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This worksheet is designed to give you practice using dimensional analysis as a problem solving tool. Good technique includes:

1. Show all of your work – even when you know a shortcut!

2. Write equality statements if you need clarification.

3. When writing units – be thorough. If the unit is “dozen eggs” – don’t just write “dozen” or “eggs”. Likewise, if it is “dozen hydrogen molecules” don’t just write “dozen hydrogen” – that could mean “hydrogen atoms” or “hydrogen molecules”.

4. Write out the string of calculations with the correct conversion factors, in the correct order, including complete units.

5. Go through the process of cross canceling units to prove that your conversion works.

PART 1 Dimensional Analysis and the metric system. Memorize the prefixes of the metric system so you can write the correct equivalence statements without assistance.

Write an equivalence statement between the two given measurements.

A) 1 Km = ____1000_ m or 1 m = ___0.001______ Km

B) 1 µm = ___1 x 10 -6_____m or 1 m = ___1,000,000_ µm

C) 1 nL = __110 -9_____L or 1 L = __1,000,000,000 nL

D) 1 ft = ___12________ in or 1 yard = _____3_____ ft

E) 1 cm = .39 in and 1 in = ___2.54________cm

F) 1 km = .62 mi and 1 mi = __3.11______ km

Perform the following conversions using the equivalence statements you wrote in #1. SHOW YOUR WORK IN THE SPACE PROVIDED UNDER EACH LETTER. WRITE SMALL!!

A) 75 Km = ? m

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|75 Km |

|1000 m |

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|75,000 m |

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|1 |

|1 Km |

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|= ____75,000 m________________ (rewrite final answer here – be sure to include units & sig figs) |

B) 26,800,000 µm = ? m

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|26,800,000 um |

|1 m |

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|26.8 m |

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|1 |

|1,000,000 um |

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|= (rewrite final answer here – be sure to include units & sig figs) |

C) 11,200,000,000 µm = ? L

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|11,200,000,000 µm |

|1 L |

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|11200L |

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|1 |

|1,000,000 µm |

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|= ____1.12 x 104 L______________ (rewrite final answer here – be sure to include units & sig figs) |

D) 37.687 yds = ? in

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|37.687 yards |

|3 feet |

|12 inches |

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|1,356.732 in |

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|1 |

|1 yard |

|1 foot |

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|= ___1, 356. 7 inches_______ (rewrite final answer here – be sure to include units & sig figs) |

E) .00898 km = ? cm

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|0.00898 km |

|1000 m |

|100 cm |

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|1 |

|1 km |

|1 m |

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|= ___898 cm ______________ (rewrite final answer here – be sure to include units & sig figs) |

F) 89,800,000,000,000 µm = ? km

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|89,800,000,000,000 µm |

|1 m |

|1 km |

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|1 |

|1,000,000 µm |

|1000 m |

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|= ___89,800 km__________________ (rewrite final answer here – be sure to include units & sig figs) |

Metric to Metric Conversions

(Note: NEVER use conversion factors to determine significant figures. Always use the original numbers given in the problem to determine significant figures of the final answer)

1. How many centimeters are in 4.12 meters? 412cm

2. How many kilometers are in 8432 meters? 8.432 km

3. Change 3.24 kilometers into meters. 3240 m

4. How many millimeters are in 25 meters? 25,000 mm

5. How many millimeters are in 251 centimeters? (2 step problem) 2,510 mm

English to English and Metric to English Conversions

1. How many inches are in 7.86 ft? 94.3 in

2. How many feet are in 256 inches? 21.3 ft

3. How many quarts are in 25.78 gallons? 103.1 qts

4. How many centimeters are in 12.0 inches? 30.5 cm

5. Change 25,678 pounds into tons. 12.839 tons

6. How many millimeters are in 1.12 feet? (multi-step problem) 341 mm

Part 2 Varied Applications

You visit the store and find a banana sale. On average 8 bananas weigh one pound. The sale price for bananas is four pounds for \$5.00. You decide to treat your advising group to 25 banana splits, 1 banana per split.

A) Write the equivalence statements.

_______8_______bananas = 1 lb ______4________lbs = _______5.00______ dollars

B) How much will it cost you to buy 25 bananas?

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|25 bananas |

|1 lb |

|5.00 dollars |

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|3.90625 dollars |

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|(given info above) |

|8 bananas |

|4 lbs |

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|= _____\$3.90___________ (rewrite final answer here – be sure to include units & sig figs) |

4) The average length of a hotdog is 12.0 cm. The distance to the moon is 3.84 x 108 meters. How many hotdogs must you line up to reach the moon?

A) Write the equivalence statements

____1__ hot dog = ___12.0 cm _______1__ m = _____100__cm

distance to moon = ______3.84 E8_____ m

B) How many hotdogs must you line up to reach the moon?

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|3.84 x 108 m |

|100 cm |

|1 hotdog |

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|3200000000 htdg |

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|(given info above) |

|1 m |

|12.0 cm |

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|= ______3.20 x 109 hotdogs______ (rewrite final answer here – be sure to include units & sig figs) |

5) The distance to the sun is 1.496 x 1011 meters. The speed of light is 3.00 x 108 m/s. A minute is defined to be 60 seconds.

A) Write the equivalence statements

Distance to sun = _____1.496 x 1011 __ m Speed of light = ____3.00 x 108 ______m/s

__1____minute = _____60______seconds

B) How long does it take for light to get from the sun to the earth?

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|1.496 x 1011 m |

|1 s |

|1 min |

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|8.31111111 min |

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|(given info above) |

|3.00 x 108 m |

|60 s |

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|= _____8.311 minutes_________ (rewrite final answer here – be sure to include units & sig figs) |

6) If the density of mercury is 13.6 g/ml, find the A) mass of 43.0 ml of mercury, B) volume of 43.0 grams of mercury. (Hint: use the density of mercury to write an equivalence statement between mass and volume.)

________13.6 g mercury __= __1 mL mercury___

A) Mass of 43.0 mL of mercury

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|? = |

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|= __________58.5g mercury______ |

B) Volume of 43.0 grams of mercury

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|? = |

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|= ____3.16 mL mercury |

7) You exhale about .0800 liters of CO2 (carbon dioxide) gas in a single breath. 22.4 liters of CO2 contain 6.022 x 1023 molecules. 6,022 x 1023 CO2 molecules have a mass of 44.0 grams. Find the A) number of carbon dioxide molecules you exhale in each breath. Find the mass of the CO2 you exhale in a single breath.

________________= ________________ ________________= ________________

A)

|.0800 L of CO2 |6,022 x 1023 CO2 molecules |

|Single breath |22.4 L of CO2 |

= ______2.15 x 1022 molecules of carbon dioxide

B)

|2.15 x 1022 molecules of of CO2 |44.0 grams CO2 |

|Single breath |6,022 x 1023 CO2 molecules |

= ______1.57 g of carbon dioxide

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43.0 ml of mercury

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