{hour} = 3600 \ \text{seconds}$ $5 \ \text{years} \times \frac{365.25 \ \text{days}}{1 \ \text{year}} \times \frac{24 \ \text{hours}}{1 \ \text{day}} \times \frac{3600 \ \text{seconds}}{1 \ \text{hour}}$ $= 5 \times 365.25 \times 24 \times 3600 = 157,788,000 \ \text{seconds}$ Answer: 157,788,000 seconds

---

3. Density Calculation

The density of a substance is given as 0.8 grams per cubic centimeter (g/cm³). Convert this density to kilograms per cubic meter (kg/m³) using dimensional analysis.

Solution: $\text{Given: } 0.8 \ \text{g/cm}^3$ Using conversion factors: $1 \ \text{kg} = 1000 \ \text{g}, \quad 1 \ \text{cm} = 0.01 \ \text{m} \quad \Rightarrow \quad 1 \ \text{cm}^3 = (0.01)^3 \ \text{m}^3 = 10^{-6} \ \text{m}^3$ $0.8 \ \text{g/cm}^3 \times \frac{1000 \ \text{g}}{1 \ \text{kg}} \times \frac{1 \ \text{cm}^3}{10^{-6} \ \text{m}^3}$ $= 0.8 \times 1000 \times 10^6 = 800,000 \ \text{kg/m}^3$ Answer: 800,000 kilograms per cubic meter (kg/m³)

---

4. Force Conversion

A force is measured as 25 Newtons (N). Convert this force to dynes (dyn) using dimensional analysis. Recall that $1 \ \text{N} = 10^5 \ \text{dyn}$.

Solution: $\text{Given: } 25 \ \text{N}$ Using the conversion factor: $1 \ \text{N} = 10^5 \ \text{dyn}$ $25 \ \text{N} \times 10^5 \ \text{dyn/N} = 25 \times 10^5 = 2.5 \times 10^6 \ \text{dyn}$ Answer: $2.5 \times 10^6$ dynes

---

5. Pressure Conversion

The atmospheric pressure at sea level is approximately $101,325 \ \text{Pa}$. Convert this pressure into atmospheres (atm) using dimensional analysis. Recall that $1 \ \text{atm} = 101,325 \ \text{Pa}$.

Solution: $\text{Given: } 101,325 \ \text{Pa}$ Using the conversion factor: $1 \ \text{atm} = 101,325 \ \text{Pa}$ $101,325 \ \text{Pa} \times \frac{1 \ \text{atm}}{101,325 \ \text{Pa}} = 1 \ \text{atm}$ Answer: 1 atmosphere (atm)

---

Do you want details or have any questions?

---

5 Related Questions:

1. How do you convert miles per hour (mph) to meters per second (m/s)?
2. Convert a temperature change of 5°C to a change in Fahrenheit using dimensional analysis.
3. How can dimensional analysis be used to check the correctness of a physics equation?
4. Convert 2 gallons to liters, given that 1 gallon = 3.785 liters.
5. How do you convert energy from joules to calories using dimensional analysis?

Tip:

Always check that the dimensions on both sides of an equation match when using dimensional analysis to verify correctness.