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GSCI 1010
SOLUTIONS: MIDTERM EXAM #1
28 OCTOBER 2009
Part A: Do all parts of the following question (15 parts @ 4 marks each = 60 marks).
1.
For each of the following, indicate whether the statement is true or false, and briefly explain or
justify your answer.
(a)
Even though scientific investigation can be very expensive, humankind should continue
exploring the universe and trying to understand the fundamental building blocks of
matter, no matter how much it costs.
1
3
(b)
One of the four aspects of the physical world can be derived from the other three.
1
3
(c)
TRUE or FALSE (matter of opinion, and could be argued either way)
Because scientific research is so expensive, the question of affordability is a
legitimate one, with scientists always arguing for more funding while many
others question the value of making such huge investments in theoretical
research when there are other pressing problems (hunger, AIDS, climate change)
also requiring attention. That said, it is a fundamental part of human nature to be
curious about the origin and nature of our incredible universe and of matter
itself. So, continuing significant levels of exploration in these areas is both
normal and inevitable.
TRUE
The four aspects of the physical world are space, time, matter and
energy (not to be confused with length, time, mass and electric charge, which
are the four fundamental properties). The properties associated with space, time
and matter are length, time and mass, while the fourth aspect, energy, has
dimensions of mass(length/time)2, and can thus be derived from the other three.
Eratosthenes didn’t need trigonometry to determine the radius of the Earth.
1
3
TRUE
Eratosthenes determined the Earth’s radius around 250 BC, and required only
knowledge of the number  and its relationship to the circumference of a circle.
It was another 100 years before Hipparchus developed his table of chords, a precursor to trigonometry, and about 300 years after that, around 150 AD, when
trigonometry was developed.
2
(d)
Each planet in our solar system travels at its own unique constant speed as it orbits the
Sun.
1
3
(e)
For the right-angled triangle shown, the only way to
determine that cos 60  sin 30 is to use a calculator.
1
3
(f)
FALSE
Let  = 60 be the angle at A and  be the angle at C.
Since the sum of all three angles is 180, it follows
that  = 30. Now, from the diagram,
adj c
opp c
and sin 30 
cos 60 

 .
hyp b
hyp b
These are the same. So, without using a calculator,
it follows that cos 60  sin 30 .
b
a
c
For the right-angled triangle shown, if b = 12.00 cm, then
c = 6.00 cm.
1
3
(g)
FALSE
As Kepler first suggested hundreds of years ago, and has since been confirmed,
the planets in our solar system travel in slightly elliptical, rather than perfectly
circular, orbits (Kepler’s 1st Law). This means that they must speed up slightly
when they are closest to the Sun and slow down a little when they are furthest
from the Sun (Kepler’s 2nd Law).
TRUE
Since the angle at A is 60, then (using a calculator)
c adj

 cos 60  0.500 .
b hyp
Given that b = 12.00 cm, it follows that c = 0.500(12.00) = 6.00 cm.
Since the Earth’s radius is about 3.66 times larger than the Moon’s radius, the Earth’s
volume is 3.66 times greater than the volume of the Moon.
1
3
FALSE
When a length increases by a factor of k, a corresponding area and a
corresponding volume increase by factors of k2 and k3, respectively. Since the
Earth’s radius is greater than the Moon’s by a factor of 3.66, the Earth’s volume
will be greater than the Moon’s by a factor of 3.66 3  49 times.
3
(h)
A single 4.0 microgram dose of a certain drug can be expressed in scientific notation as
4.0  10-6 kg.
1
3
(i)
3
1000
1
3
(j)
of a millimeter and 5000 nanometers have the same order of magnitude.
TRUE
We know 1 mm = 1  10-3 m and 1 nm = 1  10-9 m.
3
So, 1000
mm = 0.003 mm = 0.003(1  10-3) = 3  10-6 m
and 5000 nm = 5000(1  10-9) = 5  10-6 m,
both of which have the same order of magnitude, namely 10-6 m.
Taking the Earth’s circumference to be about 40,000 km, each 1 change in latitude
(along the same longitude) would represent a distance of about 75 km.
1
3
(k)
FALSE
By definition, 1 microgram (or 1 g) is 1  10-6 g.
But 1 kg = 1,000 g, i.e. 1 g = 1  10-3 kg.
It follows that 4.0 g = 4.0  10-6 g
= 4.0  10-6 g( 1  10-3 kg)
= 4.0  10-9 kg, in scientific notation.
FALSE
Since there are 360 in a full circle, each 1 change in latitude corresponds
to 1/360 of the Earth’s circumference, which is
1
40,000  111.1 km , not 75 km.
360
When we look at the universe from Earth, we are seeing some objects as they actually
appeared billions of years before our solar system even began to form (an estimated 5
billion years ago).
1
3
TRUE
The most distant objects discovered so far in the known universe are roughly 15
billion light years away. This means that the light reaching Earth right now from
these objects must have left the source about 15 billion years ago! Since the age
of our own solar system (and hence of the planet Earth) is only about 5 billion
years, we are witnessing these distant objects as they would have appeared
around 10 billion years before the formation of our own solar system began.
4
(l)
If the Sun suddenly burned out at 1:00 PM, there would still be daylight on Earth
at 1:07 PM.
1
3
(m)
Our standard 24-hour day is the time it takes the Earth to complete exactly one
rotation (i.e. 360) on its axis.
1
3
(n)
FALSE
Our 24-hour “day” is actually a “solar day”, which is the interval of time
between two successive moments when the Sun is directly overhead (i.e. from
exactly noon one day to exactly noon the next day). But this requires that the
Earth actually rotates almost 361 (360 to complete one rotation plus nearly 1
more, since the Earth completes its 360 orbit of the Sun in 365 days).
To get from a solid state to a gas (or from a gas to a solid), all matter must go through
an intermediate liquid phase.
1
3
(o)
TRUE
The distance from the Sun to the Earth is known to be 1 AU = 1.50  1011 m, and
light travels with speed c = 3.00  108 m/s. Therefore, the last pulse of light to
leave the Sun at 1:00 PM sharp will reach Earth after a time interval t given by
d 1.50  1011
t 
 500 s.
c 3.00  10 8
Since 500 s = 8 min, 20 s, light will still be reaching Earth at 1:07 PM.
FALSE
Specific examples exist where a change from solid to gas or gas to solid can
occur with no intermediate liquid phase. Ice cubes left in a freezer for a long
period of time eventually “evaporate” and disappear without ever melting into
water, and when carbon dioxide gas is cooled sufficiently, it “freezes” and
forms “dry ice” without first becoming a liquid.
Chemical elements with different atomic numbers can have the same mass number.
1
3
TRUE
Having different atomic numbers means they are different elements, with
different locations in the Periodic Table. But an element with a lower atomic
number can have the same mass number as a second element of higher atomic
number if it has more neutrons than the second element. A specific example of
this is found with 146C (carbon-14) and 147 N (nitrogen). These have distinct
atomic numbers (6 and 7) but the same mass number (14), since carbon-14 has
8 neutrons while nitrogen has only 7 neutrons.
5
Part B: Do any four questions in this section (10 marks each, as indicated).
2.
(a)
List any four of the primary “branches” of physical science (4 marks).
4
(b)
Which branch in (a) is considered the most fundamental, and why (3 marks)?
3
(c)
The most fundamental is physics, since theoretical physics is required to explain
aspects of each of the other four branches (e.g. electromagnetic forces for
chemical bonding; gravity for stellar evolution; and fluid dynamics for
geological or atmospheric movement).
Which branch in (a) do you find the most interesting or intriguing, and why (3 marks)?
3
3.
Any four of the following:
physics; chemistry; astronomy; geology; or meteorology
(earth studies is an acceptable substitute for either of the last two)
Any one of the five could be named for 1 mark. The explanation (2 marks)
is graded on its merit.
It has often been said that the gain of control over fire is the single most important milestone in
the development of humankind, and hence in the development of science and technology.
Explain how any two of the following are an ultimate result of our ability to control fire
(5 marks each, graded for writing as well as for content): (i) a ceramic coffee mug; (ii) a cotton
t-shirt; (iii) the lens in a camera; and (iv) a CD or DVD disc.
(i)
a ceramic coffee mug
5
(ii)
Ceramics involve molding objects from clay and then “firing” them in a kiln
or oven at high temperature, to harden them. Originally, these temperatures
were achieved directly through fire. Today, kilns often operate on electricity, but
all electric power is indirectly produced with the help of fire. Coal-burning is
often used to generate the electricity, but even when this is not the case, the
wiring and most other mechanical parts involved are made of copper, aluminum
or steel, all of which are produced by using fire to extract the metal from the ore
and then to cast or shape it.
a cotton t-shirt
5
The cotton in a t-shirt must undergo significant processing, even if it is initially
picked by hand in the field. Transporting it to the processing site is via truck or
train (i.e. heavy metal machinery as in (i) which burn fossil fuels. The machines
(cotton gins) which turn raw cotton into thread likewise have many metal parts
and run on electricity (as in (i)). The large industrial looms which produce cotton
textiles from the thread are also electrical and have metal parts, and the same is
true of the sewing machines that are ultimately used to make the t-shirts.
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3.
(iii) the lens in a camera
5
Camera lenses are manufactured from glass. First the raw material (silicon
oxide, essentially in the form of sand) is converted to liquid glass by using a very
hot fire. The liquid glass is poured into molds of roughly the right shape, which
are made of metal (as in (i)), and allowed to cool. Next, highly sophisticated
metal machinery (lathes, etc.), which operate on electricity (as in (i)) are used to
first grind and then polish the lenses to extremely precise dimensions.
(iv) a CD or DVD disc
5
4.
CDs and DVDs are manufactured from plastic. Plastics are produced as part of
the petroleum refinement process. Refineries use an extremely hot fire to heat
crude oil until it vaporizes. The various byproducts (motor oil, gasoline, jet
engine fuel, plastic, etc.) can then be separated, since they all condense out of the
vapor at different temperatures. The refinery itself includes a gigantic metal
boiler with countless metal pipes (as in (i)) that carry the various products to
different areas for further refinement. The plastic to be used for CDs/DVDs is
later rolled into thin sheets using massive metal rollers before the circular discs
are punched out by yet another machine.
From the far and near shores of a river 200.0 m wide
(refer to the diagram), the angles of inclination to the
top of a nearby cliff are 50 and 65, respectively.
What is the height h of the cliff above the river
(10 marks)?
2
2
2
2
2

From the diagram:
angle CAB =  = 50
angle ABC =  = 180 – 65 = 115
angle ACB =  = 180 – 115 – 50 = 15
side AB = c = 200.0 m
b
β
a
c
, so

sin  sin 
c sin  200.0 sin 50
a

 592.0 m.
sin 
sin 15
By the sine law,
Then, from the right-angled triangle BCD, sin 65 
opp h
 ,
hyp a
i.e. h  a sin 65  592.0 sin 65  536.5 m.
So, the height of the cliff above the river is h = 536.5 m.
a
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5.
(a)
In physics, the standard unit of energy is called the Joule (symbol J), a unit which is
derived from the standard properties. Use Einstein’s famous equation relating mass and
energy, E  mc 2 (where E is energy, m is mass and c is the speed of light), to determine
the total energy in Joules associated with 2.00 kg of dark chocolate (4 marks).
Using m = 2.00 kg and c = 3.00  108 m/s, we can calculate
4
(b)
E  mc 2  2.003.00  108 
 1.80  1017 J .
(This is 180,000 trillion Joules!)
2
Books sitting on a library shelf may appear to be at rest, but they are actually moving
very quickly and in several different ways. Identify three distinct motions (6 marks).
The library shelf may seem to be at rest, but it is attached to planet Earth.
Therefore, the books are undergoing (any three of) the following motions:
6.
(a)
2
2
2
(i)
(ii)
(iii)
2
(iv)
If the Earth’s radius is taken to be 6.37  106 m, calculate its circumference and confirm
that the answer is consistent with the circumference given in question 1.(j) (3 marks).
3
(b)
rotating with the Earth once every 24 hours;
orbiting the Sun once every 365.25 days;
as part of the solar system, rotating around the centre of a
spiral galaxy known as the Milky Way; and
as part of that galaxy, hurtling through the universe.
We know that the circumference of a circle is C  2r (r = radius). It follows
that:
C  2r  2 6.37  10 6
 4.00  10 7 m ,
which is 4.00  104 km, or 40,000 km.


If North Bay’s latitude is 46 N, how far due north of the equator is Nipissing University
in kilometers (4 marks)?
2
2
Since North Bay is 46 north of the equator, its latitude represents 46/360 of
the Earth’s entire circumference. Hence, its distance north of the equator is
46
40,000  5,111 km .
360
8
6.
(c)
By what factor is the surface area of the Moon smaller than the Earth’s surface area
(3 marks)? See question 1.(g).
1
2
7.
(a)
The Earth’s radius is given in 1.(j) to be 3.66 times larger than the Moon’s
radius. When a length increases by a factor of k, a corresponding area and a
corresponding volume increase by factors of k2 and k3, respectively. It follows
that the Earth’s surface area ( 4r 2 ) will be greater than the Moon’s surface area
by a factor of 3.66 2  13.4 times, i.e. the Moon’s surface area is 13.4 times
smaller than the Earth’s.
Alpha Centauri, a nearby star, is 4.16  1016 m from Earth. Express this distance:
(i)
(ii)
in astronomical units, or AU (2 marks); and
in light years, or LY (2 marks).
(i)
Since 1 AU = 1.50  1011 m, the distance of Alpha Centauri from
Earth is
4.16  1016
 2.77  10 5 AU (about 277,000 astronomical units).
11
1.50  10
(ii)
We also know that 1 LY = 9.46  1015 m. Therefore, the distance of
Alpha Centauri is
4.16  1016
 4.40 light years.
9.46  1015
2
2
(b)
Suppose a star which is 300 LY from Earth explodes at this moment. In what year will
we become aware of the explosion (2 marks)?
2
(c)
Since the star is 300 LY from Earth, it will take 300 years for any evidence
of the explosion to reach us. Therefore, we won’t know about the explosion
until October 28, 2309.
Explain how a second star which explodes exactly 100 years after the star in (b), could
appear to observers here on Earth to explode at same moment (4 marks). (This is an
example of the fact that “the simultaneity of events is relative”.)
2
2
The second star won’t explode until 100 years from now, i.e. on October 28,
2109. Evidence of the second explosion would therefore reach Earth at the same
time as knowledge of the first explosion only if the second star is 200 LY from
Earth. (In other words, although the two explosions actually occurred 100 years
apart, they would appear to happen simultaneously here on Earth.)
9
8.
(a)
When Venus is at its closest point to Earth, a radio beam sent from Earth and reflected
from its surface returns to Earth about 300 s later. How far is Venus from Earth at its
nearest point, in meters (5 marks)?
2
1
2
(b)
Since the radio beam takes 300 s to travel to Venus and return to Earth, the
time for a one-way journey is 150 s. Because we know the speed of light
(c = 3.00  108 m/s), we can find the nearest distance d of Venus from Earth, i.e.
d  ct  3.00  108 150  4.50  1010 m.
Given that Venus is closer to the Sun than Earth is, use your answer for (a) to determine
the approximate radius of Venus’s orbit, i.e. its distance from the Sun:
(i)
(ii)
in meters (3 marks); and
in astronomical units, or AU (2 marks).
(i)
The radius of the Earth’s orbit around the Sun is 1 AU = 1.50  1011 m,
and from (a), Venus is about 4.50  1010 m closer to the Sun. Therefore
the radius of Venus’s orbit is approximately
r  1.50  1011  4.50  1010  1.05  1011 m .
(ii)
Converting the above radius to AU,
1.05  1011
 0.70 AU .
1.50  1011
1
2
2
9.
(a)
Which of the following are pure substances and which are mixtures: (i) NaCl (table
salt); (ii) air; (iii) O2 (oxygen gas) and (iv) seawater, i.e. saltwater (4 marks).
1
1
1
1
(b)
(i)
(ii)
(iii)
(iv)
NaCl – pure substance
air – mixture
O2 – pure substance
seawater – mixture
Suggest how one of the mixtures in (a) can be physically separated (2 marks).
2
2
Air can be separated by cooling it, since the temperatures at which the various
constituent gases will change to a liquid or a solid state are all different.
OR
Seawater can be separated by boiling it or leaving it to evaporate; in either case
the water is converted to a gas (water vapour) and the salt is left in the container
as a residue.
10
9.
(c)
Determine the number of neutrons in the nucleus of each of the following atoms
(2 marks each): (i)
2
(i)
14
6
2
(ii)
56
26
14
6
C
C
(carbon-14, a radioactive isotope); and (ii)
has 14 – 6 = 8 neutrons in its nucleus; and
Fe has 56 – 26 = 30 neutrons in its nucleus.
56
26
Fe
(iron).