Page 1 of 6 from a dimensional analysis practice worksheet for science and engineering courses.

Let's solve each of these dimensional analysis problems step by step, showing all the work and explaining the reasoning.

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Problem 1:

When one gram of gasoline burns in a car’s engine, the amount of energy given off is approximately $1.03 \times 10^7$ cal. Express this quantity in joules (J). (Use 1 cal = 4.184 J)

#### Step-by-step solution:

We are given:
- Energy = $1.03 \times 10^7$ cal
- Conversion factor: $1 \text{ cal} = 4.184 \text{ J}$

We want to convert calories to joules:

$$
\text{Energy in J} = (1.03 \times 10^7 \text{ cal}) \times \left( \frac{4.184 \text{ J}}{1 \text{ cal}} \right)
$$

Now multiply:

$$
= 1.03 \times 10^7 \times 4.184 = 4.30952 \times 10^7 \text{ J}
$$

Now round to significant figures:
- $1.03 \times 10^7$ has 3 significant figures
- $4.184$ has 4 significant figures, but since we're multiplying, use the least number of sig figs → 3

So:
$$
\boxed{4.31 \times 10^7 \text{ J}}
$$

✔ Answer: $4.31 \times 10^7$ J

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Problem 2:

The pressure reading from a barometer is 742 mm Hg. Express this reading in kilopascals, kPa. (Use 760 mm Hg = 1.013 × 10⁵ Pa)

#### Step-by-step solution:

We are given:
- Pressure = 742 mm Hg
- Conversion: $760 \text{ mm Hg} = 1.013 \times 10^5 \text{ Pa}$

We want to convert mm Hg to Pa, then to kPa.

First, set up the conversion:

$$
742 \text{ mm Hg} \times \left( \frac{1.013 \times 10^5 \text{ Pa}}{760 \text{ mm Hg}} \right)
$$

Now calculate:

$$
= \frac{742}{760} \times 1.013 \times 10^5
$$

$$
= 0.976315789 \times 1.013 \times 10^5
$$

$$
= 98,806.3 \text{ Pa} \quad \text{(approx)}
$$

Now convert to kPa: divide by 1000

$$
98,806.3 \text{ Pa} = 98.8063 \text{ kPa}
$$

Now consider significant figures:
- 742 has 3 sig figs
- 760 and 1.013×10⁵ are exact or have more precision → so final answer should have 3 sig figs

So:
$$
\boxed{98.8 \text{ kPa}}
$$

✔ Answer: 98.8 kPa

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Problem 3:

How many megayears is equivalent to $6.02 \times 10^{21}$ nanoseconds (ns)?

#### Step-by-step solution:

We need to convert nanoseconds → seconds → years → megayears

Recall:
- $1 \text{ ns} = 10^{-9} \text{ s}$
- $1 \text{ year} = 365.25 \text{ days}$ (accounting for leap years)
- $1 \text{ day} = 24 \text{ hours}$
- $1 \text{ hour} = 3600 \text{ seconds}$
- $1 \text{ megayear} = 10^6 \text{ years}$

Start with:

$$
6.02 \times 10^{21} \text{ ns} \times \left( \frac{10^{-9} \text{ s}}{1 \text{ ns}} \right) = 6.02 \times 10^{12} \text{ s}
$$

Now convert seconds to years:

$$
\text{Seconds per year} = 365.25 \times 24 \times 3600 = 31,557,600 \text{ s/year} \approx 3.15576 \times 10^7 \text{ s/year}
$$

Now:

$$
\text{Years} = \frac{6.02 \times 10^{12} \text{ s}}{3.15576 \times 10^7 \text{ s/year}} = 1.907 \times 10^5 \text{ years}
$$

Now convert to megayears:

$$
1.907 \times 10^5 \text{ years} = 0.1907 \text{ megayears}
$$

Now check significant figures:
- $6.02 \times 10^{21}$ has 3 sig figs
- So final answer should have 3 sig figs

$$
\boxed{0.191 \text{ megayears}}
$$

✔ Answer: 0.191 megayears

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Problem 4:

The average student is in class 330 min/day.

#### a. How many hours/day is the average student in class?

Convert minutes to hours:

$$
330 \text{ min} \times \left( \frac{1 \text{ hr}}{60 \text{ min}} \right) = 5.5 \text{ hours}
$$

✔ Answer: 5.5 hours/day

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#### b. How many seconds is the average student in class per week?

We know:
- 330 min/day
- 7 days/week
- 60 seconds/minute

So:

$$
330 \text{ min/day} \times 7 \text{ days/week} = 2310 \text{ min/week}
$$

Now convert to seconds:

$$
2310 \text{ min} \times 60 \text{ s/min} = 138,600 \text{ seconds}
$$

✔ Answer: 138,600 seconds/week

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#### c. Calculate how many seconds you are in class a day (pick any weekday you have class). Show any calculations you have to do using dimensional analysis below. Answers will vary between you and your classmates.

Since this is personal, let’s assume you have 5 classes per day, each 50 minutes long.

Then total time in class per day:

$$
5 \text{ classes} \times 50 \text{ min/class} = 250 \text{ min/day}
$$

Now convert to seconds:

$$
250 \text{ min} \times 60 \text{ s/min} = 15,000 \text{ seconds/day}
$$

✔ Example Answer: 15,000 seconds/day (depending on actual schedule)

> Note: Your answer may differ based on your actual class schedule.

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✔ Final Answers Summary:

1. $4.31 \times 10^7$ J
2. 98.8 kPa
3. 0.191 megayears
4a. 5.5 hours/day
4b. 138,600 seconds/week
4c. Example: 15,000 seconds/day (based on 5 × 50-min classes)

Let me know if you'd like to adjust Problem 4c based on your actual class schedule!