Chapter 1 Problems
1.3 The micrometer is often called the micron. (a) How man microns make up 1 km? (b) What fraction of a centimeter equals 1µm? (c) How many microns are in 1.0 yard
We begin by calculating 1 km in microns d=1.0km ⋅1000m
1km ⋅ 1micron
1×10−6m =1×109microns Now calculate a distance of 1 micron in cm.
d=1micron ⋅1×10−6m
1micron ⋅100cm
1m =1×10−4cm Finally we compute a distance of 1 yard in microns.
d=1yd ⋅3 ft
1yd⋅0.3048m
1 ft ⋅ 1micron
1×10−6m = 9.144 ×105microns
1.5 The Earth is approximately a sphere of radius 6.37 x 106 m. (a) What is is its circumference in kilometers? (b) What is its surface area in square kilometers? (c) What is its volume in cubic kilometers?
To do all three sections of this problem, we can first convert the radius to kilometers.
r= 6.37 ×106m⋅ 1km
1000m = 6.37 ×103km
(a) The formula for circumference can be found in most calculus books (and in Appendix E of your Physics text). We will assume that we are finding the circumference of the equator.
c = 2π r = 2π ⋅ 6.37 ×103km= 4.002 ×104km (b) The surface area of a sphere is:
a= 4π r2 = 4π ⋅ (6.37 ×103km)2 = 5.099 ×108km2 (c) The volume of a sphere is:
v =4
3π r3 = 4
3π ⋅(6.37 ×103km)3 =1.083 ×101 2km3
1.7 Antarctica is roughly semicircular, with a radius of 2000 km. The average thickness of its ice cover is 3000m. How many cubic centimeters of ice does Antarctica contain? (Ignore the curvature of the earth) (Corrected)
r=2000m
3000m We find the volume in cubic meters first and then convert.
Volume= (Area of half circle) ⋅ (depth)
= π r2 2 ⋅ d
=π (2000 ×103m)2
2 ⋅ 3000m =1.885 ×1016m3
=1.885 ×1016m3⋅ (100cm 1m )3
=1.885 ×1022cm3
1.9 Hydraulic engineers often use, as a unit of volume of water, the acre-foot, defined a the volume of water that will cover 1 acre of land to a depth of 1 ft. A severe thunderstorm dumps 2.0 inches of rain in 30 min. on a town of area 26 km2. What volume of water in acre-feet, fell on the town?
We first convert the depth to feet and the area to acres.
d = 2.0in ⋅ 1ft
12in = 0.1667 ft a= 26km2⋅(1000m
1km )2⋅2.471acres
104m2 = 6.4245 ×103acres Volume = a ⋅ d = 6.4245 ×103acres⋅ 0.1667 ft
=1.071 ×103acre ft
1.10 The fastest growing plan on record is Hesperoyucca whipplei that grew 3.7 m in 14 days.
What was its growth rate in micrometers/second r= 3.7m
14days⋅ 1micron
1.0×10−6m⋅1day
24hrs⋅ 1hr
3600s= 3.059microns s
1.11 A fortnight is a charming English measure of time equal to 2.0 weeks (the word is a
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.
# micros =1 fortnight ⋅ 14days
1 fortnight⋅24h
1d ⋅3600s 1h ⋅ 1µs
10−6s = 1.2096 ×1012
1.12 Enrico Fermi once pointed out that that standard lecture period (50 min.) is close to 1 microcentury. How long is a microcentury in minutes and what is the percent difference from Fermi’s approximation?
# s= 1×10−6century⋅ 100yr
1century⋅ 365.25days
1yr ⋅ 24hrs
1day ⋅ 60min 1hr
= 52.596min
%error= 52.596min − 50min
52.596min ×100 = 4.9%
1.14 Time standards are now based on atomic clocks. A promising second standard is based on pulsar, which are rotating neutron stars (highly compact stars consisting only of neutrons). Some rotate at a rate that is highly stable, sending out a radio beacon that sweeps briefly across Earth once with each rotation, like a light house beacon. Pulsar PSR 1937+21 is an example; it rotates once every 1.557 806 448 872 75 ± 3 ms, where the trailing ±3 indicates the uncertainty in the last decimal place (it does not mean ±3 ms). (a) How many times does PSR 1937+21 rotate in 7.00 days? (b) How much time does the pulsar to rotate 1.0x106 times and (c) what is the associated uncertainty?
(a) We compute the number of rotations No.of rotations= (7days ⋅ 24hrs
1day ⋅ 3600s
1hr ⋅ 1000ms
1s )⋅ 1rotation 1.57780644887275ms
= 3.83317 ×108rotations (b) We can compute the time:
t= 1×106rotations⋅1.57780644887275ms 1rotation
= 1577.80644887275s
(c) A quick and dirty computation of the uncertainty os just to take the uncertainty per rotation and multiply by the number of rotations
Δt = 1 ×106⋅ 3 ×10−14ms= 3 ×10−8ms= 3 ×10−11s
This might not be quite right. For a product of numbers with uncertainties, like r is the product of two numbers, a and b, each with uncertainties..
r= ab Δ r
r = Δ a a
⎛
⎝⎜ ⎞
⎠⎟
2
+ Δ b b
⎛
⎝⎜ ⎞
⎠⎟
2
If we apply this equation here, where the total time T for N turns is the product of N turns and the time per turn.
T = N ⋅t ΔT
T = Δ N N
⎛⎝⎜ ⎞
⎠⎟
2
+ Δt t
⎛⎝⎜ ⎞
⎠⎟
2
ΔT
1577.8s = 0
1×106
⎛⎝⎜ ⎞
⎠⎟
2
+ 3×10−14ms 1.577 ms
⎛
⎝⎜
⎞
⎠⎟
2
ΔT =1577.8s ⋅ 3×10−14ms 1.577 ms
⎛
⎝⎜
⎞
⎠⎟
= 3× 10−11s
The uncertainty IS our quick calculation--because there was no uncertainty in the number of turns.
1.18 Because Earth’s rotation is gradually slowing, the length of each day increases: The day at the end of the 1.0 century is 1.0 ms longer than the day at the start of the century. In 20 centuries, what is the total of the daily increases in time (that is, the sum of the gain on the first day, the gain on the second day)
See class notes.
1.19 Suppose that, while lying on the beach near the equator watching the Sun set over a calm ocean, you start a stopwatch just as the top of the Sun disappears. You then stand, elevating your eyes by a height h=1.7m, and stop the watch when the top of the Sun again disappears If the elapsed time is 11.1s, what is the radius of the earth.
Getting the picture is critical for doing this problem. To proceed, we find the angle that the earth turns through in 11.1 s.
re θ
re
h
This is the angle in the picture.
θ
11.1s= 360°
24hrs⋅ 60min/ hr ⋅ 60sec/min θ = 4.625 ×10−2°
We now use some geometry to find h cosθ = re
re+ h (re+ h)cosθ = re
hcosθ = re− recosθ re = hcosθ
1− cosθ = 1.7m⋅ cos(4.625 ×10−2)
1− cos(4.625 ×10−2) = 5.22 ×106m
This isn’t a great value, but not bad for lying on the beach.
1.20. Gold, which as a mass of 19.32 g for each cubic centimeter of volume, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber. (a) If 1.000 oz of gold, with a mass of 27.63 g is pressed into a leaf of 1.000 microns thickness, what is the area of the leaf? (b) If, instead, the gold is drawn out into a cylindrical fiber of radius of 2.5 microns, what is the length of the fiber.
In this problem, the volume of the gold remains constant. It simply gets “reshaped” into different
shapes. We compute the volume first...
V = 27.63g⋅ 1cm3
19.32g=1.43cm3 a) In the case of a thin sheet of leaf, the volume is
V = Area ⋅ thickness Area= V
thickness = 1.43cm3
1×10−4cm =1.43×104cm2 (b) In the case of a cylinder, the volume is
V=π r2l l= V
π r2 = 1.43cm3
π (2.5 ×10−4cm)2 = 7.28 ×106cm
1.24 One cubic centimeter of a typical cumulus cloud contains 50 to 500 water drops, which have a typical radius of 10 microns. For that range, give the lower value and the higher values, respectively, for the following. (as) 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 mass of 1000 kg per cubic meter of volume. How much mass does the water in the cloud have?
We will calculate the lower number first. The higher values will be 10 times larger since the number of drops is 10 times higher.
a) We are told that the cloud is a cylinder. We begin by calculating the volume of the cloud in cubic cm. This will require us to convert the cloud dimensions to cm.
rcloud= 1km ⋅1000m
1km ⋅100cm
1m = 1× 105cm hcloud= 3km ⋅1000m
1km ⋅100 cm
1m = 3× 105cm
Vcloud=π r2h=π ⋅(1 ×105cm)2⋅ 3× 105cm= 9.42 × 1015cm3 Vwater in cloud= Vcloud ⋅ # drops
cm3of cloud⋅Vdrop Vwater in cloud= 9.42 ×1015cm3⋅50drops
cm3 ⋅4
3π (10 × 10−6m)3
= 1972.9 m3 The upper limit would be 19729 m3
b. We can find the number of 1 L pop-bottles by converting the volume of water to L. The
number of L will be the number of bottles.
1m3 =1000L
Vwater in cloud = 1972.9m3⋅1000L
1m3 = 1.9729 ×106L The upper limit is 1.9729×107L .
c We can now compute the mass of water in the cloud.
m= Vwater incloud ⋅ρwater = 1972.9m3⋅1000kg
1m3 = 1.9729 ×106kg The upper limit is 1.9729×107kg .
1.26 A mole of atoms is atoms to the nearest order of magnitued, how many moles of atoms are in a large domestic cat. The masses of a hydrogen atom, an oxyten atom and a cargon atom are 1.0 u, 16 u, and 12 u respectively. (Hint: Cats are sometimes know to kill a mole).
We begin by estimating the mass of a large cat to be 10 kg. We also take the cat to be made entirely out of water (we’ll ignore the carbon). Since each mole of water has a mass of 18 g (16 for O, 1g for each H), we can compute
mcat = # moles ⋅ 18g mole
# moles= mcat
18g / mole = 10, 000g 18g / mole
= 555 moles
≈ 1
2kilomole
1.30 Water is poured inot a container that has a leak. The mass m of the ater is given as a function of time t by
m= 5.00t0.8− 3.00t + 20.00
with t ≥ 0 , m in grams, and t in seconds. (a) AT what time is the water mass greatest, and (b)what is that greatest mass? In kilograms per minute, what is the rate of mass change at (c) t=2.00s and (d) t=5.00s.
If we plot this equation
We can see that the maximum mass occurs at t≅ 4.25s . The maximum mass can also be read off the graph. The maximum mass is m≅ 23.15g
We can use calculus to do this as well. The rate of change of the mass is given by the derivative.
dm dt = d
dt(5.00t0.8− 3.00t + 20.00)
= 0.8 ⋅5.00 t0.8−1− 3.00
= 4.00t−0.2− 3.00
The maximum mass occurs in this problem when the rate goes to zero (the slope of the mass curve goes to zero.
dm
dt = 4.00t−0.2− 3.00 0= 4.00t−0.2− 3.00 t−0.2= 3.00
4.00 t0.2 = 4.00 3.00 t= 4.00
3.00
⎛⎝⎜ ⎞
⎠⎟
5
= 4.214s
Our estimate from the graph was pretty good. Now we can find the mass by substituting in this time.
m= 5.00t0.8− 3.00t + 20.00 m(4.214s)= 23.16g
Again, our estimate from the graph was very good. We can find the rate of change at 2 s and 5s from the derivative that we have computed.
dm
dt = 4.00t−0.2− 3.00 At t= 2.0s dm
dt = 4.00t−0.2− 3.00 = 0.4822g / s At t= 5.0s dm
dt = 4.00t−0.2− 3.00 = −0.1009g / s 1.46 What mass of water fell on the town in problem 9
Let’s calculate the volume of water in MKS units and then find mass.
d = 2in ⋅2.54cm 1in ⋅ 1m
100cm = 5.08 ×10−2m A= 26 km2 ⋅ 1000m
1km
⎛
⎝⎜
⎞
⎠⎟
2
= 2.6 × 107m
V= A ⋅d = 2.6 ×107m⋅ 5.08 ×10−2m= 1.32 ×106m3 m= V ⋅ρ=1.32 ×106m3⋅ 1000kg
1m3 = 1.32 ×109kg
