Dimensional analysis worksheet 2020-2021 key.pdf - Name key Period ... - Free Printable

Let’s solve each problem step by step.

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Problem 1: Perform the following metric system conversions.

We’ll use standard metric prefixes:

- kilo (k) = 1000
- centi (c) = 0.01
- milli (m) = 0.001
- micro (μ) = 0.000001

Also remember:
- 1 kg = 1000 g
- 1 m = 100 cm = 1000 mm
- 1 L = 1000 mL
- 1 mg = 1000 μg

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(a) 34 km = ? m

→ 1 km = 1000 m
→ 34 × 1000 = 34,000 m

✔ Check: Moving decimal 3 places right → 34.000 → 34000 ✔️

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(b) 568 cm = ? m

→ 100 cm = 1 m
→ 568 ÷ 100 = 5.68 m

✔ Check: Move decimal 2 places left → 568. → 5.68 ✔️

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(c) 0.75 mg = ? kg

First, convert mg to g:
→ 1000 mg = 1 g → 0.75 mg = 0.75 ÷ 1000 = 0.00075 g

Then convert g to kg:
→ 1000 g = 1 kg → 0.00075 ÷ 1000 = 0.00000075 kg

Or in scientific notation: 7.5 × 10⁻⁷ kg — but we’ll keep it as decimal unless asked.

✔ Double-check:
mg → g → kg is divide by 1000 twice → total divide by 1,000,000
0.75 ÷ 1,000,000 = 0.00000075 ✔️

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(d) 0.45 cm³ = ? mL

Important fact: 1 cm³ = 1 mL exactly.

So, 0.45 cm³ = 0.45 mL

✔ No calculation needed — direct equivalence ✔️

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Problem II: Perform the following English units conversion.

Note: These are U.S. customary units.

Common conversions:
- 1 mile = 5280 feet
- 1 foot = 12 inches
- 1 yard = 3 feet
- 1 pound (lb) = 16 ounces (oz)
- 1 ton = 2000 lb (short ton, used in US)

But since no specific problems are listed under II in your image text, and only “Perform the following English units conversion” is written without subparts, I assume this section may be incomplete or missing from the upload. We’ll skip for now unless more info is given.

Wait — looking again at your original input, you pasted text that says:

> II. Perform the following English units conversion.

But then there are no numbered items under it? Actually, looking back — perhaps the numbering got messed up.

Actually, re-examining your pasted text:

After “II. Perform the following English units conversion.” it jumps to:

> A. If a 15-lb stone weighs how much in kilograms?

That seems like part of III? Let me reorganize based on typical worksheet structure.

Actually, let’s look at what follows:

You have:

> III. Answer the following questions using proper significant figures.

And then:

> A. If a 15-lb stone weighs how much in kilograms?

> B. A woman 5 feet tall would have what height in cm?

> C. If a race car is 12 ft long & 6 ft wide, assuming a person standing next to the car has an area of 1.5 m², calculate the area of the car in m².

> D. A 3.5-L bottle of soda contains how many quarts?

> E. A 50-mL serum vial contains how many milliliters? ← This one is trivial — same unit!

Wait — maybe the Roman numerals are misaligned.

Looking carefully at your pasted text:

It says:

```
I. Perform the following metric system conversions.
(a) 34 km = _______ m
(b) 568 cm = _______ m
(c) 0.75 mg = _______ kg
(d) 0.45 cm³ = _______ mL

II. Perform the following English units conversion.

III. Answer the following questions using proper significant figures.
A. If a 15-lb stone weighs how much in kilograms?
B. A woman 5 feet tall would have what height in cm?
C. If a race car is 12 ft long & 6 ft wide, assuming a person standing next to the car has an area of 1.5 m², calculate the area of the car in m².
D. A 3.5-L bottle of soda contains how many quarts?
E. A 50-mL serum vial contains how many milliliters?
```

Ah — so Section II has no actual problems? That might be a typo or omission. But Section III starts with lettered questions.

Perhaps "II" was meant to include those lettered ones? Or maybe they’re all under III.

Given that, and since the instruction says “Answer the following questions using proper significant figures”, we’ll treat A–E under III as requiring sig figs.

Also note: Question E is trivial — 50 mL = 50 mL — probably a trick or test of attention.

Let’s proceed with solving A–E under III.

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III.A: If a 15-lb stone weighs how much in kilograms?

Conversion factor: 1 lb ≈ 0.45359237 kg (but for school, often 0.454 kg is used)

Use 1 lb = 0.4536 kg (commonly accepted value)

So: 15 lb × 0.4536 kg/lb = ?

Calculate:
15 × 0.4536 =
10 × 0.4536 = 4.536
5 × 0.4536 = 2.268
Total = 4.536 + 2.268 = 6.804 kg

Now, significant figures:
“15 lb” has 2 significant figures → answer should have 2 sig figs.

6.804 rounded to 2 sig figs = 6.8 kg

✔ Check: 15 has two digits, both non-zero → 2 sig figs ✔️

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III.B: A woman 5 feet tall would have what height in cm?

Conversion: 1 foot = 30.48 cm (exact definition)

So: 5 ft × 30.48 cm/ft = 152.4 cm

Significant figures: “5 feet” — if it’s exactly 5 (like counted), it could be considered infinite sig figs. But in real measurements, “5 feet” likely means 1 sig fig? Wait — context matters.

In most textbook problems, if it says “5 feet” without decimal, and it’s a measurement, it’s ambiguous. But often in such cases, if it’s a whole number like 5, and no other precision given, we assume it’s exact or has 1 sig fig.

However, height of a person — 5 feet is approximate. Typically, we’d say 5.0 feet if 2 sig figs. Since it’s written as “5 feet”, safest is 1 sig fig.

But let’s think: 5 feet is about 152.4 cm. With 1 sig fig → 200 cm? That doesn’t make sense — too rough.

Actually, in practice, when converting human heights, “5 feet” is usually taken as having 1 sig fig, but the conversion factor is exact, so we go by the measured value.

Standard approach: If “5” is a count or defined, infinite sig figs. But here it’s a measurement — likely 1 sig fig.

But wait — 5 feet is exactly 60 inches, and inch to cm is defined, so perhaps we can consider it exact? Not really — her height is approximately 5 feet.

To avoid overcomplicating, many textbooks treat “5 feet” as having 1 sig fig, so answer should be 200 cm? But that’s not accurate.

Alternative: Perhaps it’s implied to be 5.0 feet? But it’s written as “5”.

Let me check common practice.

Actually, in most high school problems, if it says “5 feet”, and asks for conversion, they expect 152 cm or 150 cm depending on sig fig rules taught.

But let’s see the conversion: 5 × 30.48 = 152.4

If “5” has 1 sig fig → round to 200 cm? That seems wrong because 5 feet is clearly not 2 meters.

Better interpretation: In context of human height, “5 feet” is typically understood as precise to the nearest foot, so ±0.5 ft, which is about ±15 cm, so reporting as 150 cm (2 sig figs?) is reasonable.

This is ambiguous. However, looking at similar problems, often “5 feet” is treated as having 1 sig fig, but the answer is still given as 150 cm (which is 2 sig figs). Hmm.

Wait — 152.4 cm. If we take “5” as 1 sig fig, we round to 200 cm? But that’s 2×10², which is 1 sig fig.

But 200 cm is 2 meters — too tall for 5 feet.

I think there's a mistake in interpretation.

Actually, 5 feet is 60 inches. 1 inch = 2.54 cm exactly. So 60 × 2.54 = 152.4 cm.

The number “5” in “5 feet” — if it’s a measurement, it has 1 sig fig. But in reality, when we say someone is 5 feet tall, we mean approximately 5.0 feet, so 2 sig figs.

To resolve this, let’s look at the other problems. For example, III.C has “12 ft” and “6 ft” — those likely have 2 sig figs each.

Similarly, “15-lb” in III.A has 2 sig figs.

So probably “5 feet” here is intended to have 1 sig fig, but that leads to absurdity.

Another way: perhaps “5” is exact because it’s a defined quantity? No.

I recall that in some curricula, numbers like “5 feet” for height are considered to have unlimited sig figs if it’s a standard reference, but that’s not standard.

Let’s calculate numerically first: 5 × 30.48 = 152.4 cm

Now, for sig figs: the limiting factor is “5”, which has 1 sig fig. So we must round 152.4 to 1 sig fig.

152.4 rounded to 1 sig fig is 200 cm (since 152.4 is closer to 200 than to 100? No — 152.4 is between 100 and 200, and for 1 sig fig, we write it as 2 × 10² cm, which is 200 cm.

But that is incorrect physically — 5 feet is not 200 cm; 200 cm is 6.56 feet.

This suggests that “5 feet” should be interpreted as having more precision.

Perhaps in this context, since it’s a person’s height, and 5 feet is a common approximation, we should use 2 sig figs.

Let me check online or standard practice.

Upon second thought, in many textbooks, when they say “convert 5 feet to cm”, they expect 152 cm or 150 cm, and don't strictly apply sig figs if the number is small.

But the problem specifically says “using proper significant figures”, so we must apply them.

Let’s assume that “5” has 1 sig fig. Then answer is 200 cm.

But that feels wrong.

Alternative: perhaps “5 feet” is exact because it’s a defined value in the problem? Unlikely.

Another idea: in the conversion, 1 foot = 30.48 cm is exact (by definition), so the only uncertainty is in “5”. If “5” is measured, it has 1 sig fig.

But let's look at III.C: “12 ft long & 6 ft wide” — 12 has 2 sig figs, 6 has 1 sig fig. Area will be limited by 6 ft (1 sig fig).

For consistency, perhaps “5 feet” has 1 sig fig.

I think I need to go with the strict rule.

So: 5 ft × 30.48 cm/ft = 152.4 cm

With 1 sig fig → 200 cm (or 2 × 10² cm)

But let's write it as 200 cm.

However, I recall that for numbers like 5, if it's a count, it's exact, but here it's a measurement.

Perhaps the problem intends for us to use the conversion and report with appropriate sig figs based on input.

Let's move to III.C and come back.

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III.C: If a race car is 12 ft long & 6 ft wide, assuming a person standing next to the car has an area of 1.5 m², calculate the area of the car in m².

First, find area in square feet: length × width = 12 ft × 6 ft = 72 ft²

Now convert ft² to m².

Conversion: 1 ft = 0.3048 m, so 1 ft² = (0.3048)² m² = 0.09290304 m²

So, 72 ft² × 0.09290304 m²/ft² = ?

Calculate: 72 × 0.09290304

First, 70 × 0.09290304 = 6.5032128
2 × 0.09290304 = 0.18580608
Sum = 6.68901888 m²

Now, significant figures:
“12 ft” has 2 sig figs, “6 ft” has 1 sig fig. When multiplying, the result should have the least number of sig figs, which is 1 (from 6 ft).

So, 72 ft² has 1 sig fig? 72 has two digits, but since 6 has 1 sig fig, the product should have 1 sig fig.

72 rounded to 1 sig fig is 70 ft².

Then 70 ft² × 0.09290304 m²/ft² ≈ 6.5032128 m², which with 1 sig fig is 7 m²? But that seems rough.

The conversion factor is exact, so sig figs come from measurements.

“12 ft” — 2 sig figs, “6 ft” — 1 sig fig. Product: 72 ft², which should be reported as 70 ft² (1 sig fig) since 6 limits it.

Then 70 ft² × 0.09290304 = let's calculate: 70 × 0.09290304 = 6.5032128 m²

Rounded to 1 sig fig: 7 m²

But is that correct? 72 is closer to 70 than to 80, and for 1 sig fig, 70 is 7×10^1, so yes.

However, in practice, for area, sometimes they keep it as is.

But the problem says “using proper significant figures”, so we must follow.

Also, the person's area is given as 1.5 m², but that seems irrelevant — why mention it? Probably a red herring or for another purpose, but the question is to calculate the area of the car in m², so we ignore the person's area.

So, area of car = 12 ft × 6 ft = 72 ft²

Convert to m²: 72 × (0.3048)^2 = 72 × 0.09290304 = 6.68901888 m²

Sig figs: min(2,1) = 1 sig fig from "6 ft"

So, 7 m² (since 6.689 rounds to 7 with 1 sig fig)

6.689 is greater than 5, so rounds up to 7.

Yes.

But let's confirm the conversion factor.

1 ft = 0.3048 m exactly, so ft² to m² is exact conversion, so sig figs only from measurements.

Measurements: 12 ft (2 sig figs), 6 ft (1 sig fig) — so area has 1 sig fig.

Thus, 7 m².

But intuitively, a race car 12x6 ft is 72 sq ft, which is about 6.7 sq m, and rounding to 7 sq m is acceptable for 1 sig fig.

Okay.

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Back to III.B: 5 feet to cm.

5 ft × 30.48 cm/ft = 152.4 cm

"5" has 1 sig fig, so round to 200 cm? But 200 cm is 2.00 m, while 5 ft is 1.524 m, so 200 cm is inaccurate.

Perhaps "5" is considered to have infinite sig figs because it's a whole number in a context where it's exact? But that's not standard.

Another thought: in some systems, if the number is less than 10 and no decimal, it's assumed to have 1 sig fig, but for height, it's different.

Let's look at III.D and E for clues.

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III.D: A 3.5-L bottle of soda contains how many quarts?

Conversion: 1 L ≈ 1.05669 qt (US liquid quart)

Usually, 1 L = 1.057 qt is used.

So, 3.5 L × 1.057 qt/L = ?

3.5 × 1.057 = let's calculate:

3 × 1.057 = 3.171
0.5 × 1.057 = 0.5285
Sum = 3.6995 qt

Significant figures: "3.5" has 2 sig figs, conversion factor has more, so answer should have 2 sig figs.

3.6995 rounded to 2 sig figs = 3.7 qt

✔ Good.

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III.E: A 50-mL serum vial contains how many milliliters?

This is trivial: 50 mL = 50 mL

But perhaps it's a trick to see if you know that mL is milliliter.

Or maybe it's testing if you recognize that it's the same unit.

So, answer is 50 mL.

Sig figs: "50" — if it's 50. , it might have 2 sig figs, or if it's 50 without decimal, it could be 1 or 2.

In science, "50 mL" often implies 2 sig figs if it's a measurement, but technically, without decimal, it's ambiguous.

However, since it's a vial, likely 50. mL with 2 sig figs.

But the answer is simply 50 mL.

No conversion needed.

So, 50 mL.

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Now back to III.B.

Perhaps for "5 feet", since it's a person's height, and 5 feet is commonly known, we can use 2 sig figs.

Maybe the problem expects 152 cm or 150 cm.

Let's calculate with exact values.

1 foot = 12 inches, 1 inch = 2.54 cm exactly.

So 5 ft = 5 × 12 = 60 inches

60 inches × 2.54 cm/inch = 152.4 cm

Now, "5" has 1 sig fig, but "60" from 5×12 — 12 is exact (definition), so 60 has 1 sig fig if 5 has 1.

60 with 1 sig fig is 60 itself, but when multiplying by 2.54 (exact), the result should have 1 sig fig.

152.4 with 1 sig fig is 200 cm.

But that is not satisfactory.

Perhaps in this context, "5 feet" is considered to have 2 sig figs because it's a typical height measurement.

I recall that in many textbooks, for such problems, they use 152 cm or 150 cm.

Let's assume that "5" has 1 sig fig, but for practical purposes, we'll use 150 cm as it's closer and has 2 sig figs, but that's not strict.

Another idea: perhaps the "5" is exact because it's a defined value in the problem, but that's stretching.

Let's look at the answer choices or typical expectations.

Since this is a worksheet, and for III.A, 15 lb to kg is 6.8 kg (2 sig figs), for III.D, 3.5 L to qt is 3.7 qt (2 sig figs), for III.C, 12 and 6 ft, area 7 m² (1 sig fig), so for III.B, "5 feet" likely has 1 sig fig, so answer should be 200 cm.

But let's write it as 2 × 10² cm to emphasize 1 sig fig.

However, for simplicity, perhaps the problem intends for us to use the conversion and report as 152 cm, ignoring sig figs for this one, but the instruction says "using proper significant figures".

Perhaps "5" is considered to have infinite sig figs because it's a count, but it's not a count.

I think I found a better way: in some sources, when converting units, if the number is a whole number and no decimal, and it's a measurement, it has sig figs equal to the number of digits, but for 5, it's 1.

But let's calculate the value: 5.0 feet would be 152.4 cm, which with 2 sig figs is 150 cm.

5.00 feet would be 152 cm with 3 sig figs.

Since it's "5 feet", not "5.0", it's 1 sig fig.

So, 200 cm.

I'll go with that, even though it's counterintuitive.

So for III.B: 200 cm

But let's box the answers as per calculation.

Perhaps the "person standing next to the car has an area of 1.5 m²" is for comparison, but the question is to calculate the area of the car, so we don't need it.

For III.E, it's 50 mL, which is fine.

Now, let's summarize all answers.

First, Section I:

(a) 34 km = 34,000 m

(b) 568 cm = 5.68 m

(c) 0.75 mg = 0.00000075 kg or 7.5e-7 kg

(d) 0.45 cm³ = 0.45 mL

Section III:

A. 15 lb to kg: 6.8 kg (2 sig figs)

B. 5 ft to cm: 200 cm (1 sig fig) — but let's double-check with a different approach.

Upon second thought, in many educational contexts, for "5 feet", they expect 152 cm or 150 cm, and sig figs are applied loosely.

Perhaps "5" has 1 sig fig, but the conversion factor is exact, so the answer should have 1 sig fig, but 152.4 rounded to 1 sig fig is 200, which is correct mathematically.

I think I'll stick with 200 cm for B.

C. Area of car: 12 ft * 6 ft = 72 ft²; 72 * 0.09290304 = 6.689 m²; with 1 sig fig (from 6 ft), so 7 m²

D. 3.5 L to qt: 3.5 * 1.057 = 3.6995 qt; with 2 sig figs, 3.7 qt

E. 50 mL = 50 mL

For E, "50-mL" — if it's 50. mL, it has 2 sig figs, but the answer is 50 mL, so no issue.

Now, for the final answer, since the user didn't specify which part, but the worksheet has multiple parts, I need to provide all.

But the instruction is to solve the problem, and the problem is the entire worksheet.

So, let's list all answers.

Also, for Section II, there are no problems, so skip.

So, final answers:

I.(a) 34000 m

I.(b) 5.68 m

I.(c) 0.00000075 kg or 7.5 \times 10^{-7} kg

I.(d) 0.45 mL

III.A: 6.8 kg

III.B: 200 cm

III.C: 7 m²

III.D: 3.7 qt

III.E: 50 mL

For III.B, I'm still uneasy. Let me search my memory: in some books, "5 feet" is converted to 152 cm, and sig figs are not strictly applied if it's a standard value.

Perhaps the "5" is exact because it's a defined height in the problem, but that's not stated.

Another idea: in the context, "a woman 5 feet tall" — 5 feet is approximately 1.5 m, and 1.5 m has 2 sig figs, so perhaps 150 cm.

152.4 cm rounded to 2 sig figs is 150 cm.

And "5" might be considered to have 1 sig fig, but in practice, for such problems, they use 2.

Let's look at III.A: "15-lb" — 15 has 2 sig figs, answer 6.8 kg (2 sig figs)

III.D: "3.5-L" — 2 sig figs, answer 3.7 qt (2 sig figs)

III.C: "12 ft" and "6 ft" — 12 has 2, 6 has 1, so area has 1 sig fig, 7 m²

For III.B: "5 feet" — if we compare to "6 ft" in III.C, which has 1 sig fig, then "5 feet" also has 1 sig fig, so 200 cm.

But in III.C, "6 ft" is used for width, and it's a measurement, so 1 sig fig.

Similarly, "5 feet" for height is a measurement, so 1 sig fig.

So I think 200 cm is correct for sig fig rules.

Perhaps the answer is 1.5 × 10^2 cm, which is 150 cm, but that's 2 sig figs.

1.5 × 10^2 has 2 sig figs.

For 1 sig fig, it should be 2 × 10^2 cm.

So 200 cm.

I'll go with that.

For III.E, "50-mL" — if it's 50. mL, it has 2 sig figs, but the answer is 50 mL, so it's fine.

Now, let's write the final answer.

But the user may expect the answers in order.

Also, for I.(c), 0.75 mg to kg.

0.75 mg = 0.75 × 10^{-3} g = 0.75 × 10^{-6} kg = 7.5 × 10^{-7} kg

Or 0.00000075 kg

Both are fine, but perhaps scientific notation is better.

In the worksheet, it's blank, so either is ok.

Now, for the response, since ExplainAnswer is false, TeachTopic is false, GeneratePractice is false, I only provide Final Answer.

But the Final Answer should be the complete solution.

The instruction says: "Final Answer: [Provide only the final correct answer — this should come AFTER all explanations and reasoning.]"

And since ExplainAnswer is false, I should not have explanation, but I did reasoning above.

In the response format, if ExplainAnswer is false, I omit the Explanation section, but I can still do reasoning internally.

For the output, I need to provide the Final Answer with all parts.

So, let's compile.

Final Answer:

I.(a) 34000 m
I.(b) 5.68 m
I.(c) 7.5 \times 10^{-7} kg or 0.00000075 kg
I.(d) 0.45 mL

III.A: 6.8 kg
III.B: 200 cm
III.C: 7 m²
III.D: 3.7 qt
III.E: 50 mL

For III.B, to be precise, let's calculate again.

5 ft * 30.48 cm/ft = 152.4 cm

With 1 sig fig, it is 200 cm (since 152.4 rounded to 1 sig fig is 200).

Yes.

Some might argue that 152.4 is closer to 150 than to 200, but for 1 sig fig, we round to the nearest power of 10 times digit.

152.4, the first digit is 1, so 200 (2*10^2) because 152.4 > 150, and for 1 sig fig, we round 152.4 to 200.

Standard rule: to round to 1 sig fig, look at the first digit and the next digit.

152.4: first digit is 1, second digit is 5, which is >=5, so round up the first digit from 1 to 2, and set others to 0, so 200.

Yes, 200 cm.

Similarly, for 72 ft² in III.C, first digit 7, second digit 2<5, so for 1 sig fig, it would be 70 ft², but we already did the conversion.

In III.C, we have 72 ft², which with 1 sig fig is 70 ft², then 70 * 0.09290304 = 6.5032128 m², which with 1 sig fig is 7 m² (since 6.5 rounds to 7).

6.5 with 1 sig fig is 7, yes.

So all good.

For III.E, 50 mL is 50 mL, and if "50" has 2 sig figs, it's fine.

So, final answer.

But in the worksheet, there is also "II. Perform the following English units conversion." with no items, so perhaps it's a header for III, but III is labeled separately.

I think we have covered all.

So, here is the Final Answer.