Problems #1-10

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

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

**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/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

5) Here's a dimensional analysis set-up for converting to miles per hour (mph)

100 m 39.3701 inches 1 foot 1 mile 3600 s –––––– x –––––––––––– x ––––––– x ––––––– x ––––––– = 23.37 mph 9.572 s 1 m 12 inches 5280 feet 1 hr

6) From the last factor to the first:

change seconds to hours

change feet to miles

change inches to feet

change meters to inches

meters per second is the unit to be changed to miles per hour (mph)

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.

1) Here's a solution in a step-wise manner:

40.0 mi 1 km ––––––– x –––––––– = 64.3708 km/gal 1 gal 0.6214 mi

64.3708 km 1 gal –––––––––– x ––––––– = 17.005 km/L 1 gal 3.7854 L

142 km 1 L 142 km ––––––– x –––––––– = 8.35 L -----> another way -----> –––––––––– = 8.35 L 1 17.005 km 17.005 km/L

2) Here's a solution in a dimensional analysis manner:

142 km 0.6214 mi 1 gal 3.7854 L ––––––– x ––––––––– x –––––– x ––––––– = 8.35 L 1 1 km 40.0 mi 1 gal

**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 |