02 Units of Measurement Problem Set

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UNITS OF MEASUREMENT | PRACTICE PROBLEMS Complete the following to reinforce your understanding of the concept covered in this module.

PROBLEM 1: A truck is measured to be traveling at a velocity of 0.035 kilometers per second on a highway in Houston, TX. If the speed limit is 70 miles per hour, the number of miles per hour the truck is traveling over the speed limit is most close to: A. 2 B. 4 C. 6 D. 8

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SOLUTION 1: The TABLE of CONVERSION FACTORS can be referenced under the SUBJECT of UNITS on page 1 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. In this problem, we are given the VELOCITY of a TRUCK in METRIC UNITS, and are looking to COMPARED this value, to the posted SPEED LIMIT given in UCSCS units. As we are looking to SOLVE for the DIFFERENCE of the VELOCITY in terms of UCSC units, we will CONVERT the TRUCK VELOCITY to USCS units using DIMENSIONAL ANALYSIS and CONVERSION FACTORS. A CONVERSION FACTOR is an EQUALITY or NUMERICAL VALUE that can be used to CONVERT between UNITS from one UNIT to another. Looking at the TABLE of CONVERSION FACTORS in the REFERENCE HANDBOOK, we can use DIMENSIONAL ANALYSIS to CONVERT from METERS per SECOND to units of MILES per HOUR. As we are given a VELOCITY, we should immediately realize that VELOCITY is comprised of two BASE UNITS, representing the dimensions of LENGTH and TIME:

π‘‰π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ =

πΏπ‘’π‘›π‘”π‘‘β„Ž π‘‡π‘–π‘šπ‘’

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Three Steps of Dimensional Analysis: 1. Write down the GIVEN MEASUREMENT with the NUMERICAL VALUE and associated UNITS. As we are working with UNITS of LENGTH and TIME, we will need to convert each ONE separately, and make sure. Let’s write out the GIVEN base units of the truck’s VELOCITY in METRIC UNITS, and the posted SPEED LIMIT given in UCSC units:

π‘€π‘’π‘‘π‘Ÿπ‘–π‘ π‘ˆπ‘›π‘–π‘‘π‘ : π‘‰π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ =

π‘ˆπΆπ‘†πΆ π‘ˆπ‘›π‘–π‘‘π‘ : π‘‰π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ =

πΏπ‘’π‘›π‘”π‘‘β„Ž π‘˜π‘š = π‘‡π‘–π‘šπ‘’ 𝑠

πΏπ‘’π‘›π‘”π‘‘β„Ž π‘šπ‘–π‘™π‘’π‘  = π‘‡π‘–π‘šπ‘’ β„Žπ‘œπ‘’π‘Ÿ

2. MULTIPLY the MEASUREMENT by one or more CONVERSION FACTORS. The UNIT in each DENOMINATOR must CANCEL OUT units of the original measurement, leaving the DESIRED UNITS in the NUMERATOR. a. The FIRST CONVERSION will be CONVERTING from KILOMETERS (π‘˜π‘š) to MILES (mi), which will require us to use MULTIPLE CONVERSION FACTORS:

πΆπ‘œπ‘›π‘£π‘’π‘Ÿπ‘ π‘–π‘œπ‘› πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿ (π‘˜π‘š β†’ 𝑓𝑑):

1 π‘˜π‘š 3,281 𝑓𝑑

πΆπ‘œπ‘›π‘£π‘’π‘Ÿπ‘ π‘–π‘œπ‘› πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿ (π‘šπ‘– β†’ 𝑓𝑑):

1 π‘šπ‘– 5,280 𝑓𝑑

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b. The SECOND CONVERSION will be CONVERTING from SECONDS (𝑠) to HOURS (hr), which will also require us to use MULTIPLE CONVERSION FACTORS:

πΆπ‘œπ‘›π‘£π‘’π‘Ÿπ‘ π‘–π‘œπ‘› πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿ (β„Žπ‘Ÿ β†’ min):

πΆπ‘œπ‘›π‘£π‘’π‘Ÿπ‘ π‘–π‘œπ‘› πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿ (min β†’ s) :

1 β„Žπ‘Ÿ 60 min

1 min 60 s

3. Perform the CALCULATION and report the answer to the proper SIGNIFICANT FIGURES based on the MEASUREMENTS given in the problem statement:

(0.035

π‘˜π‘š 3,281 𝑓𝑑 1 π‘šπ‘– 60 𝑠𝑒𝑐 60 min π‘šπ‘– )Γ—( )Γ—( )Γ—( )Γ—( ) = 78.29 𝑠 1 π‘˜π‘š 5,280 𝑓𝑑 1 π‘šπ‘–π‘› 1 β„Žπ‘Ÿ β„Žπ‘Ÿ

Now that we have the VELOCITY of the TRUCK in UCSC units, we can calculate the difference in VELOCITY with respect to the SPEED LIMIT as:

π›₯π‘‰π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ = 78

π‘šπ‘– π‘šπ‘– π‘šπ‘– βˆ’ 70 =8 β„Žπ‘Ÿ β„Žπ‘Ÿ β„Žπ‘Ÿ

Therefore, the correct answer choice is D. πŸ–

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PROBLEM 2: A mythical element is found to have a boiling point of 298 °𝐹. Given this measurement, the boiling point on the absolute temperature scale in units of Kelvin is most close to: A. 289 B. 375 C. 421 D. 529

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SOLUTION 2: The FORMULAS for the UNIT CONVERSIONS OF TEMPERATURE can be referenced under the SUBJECT of UNITS on page 1 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. In this problem we are given the BOILING POINT of an ELEMENT, and asked to CONVERT it from the FAHRENHEIT TEMPERATURE SCALE to the KELVIN ABSOLUTE TEMPERATURE SCALE. The FAHRENHEIT TEMPERATURE SCALE is based on 32°𝐹 for the FREEZING POINT of water and 212°𝐹 for the BOILING POINT of water. A degree on the FAHRENHEIT scale represents 1/180 of the temperature range between the freezing and boiling points of water. As we are given the TEMPERATURE in DEGREES FAHRENHEIT, we must first CONVERT the TEMPERATURE to DEGREES CELSIUS, which we can then CONVERT to KELVIN. The CELSIUS TEMPERATURE SCALE is based on 0°𝐢 for the FREEZING POINT and 100°𝐢 for the BOILING POINT of water. A degree on the CELSIUS scale represents 1/100 of the temperature range between the freezing and boiling points of water.

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The formula for converting a FAHRENHEIT temperature to degrees CELSIUS is expressed using the following expression: 𝑇𝐢 = (𝑇𝐹 βˆ’ 32)/1.8 Where: β€’ 𝑇𝐹 is the temperature given in degrees Fahrenheit(°𝐹) β€’ 𝑇𝐢 is the temperature given units of degrees Celsius (°𝐢) Plugging in the VALUE for the TEMPERATURE in DEGREES FAHRENHEIT, we find the TEMPERATURE in DEGREES CELSIUS is:

𝑇𝐢 =

(298°𝐹 βˆ’ 32) = 147.78°𝐢 1.8

Now that we have the TEMPERATURE in DEGREES CELSIUS, we need to CONVERT it to the KELVIN ABSOLUTE TEMPERATURE SCALE. The KELVIN TEMPERATURE SCALE is the absolute temperature scale used in the International Engineering Systems and SI System of units. The Kelvin scale of temperature uses degrees CELSIUS from ABSOLUTE ZERO, and is based on 0 K being defined as absolute zero.

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The formula for converting a CELSIUS temperature to the Absolute Temperature Scale in units of KELVIN (𝐾) is expressed using the following expression: 𝑇𝐾 = 𝑇𝐢 + 273.15

π›₯𝑇𝐾 = π›₯𝑇𝐢

Where: β€’ π‘‡π‘˜ is the temperature given in units of Kelvin (𝐾) β€’ 𝑇𝐢 is the temperature given units of degrees Celsius (°𝐢) Plugging in the VALUE for the TEMPERATURE in DEGREES CELSIUS, we find the TEMPERATURE in KELVIN is: 𝑇𝐾 = 147.78°𝐢 + 273.15 = 420.93 𝐾

Therefore, the correct answer choice is C. πŸ’πŸπŸ

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PROBLEM 3: A silver bullet is measured to have a mass of 3 kilograms. Given the bullet is made of 100% silver, the volume of silver in the bullet is calculated in cubic inches (𝑖𝑛3 ) is most close to: A. 17.4 B. 21.8 C. 23.5 D. 29.4

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SOLUTION 3: The TOPIC of DENSITY can be referenced under the SUBJECT of FLUID MECHANICS on page 103 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. In this problem we are looking to SOLVE for the VOLUME of the SILVER in the BULLET, given the MASS of the BULLET and ASSUMPTION that the BULLET is COMPRISED entirely of SILVER. DENSITY (𝜌) is RATIO of the MASS of a SUBSTANCE to the VOLUME occupied by that SUBSTANCE. As we are told the bullet is comprised of the element SILVER, we can look up the DENSITY for SILVER in the Reference Handbook, and relate that to the GIVEN MASS to SOLVE for the VOLUME of SILVER in the BULLET. The TABLE for PROPERTIES of METALS can be referenced under the SUBJECT of MATERIALS SCIENCE on page 61 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.

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Referencing the TABLE for the PROPERTIES of METAL in the REFERENCE HANDBOOK, we find the DENSITY of SILVER is given as: 𝜌 = 10,500 π‘˜π‘”/π‘š3 Made with

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The FORMULA for DENSITY can be referenced under the SUBJECT of FLUID MECHANICS on page 103 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The FORMULA for DENSITY calculates DENSITY as the QUOTIENT of the MASS (π‘š) divided by the VOLUME (𝑉) or the INVERSE of the SPECIFIC VOLUME (𝑣) and is given as: π›₯π‘š π‘š 1 1 = = π‘œπ‘Ÿ 𝜈 = π›₯𝑉→0 π›₯𝑉 𝑉 𝜈 𝜌

𝜌 = lim Where:

β€’ 𝜌 is the DENSITY β€’ m is the MASS β€’ V is the VOLUME β€’ v is the SPECIFIC VOLUME Let’s re-write the FORMULA for DENSITY to ISOLATE the term for VOLUME (V):

𝑉=

π‘š 𝜌

Plugging in the GIVEN MASS and REFERENCE DENSITY for SILVER, we CALCULATE the VOLUME of the BULLET as:

𝑉=

3 π‘˜π‘” = 0.000286 π‘š3 3 10,500 π‘˜π‘”/π‘š

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In the USCS system of units, LENGTH is measured using BASE UNITS of INCHES given as "𝑖𝑛". Therefore, we will need to CONVERT the BASE UNITS of VOLUME from CUBIC METERS (π‘š3 ) to CUBIC INCHES (𝑖𝑛3 ) to CALCULATE the VOLUME of the GOLD in USCS UNITS. The TABLE of CONVERSION FACTORS can be referenced under the SUBJECT of UNITS on page 1 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A CONVERSION FACTOR is an EQUALITY or NUMERICAL VALUE that can be used to CONVERT between UNITS from one UNIT to another. Looking at the TABLE of CONVERSION FACTORS in the REFERENCE HANDBOOK, we can use DIMENSIONAL ANALYSIS to CONVERT from CUBIC METERS (π‘š3 ) to LITERS (𝐿) and then from LITERS (𝐿) to CUBIC INCHES (𝑖𝑛3 ). Three Steps of Dimensional Analysis: 1. Write down the GIVEN MEASUREMENT with the NUMERICAL VALUE and associated UNIT. As we are working with UNITS of VOLUME, we need to take note that all UNITS are to THIRD POWER. π‘‰π‘œπ‘™π‘’π‘šπ‘’ = 0.000286 π‘š3

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2. MULTIPLY the MEASUREMENT by one or more CONVERSION FACTORS. The UNIT in each DENOMINATOR must CANCEL OUT units of the original measurement, leaving the DESIRED UNITS in the NUMERATOR. a. The FIRST CONVERSION will be CONVERTING from CUBIC METERS (π‘š3 ) to LITERS (𝐿):

πΆπ‘œπ‘›π‘£π‘’π‘Ÿπ‘ π‘–π‘œπ‘› πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿ:

1,000 𝐿 1 π‘š3

b. The SECOND CONVERSION will be CONVERTING from LITERS (𝐿) to CUBIC INCHES (𝑖𝑛3 ):

πΆπ‘œπ‘›π‘£π‘’π‘Ÿπ‘ π‘–π‘œπ‘› πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿ:

1𝐿 61.02 𝑖𝑛3

3. Perform the CALCULATION and report the answer to the proper SIGNIFICANT FIGURES based on the MEASUREMENTS given in the problem statement: 1,000 𝐿 61.02 𝑖𝑛3 0.000286 π‘š Γ— ( )Γ—( ) = 17.4343 𝑖𝑛2 = 17.4 𝑖𝑛2 3 1π‘š 1𝐿 3

Therefore, the correct answer choice is A. πŸπŸ•. πŸ’

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