Dimensional Analysis Worksheet 2 Answer Key

Dimensional Analysis Worksheet 2 Answer Key – A Complete Guide

When students tackle dimensional analysis problems, they often feel overwhelmed by the sheer number of unit conversions and the need for meticulous bookkeeping. Also, a well‑structured answer key can transform that anxiety into confidence. Below is a comprehensive walkthrough of Dimensional Analysis Worksheet 2, complete with detailed solutions, explanations, and tips for mastering the technique Not complicated — just consistent..

Introduction

Dimensional analysis, also known as the unit‑in‑unit‑out method, is a powerful tool for converting between different measurement systems. In Worksheet 2, problems range from simple length conversions to more involved multi‑step calculations involving speed, volume, and mass. Understanding how to read the problem, set up conversion factors, and simplify the calculation is essential for success in physics, chemistry, engineering, and everyday life Less friction, more output..

This guide will:

1. Present the full answer key for each problem.
2. Explain the reasoning behind every step.
3. Offer shortcuts and common pitfalls to avoid.
4. Provide a quick‑reference cheat sheet for future use.

Problem 1 – Converting Length Units

Question:
Convert 12.5 miles to centimeters.

Answer

1. Start with the given value.
   (12.5\ \text{miles})

2. Use conversion factors.

   • 1 mile = 1,609.34 meters
   • 1 meter = 100 centimeters

3. Set up the calculation.
   [ 12.5\ \text{mi} \times \frac{1,609.34\ \text{m}}{1\ \text{mi}} \times \frac{100\ \text{cm}}{1\ \text{m}} ]

4. Cancel units.
   The miles cancel, leaving only centimeters.

5. Multiply the numbers.
   [ 12.5 \times 1,609.34 \times 100 = 12,5 \times 160,934 = 2,011,675\ \text{cm} ]

Result:
(\boxed{2{,}011{,}675\ \text{cm}})

Problem 2 – Speed Conversion

Question:
A car travels 60 km/h. Express this speed in m/s Worth knowing..

Answer

1. Given speed.
   (60\ \text{km/h})

2. Conversion factors.

   • 1 km = 1,000 m
   • 1 hour = 3,600 s

3. Set up the expression.
   [ 60\ \frac{\text{km}}{\text{h}} \times \frac{1{,}000\ \text{m}}{1\ \text{km}} \times \frac{1\ \text{h}}{3{,}600\ \text{s}} ]

4. Cancel units.
   km and h cancel It's one of those things that adds up..

5. Compute.
   [ 60 \times \frac{1{,}000}{3{,}600} = 60 \times 0.27778 \approx 16.667\ \text{m/s} ]

Result:
(\boxed{16.67\ \text{m/s}}) (rounded to two decimals)

Problem 3 – Volume to Mass

Question:
A liquid has a density of 0.8 g/mL. What is the mass of 250 mL of this liquid?

Answer

1. Given data.
   Density: (0.8\ \text{g/mL})
   Volume: (250\ \text{mL})

2. Set up the multiplication.
   [ 250\ \text{mL} \times 0.8\ \frac{\text{g}}{\text{mL}} ]

3. Cancel mL.
   mL cancels, leaving grams.

4. Multiply.
   (250 \times 0.8 = 200\ \text{g})

Result:
(\boxed{200\ \text{g}})

Problem 4 – Multi‑Step Conversion

Question:
Convert 5.0 liters of water to pounds (lb). (Use 1 L = 0.264172 US gallons, 1 gallon = 8.3454 lb)

Answer

1. Start with liters.
   (5.0\ \text{L})

2. First conversion: liters to gallons.
   [ 5.0\ \text{L} \times \frac{0.264172\ \text{gal}}{1\ \text{L}} = 1.32086\ \text{gal} ]

3. Second conversion: gallons to pounds.
   [ 1.32086\ \text{gal} \times \frac{8.3454\ \text{lb}}{1\ \text{gal}} \approx 11.023\ \text{lb} ]

Result:
(\boxed{11.02\ \text{lb}}) (rounded to two decimals)

Problem 5 – Temperature and Heat Capacity

Question:
If a substance with a heat capacity of 0.9 J/K absorbs 180 J of energy, what is the temperature change in Kelvin?

Answer

1. Use the formula
   (Q = C \Delta T)
   where (Q) = energy, (C) = heat capacity, (\Delta T) = temperature change Worth keeping that in mind. Turns out it matters..

2. Rearrange for (\Delta T).
   [ \Delta T = \frac{Q}{C} ]

3. Plug in values.
   [ \Delta T = \frac{180\ \text{J}}{0.9\ \text{J/K}} = 200\ \text{K} ]

Result:
(\boxed{200\ \text{K}})

Problem 6 – Speed from Distance and Time

Question:
A runner covers 10 km in 0.5 hours. What is the average speed in m/s?

Answer

1. Convert distance to meters.
   (10\ \text{km} = 10{,}000\ \text{m})

2. Convert time to seconds.
   (0.5\ \text{h} = 0.5 \times 3{,}600\ \text{s} = 1{,}800\ \text{s})

3. Speed formula (v = \frac{d}{t}).
   [ v = \frac{10{,}000\ \text{m}}{1{,}800\ \text{s}} \approx 5.556\ \text{m/s} ]

Result:
(\boxed{5.56\ \text{m/s}}) (rounded to two decimals)

Problem 7 – Energy to Work

Question:
A machine does 250 J of work while moving an object 5 m. What is the average force applied in Newtons?

Answer

1. Use work formula (W = F \times d).
   Rearrange for force: (F = \frac{W}{d}).

2. Plug values.
   [ F = \frac{250\ \text{J}}{5\ \text{m}} = 50\ \text{N} ]

Result:
(\boxed{50\ \text{N}})

Problem 8 – Concentration Conversion

Question:
A solution contains 0.25 mol/L of a solute. How many grams of solute are present in 2 L if the molar mass is 58.44 g/mol?

Answer

1. Find moles in 2 L.
   [ 0.25\ \text{mol/L} \times 2\ \text{L} = 0.5\ \text{mol} ]

2. Convert moles to grams.
   [ 0.5\ \text{mol} \times 58.44\ \frac{\text{g}}{\text{mol}} = 29.22\ \text{g} ]

Result:
(\boxed{29.22\ \text{g}})

Problem 9 – Power Calculation

Question:
A device consumes 1500 W of power for 2 hours. What is the total energy consumed in kilowatt‑hours (kWh)?

Answer

1. Convert watts to kilowatts.
   (1500\ \text{W} = 1.5\ \text{kW})

2. Multiply by time.
   [ 1.5\ \text{kW} \times 2\ \text{h} = 3\ \text{kWh} ]

Result:
(\boxed{3\ \text{kWh}})

Problem 10 – Density Reversal

Question:
A metal block has a mass of 2.5 kg and a volume of 0.005 m³. What is its density in g/cm³?

Answer

1. Calculate density in kg/m³.
   [ \rho = \frac{2.5\ \text{kg}}{0.005\ \text{m}^3} = 500\ \text{kg/m}^3 ]

2. Convert kg/m³ to g/cm³.

   • 1 kg = 1,000 g
   • 1 m³ = 1,000,000 cm³

   [ 500\ \frac{\text{kg}}{\text{m}^3} \times \frac{1{,}000\ \text{g}}{1\ \text{kg}} \times \frac{1\ \text{m}^3}{1{,}000{,}000\ \text{cm}^3} = 0.5\ \frac{\text{g}}{\text{cm}^3} ]

Result:
(\boxed{0.5\ \text{g/cm}^3})

Common Mistakes and How to Avoid Them

Quick‑Reference Cheat Sheet

• Length: 1 in = 2.54 cm, 1 ft = 12 in, 1 mi = 1,609.34 m
• Mass: 1 kg = 1,000 g, 1 lb = 0.453592 kg
• Time: 1 h = 3,600 s, 1 min = 60 s
• Volume: 1 L = 1,000 cm³, 1 gal = 3.78541 L
• Energy: 1 kWh = 3.6 MJ, 1 W = 1 J/s
• Density: ρ = m / V (kg/m³ or g/cm³)
• Speed: v = d / t (m/s, km/h)
• Work: W = F × d (J = N·m)
• Heat: Q = C × ΔT (J = J/K × K)

Conclusion

Mastering dimensional analysis is about turning a seemingly complex web of units into a clear, step‑by‑step calculation. By systematically setting up conversion factors, carefully canceling units, and performing arithmetic accurately, students can solve any conversion problem with confidence. Use this answer key as a template for practice, and keep the cheat sheet handy for quick reference. With consistent practice, dimensional analysis will become an intuitive part of every problem‑solving toolkit.