Unit Conversions Problem Set
Introduction – HW#1
Name: _______________________
Date: ___________ Period: _____
LENGTH
1. Horses are to race over a certain English meadow for a distance of 4.0 furlongs. What is the race
distance in (a) rods and (b) chains? (1 furlong = 201.168 m, 1 rod = 5.0292 m and 1 chain = 20.117 m.)
(a) 𝑑 =
4.0 furlongs 201.168 m
1 rod
4.0 201.168 rods
×
×
=
= 160 rods
1
1 furlong 5.0292 m
5.0292
(b) 𝑑 =
4.0 furlongs 201.168 m
1 chain
4.0 201.168 chains
×
×
=
= 40 chains
1
1 furlong 20.117 m
20.117
2. The micrometer (1 µm) is often called the micron. (a) How many microns make up 1.0 km? (b) What
fraction of a centimeter is equal to 1.0 µm? (c) How many microns are in 1.0 yd?
(a) 1.0 km =
1.0 km 103 m 106 µm
×
×
= 1.0 × 109 µm
1
1 km
1m
(b) 1.0 µm =
1.0 µm 10−6 m 102 cm
×
×
= 1.0 × 10−4 cm
1
1 µm
1m
(c)
1.0 yd 3 ft 0.3048 m 106 µm
1.0 yd =
×
×
×
= 914400 µm = 9.1 × 105 µm
1
1 yd
1 ft
1m
3. Antarctica is roughly semicircular, with a radius of 2000 km. The average thickness of its ice cover is
3000 m. How much cubic centimeters of ice does Antarctica contain? (Ignore the curvature of Earth.)
Convert radius into centimeters:
Convert thickness into centimeters:
2000 km 103 m 102 cm
𝑟=
×
×
= 2 × 108 cm
1
1 km
1m
3000 m 102 cm
𝑧=
×
= 3 × 105 cm
1
1m
Plug into the formula for the volume of a semicircular cylinder:
1
𝜋
𝐴 = 𝜋𝑟 2 𝑧 = 2 × 108 cm
2
2
2
3 × 105 cm = 18.85 × 1021 cm3 = 1.9 × 1022 cm3
TIME
4. A lecture period (50 min) is close to 1 microcentury. (a) How long is a microcentury in minutes? (b)
Find the percentage difference from the approximation.
10−6 century
100 yr
365 day 24 h 60 min
1 µcenutry =
×
×
×
×
= 52.56 min
1
1 century
1 yr
1 day
1h
percentage difference =
52.6 min − 50 min
= 4.9 %
52.6 min
5. A fortnight is a charming English measure of time equal to 2.0 weeks (the word is contraction of
“fourteen nights”). That is a nice amount of time in pleasant company but perhaps a painful string of
microseconds in unpleasant company. How many microseconds are in a fortnight?
2.0 eek =
2
eek
7 day
24 h 3600 s 106 µs
×
×
×
×
= 1.21 × 1012 µs
1
1 eek 1 day
1h
1s
6. The fastest growing plant on record is a Hesperoyucca whipplei that grew in 3.7 m in 14 days. What
was its growth rate in micrometers per second?
3.7 m 106 µm 1 day
1h
×
×
×
= 3.05886 µm s = 3.1 µm s
14 day
1m
24 h 3600 s
MASS
7. Earth has a mass of 5.98 × 1024 kg. The average mass of the atoms that make up Earth is 40 u. How
many atoms are there in Earth?
5.98 × 1024 kg
1u
1 atom
×
×
= 9.0 × 1049 atoms
−2
1
1.661 × 10
kg
40 u
8. One cubic centimeter of a typical cumulus cloud contains 50 to 500 water drops, which have a
typical radius of 10 µm. For that range, give the lower value and the higher value, respectively, for the
following. (a) How many cubic meters of water are in a cylindrical cumulus cloud of height 3.0 km and
radius 1.0 km? (b) How many 1-liter pop bottles would that water fill? (c) Water has a density of 1,000
kg/m3. How much mass does the water in the cloud have?
a) Find the volume of the cloud in cubic centimeters: (using 1 km = 105 cm)
𝑉𝑐 = 𝜋𝑟 2 ℎ = 𝜋 1 × 105 cm
2
3 × 105 cm = 9.42 × 1015 cm3
Since each cm3 has 50 to 500 drops, the cloud contains between 4.71 × 101 to 4.71 × 1018 drops.
Next find the volume of water in one drop. Then multiply by the number of drops.
4
4
3
3
𝑉𝑤 = 𝜋𝑟 3 = 𝜋 10 × 10−6 m
3
= 4.19 × 10−15 m3
The volume of water in the cloud is from 2 × 103 to 2 × 104 m3
b) Since 1 m3 is equal to 1,000 L we just multiply the answer from part (a) by 1,000 and we find that
the amount of water in the cloud could fill between 2 × 106 to 2 × 10 bottles.
c) Since 1 m3 of water has a mass of 1,000 kg, we just multiply the answer from part (a) by 1,000
and we find that mass of the water in the cloud is between 2 × 106 to 2 × 10 kg.