Problems #1-10

Metric-English Conversion Problems #11-25 | Metric-English Conversion Examples |

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**Problem #1:** Convert 92.33 yd^{3} to m^{3}

**Solution:**

1) Think of 92.33 yd^{3} this way:

92.33 yd by 1 yd by 1 yd

2) We convert yard to meters with this:

1 yd = 0.9144 m

3) Substituting, we obtain:

(92.33 yd x 0.9144 m/yd) by 0.9144 m by 0.9144 m = 70.59 m^{3}

4) Another way to express the conversion is this:

92.33 yd^{3}x (0.9144 m / yd)^{3}

**Problem #2:** Convert 8.75 lb/ft^{3} to g/mL

**Solution:**

1) The first step is to change lb to g. For this, we use this conversion factor:

1 lb = 453.6 g8.75 lb/ft

^{3}times (453.6 g/lb) = 3969 g/ft^{3}

2) I'm going to use cm^{3} rather than mL. For this, we use this conversion factor:

1 foot = 30.48 cm3969 g/ft

^{3}times (1 ft / 30.48 cm)^{3}= 0.140 g/cm^{3}Since 1 cm

^{3}= 1 mL, we have 0.140 g/mL

Advice: when you need to determine mL (as in the above example), it is often much more convenient to go about determining cm^{3}. Since 1 cm^{3} equals 1 mL, in determining cm^{3}, you are determining mL.

**Problem #3:** Convert 6000 cm^{2} to in^{2}

**Solution:**

**Problem #4:** Convert 3.55 x 10^{-7} miles/min to hm/day

**Solution:**

1) Convert denominator first:

3.55 x 10^{-7}miles/min x (60 min / hr) x (24 hr / day) = 5.112 x 10^{-4}miles/day

2) Convert miles to meters, then to hectometers:

5.112 x 10^{-4}miles/day x (1609.34 m / mile) x (1 hm / 100 m) = 8.23 x 10^{-3}hm/day

**Problem #5:** The men's world record for the hundred-meter (100.0 m) dash was set by Usain Bolt in 2009. His time was 9.572 seconds (official value = 9.58 s). What was his average speed in: (a) m/s; (b) km/hr; (c) ft/s; (d) miles/hr

**Solution:**

1) m/s:

100.0 m / 9.572 s = 10.447 m/s (to four sig figs, 10.45 m/s)

2) km/hr:

10.45 m/s x (1 km / 1000 m) x (60 s / min) x (60 min / hr) = 37.62 km/hr

3) ft/s:

10.447 m/s x (3.28084 ft/ 1 m) = 34.27493548 ft/s = 34.27 ft/sSince one meter is larger than one foot, we assign the 1 to the meter in the conversion unit.

4) miles/hr:

34.27493548 ft/s x (1 mile / 5280 feet) x (60 s / min) x (60 min / hr) = 23.37 mi/hr

You may view video of the sprint here.

**Problem #6:** In America, a car's gasoline efficiency is measured in miles/gallon. In Europe, it is measured in km/L. If your car's gas mileage is 40.0 mi/gal, how many liters of gasoline would you need to buy to complete a 142 km trip? Use the following conversions: 1 km = 0.6214 mi and 1 gal = 3.7854 L

**Solution:** When I solved this problem, I did not clear the calculator at any point before rounding off the answer to three significant figures.

**Problem #7:** If 4.35 x 10^{9} gallons of rain and snow fall on the United States daily, how many kilograms of water fall on this country each hour?

**Solution:**

1) Convert to gal/hr:

4.35 x 10^{9}gal/day times (1 day / 24 hr) = 1.8125 x 10^{8}gal/hr

2) Determine weight of the gallons per hour that fall:

1.8125 x 10^{8}gal/hr times 8.329 lb/gal = 1509631250 lb/hrI used the density from the table here. I used the value for 70 °F, which is equal to about 21 °C

3) Convert pounds to kilograms:

1509631250 lb/hr times (1 kg / 2.20462 lb) = 6.84758 x 10^{8}kgTo three sig figs would be 6.85 x 10

^{8}kg

**Problem #8:** The density of balsa wood is 7.8 lb per cubic foot. What is the weight, in kg, of a piece of balsa wood 4.0 inch by 6.0 inch by 20.0 inch?

**Solution:**

1) Let us determine how many cubic inches are in our piece of balsa wood:

4.0 in x 6.0 in x 20.0 in = 480 in^{3}

2) Determine how many cubic inches are in a cubic foot:

1 ft^{3}= 1 ft x 1 ft x 1 ft = 12 in x 12 in x 12 in = 1728 in^{3}

3) How much of a cubic foot is our piece of balsa wood?

480 in^{3}divided by (1728 in^{3}/ ft^{3}) = 0.27778 ft^{3}

4) How many pounds of balsa wood do we have?

7.8 lb / ft^{3}times 0.27778 ft^{3}= 2.166684 lb

5) Convert lb to kg:

2.166684 lb times (1 kg / 2.20462 lb) = 0.9828 kgTo two sig figs, this is 0.98 kg.

Comment: we could have converted our inch values to feet by dividing each by 12 inches / foot, to get this:

0.33333 ft x 0.5 ft x 1.666667 ft = 0.027778 ft^{3}

then, go to step 4 above.

**Problem #9:** A car's engine is rated at 3.60 liters (volume in the cylinders). How many cubic inches is this? (Remember: 1 L = 1000 mL and 1 mL = 1 cm^{3})

**Solution:**

1 L = 1000 cm^{3}= 10 cm by 10 cm by 10 cm1 inch = 2.54 cm

(10 cm x 1 in / 2.54 cm) by (10 cm x 1 in / 2.54 cm) by (10 cm x 1 in / 2.54 cm)

3.937 in by 3.937 in by 3.937 in = 61.0234 in

^{3}<--- cubic inches in one liter3.60 L x (61.0234 in

^{3}/ L) = 220. in^{3}(to three sig figs)

**Problem #10:** If zinc has a density of 446 lb/ft^{3} , what is the density of zinc in g/cm^{3}?

(446 lb/ft^{3}) (453.6 g/lb) <--- converts lb to g(446 lb/ft

^{3}) (453.6 g/lb) (1 ft^{3}/ (12^{3}in^{3}) <--- converts ft^{3}to in^{3}, the unit would be g/in^{3}if we ended here(446 lb/ft

^{3}) (453.6 g/lb) (1 ft^{3}/ (12^{3}in^{3}) (1 in^{3}/ 2.54^{3}cm^{3})The answer is 7.14 g/cm

^{3}Note the style of the conversion factors, to wit:

(1 ft^{3}/ (12^{3}in^{3})The idea is this:

1 ft^{3}is a cube 12 inches on a side.1 ft

^{3}= 12 in x 12 in x 12 in = 12^{3}in^{3}

**Bonus Problem:** The speed of light in a vacuum is 2.998 x 10^{8} m/s. What is the speed of light in miles/sec?

**Solution:**

2.998 x 10^{8}m/s times (1 mile / 1609.34 m) = 186,288 mile/s (not paying any attention to sig figs)Just remember what Al (our pal) Einstein said.

Metric-English Conversion Problems #11-25 | Metric-English Conversion Examples |

Metric-English Conversion Problems #26-60 | Return to Metric Menu |