Sunshine State Scholars
Final Competition - Solutions
March 12-14, 2006
Harmful algal blooms have been documented in the Gulf of Mexico since the time of the
early Spanish explorers and are now commonly called red tides.
a. Give three mechanisms by which these blooms have adverse effects on marine
b. Give a possible reason why these blooms may actually be helpful to the
a. 1. They have direct toxic effects on fish and indirect toxic effects on species that
feed on the fish. They also have direct toxic effects on other marine life.
2. They reduce oxygen at night due to the respiration of micro algae and bacteria
that are decomposing the animals killed by the toxins. This further increases
the mortality of fish and other organisms.
3. They increase turbidity and reduce the light available for benthic algal and
higher plant (sea grass) species.
b. Although HABs often affect species of value to humans they may in fact be of
benefit to the ecosystem. The large fish kills release organics and other nutrients
to the environment more rapidly than they might be in the absence of HABs.
While this may help to sustain the HABs in the short term, it can also promote the
development of other primary producers, such as diatoms, which may be more
favorable food for herbivores and secondary producers. Secondly, HABs cause
the removal by death of many carnivores and planktivores. This can increase the
amount of available phytoplankton, which can result in increased survival of
larvae of many species. Often after a “red tide” fish stocks actually increase.
The construction of platforms in space uses a certain amount of titanium in the steel.
Titanium is often included in steel alloys to increase the strength of the alloys. However,
there is some evidence that the number of defects found in steel ingots increases with the
amount of titanium used in the alloy. The numbers of defects in 28 steel ingots were
determined for ingots with titanium concentrations ranging from 69 to 87 parts per
million (ppm). The number of defects ranged from 11 to 47. A “scatter plot” along with
the “least squares” regression line and some computer output from a regression analysis
of these data is shown below.
70 75 80 85
Predictor Coefficient Standard Error T P
Constant -125.044 6.606 -18.93 .00001
Concentration 1.948 0.088 21.98 .0001
R-Squared = 94.9%
a. What is the equation of the least squares regression line? What is the physical
meaning of each variable used in that equation?
b. What is the numerical value of the correlation coefficient for the concentration of
titanium and the number of defects in the alloys? Interpret this correlation.
c. Suppose that you want to describe the relationship between the concentration of
titanium and the number of defects, for alloys listed having numbers of defects in
the range of 70 to 80 ppm. Does the line shown in the scatter plot provide the
best description of the relationship for data in this range? Why or why not?
a. The predicted number of defects = -125.044 + 1.948 × (The concentration of
y = −125.044 + 1.948x
y = The average (or expected) number of defects
x = The concentration of titanium
b. The value of the correlation coefficient is r = 0.949 ≈ 0.974 (r is positive
because the slope of the regression line is positive). There is a strong increasing
linear relationship between the concentration of titanium and the number of
defects in the steel.
Approximately 94.9 percent of the variability in the number of defects can be
explained by a linear relationship between the concentration of titanium and the
numbers of defects found in the ingots. The number of defects increases as the
concentration of titanium increases.
c. No. The three points in the upper right-hand corner influence the equation of the
given least squares regression line. The remaining points with concentrations
between 70 and 80 would have a regression line with a smaller slope.
Light is absorbed as it passes through water. Let Iz be the light intensity at depth z
meters. The light intensity can be calculated from the equation Iz = I0e−kz . An average
value of k for coastal waters is k = 0.4 and for open ocean is k = 0.07 . This has effects
on the primary production of plants, either inside the water column or at the bottom of the
water column. The depth at which the rate of oxygen production by photosynthesis
equals the rate of oxygen consumption by respiration is the depth at which 1% of the
surface illumination remains. €
a. What are the factors that effect light absorption?
b. What are the corresponding depths in coastal and in open ocean environments at
which the rates of oxygen production by photosynthesis and oxygen consumption
by respiration are equal?
c. How does the relationship between the rates of oxygen production by
photosynthesis and oxygen consumption by respiration affect the feeding types of
animal communities in the water column?
a. Some factors affecting light penetration are the absorption and scattering by water
molecules; particles (including phytoplankton and zooplankton) in the water; and,
dissolved colored organic matter.
b. Iz = I0e−kz
ln z = −kz
€ ln( Iz I0 )
ln(0.01I0 I0 ) ln(0.01) −4.605
Coastal waters: z = − =− ≈− ≈ 11.51 meters.
€ 0.4 0.4 0.4
Open Ocean: z=− ≈− ≈ 65.79 meters
c. Herbivores, decomposers, and carnivores can exist in the zone of active
photosynthesis. Below this zone only carnivores, chemosynthetic organisms, and
decomposers can survive.
A developer is studying the possibility of constructing a building on a wedge-shaped
piece of land that she recently purchased. The land is in the shape of a sector of a circle
with radius 200 meters and a central angle of 60 degrees. In order to make the most
effective use of the land, she wants the “footprint” of the building to be as large as
possible, and for cost reasons the base of the building must be rectangular with one side
along the boundary of the parcel. What is the largest possible area for the base of the
Let l and h be the dimensions of the base of the building. To satisfy the requirements of
the developer, it must be placed on the lot as illustrated and labeled in the following
h = 200sin θ
= cot θ
€ = cot 60o =
l x 1
€ = cot θ − = cot θ −
h h 3
€ l = h cot θ − = 200sin θ cot θ −
l = 200cosθ − sin θ
Let A be the area of the base of the building.
200 1 1 π
A = lh = 200sin θ 200cos θ − sin θ = 2002 sin(2θ ) − sin2 θ 0 ≤ θ ≤
3 2 3 3
dA 2 1
= 2002 cos (2θ ) − sin θ cos θ = 2002 cos (2θ ) − sin (2θ )
dθ 3 3
€ dA 1 π π
= 0 ⇒ cos (2θ ) = sin (2θ ) ⇒ 3 = tan (2θ ) ⇒ 2θ = ⇒ θ =
dθ 3 3 6
Let Amax be the maximum area of the base of the building.
1 2π 1 π 1 3 1 1 3
Amax = 2002 sin − sin2 = 2002 × − × = 2002
2 6 3 6 2 2 3 4 6
Amax = × 3 ≈ 11547 square meters
Arsenic (III) can be removed from water through catalytic oxidation in the presence of
activated carbon and dissolved oxygen in a stirred reactor. (Nriagu, 1994) This reaction
proceeds as a first order reaction and at a pH 6 to 10 range. It has been found that 90% of
the As3+ can be removed in about 25 minutes. At this rate, how long would it take to
clean well water at 80 ppb to a safe level of 10 ppb?
€ Let C0 be the initial concentration of As3+ and let Ct be the concentration at time t.
ln = −kt
It takes 25 minutes to reduce the concentration from 100% to 10%.
ln € = −k × 25 min ⇒ k ≈ 0.092min−1
Let t final be the time required to clean the well water from 80 ppb to 10 ppb.
1 Ct 1 10
t final = − ln final ≈ − ln ≈ 22.6min
€ k C0 0.092 min−1 80
Two trains are headed towards each other on the same straight track. When they are a
distance of 2 km apart they see each other. They both immediately brake. The braking
for each train produces a constant deceleration. The first train is traveling at 100 km/hr
and it decelerates at 0.25 m/sec , the second train is traveling at 50 km/hr. What is the
minimum deceleration for the second train to just avoid a collision with the first train?
(Give your answer in m/sec2.)
In a dynamical system the change in kinetic energy is the “work” done. If the
acceleration is a constant, then the work is the force times the distance. Let v 0 be the
initial velocity and let v f be the final velocity.
1 1 2 2
F d = mad = m v 2 − m v 0 ⇒ 2ad = v 2 − v 0
f f €
v 2 = v 0 − 2ad
Let train 1 be traveling at 100 km/hr and place the origin of a coordinate system at the
position of the front of this train when the braking begins. Take the positive direction of
the coordinate system in € direction from train 1 to train 2. Train 2 is at coordinate
2000 (in meters) when the braking begins. The deceleration of train 1 is 0.25 m/sec and
the deceleration of train 2 is a2 . The initial velocity of train 1 is approximately 27.78
m/sec and the initial velocity of train 2 is approximately 13.89 m/sec. Let x1 be the
location on the coordinate system of train 1 and let x 2 be the location on the coordinate
system of train 2.
v1 ≈ 27.78 2 − 2 × 0.25 x1 ≈ 771.1− 0.5 x1
v 2 ≈ 13.89 2 − 2a2 (2000 − x 2 ) ≈ 192.9 − 2a2 (2000 − x 2 )
The two trains should meet at coordinate x with velocity zero.
x≈ ≈ 1543.46
Train 1 will stop after traveling approximately 1543.46 meters.
€ 192.9 − 2a (2000 −1543.46) ≈ 0
192.9 − 913.08 a2 ≈ 0
a2 ≈ ≈ 0.211
Train 2 must decelerate at approximately 0.211 m/sec .
In olden days learning to tell time meant looking at a disk on a wall with hands and
numbers and translating the image you saw into a time. Such phrases as “half past nine”
were staples of those days. Today digital clocks have taken over and “9:30” has replaced
“half past nine”. The following problem illustrates that from a mathematical point of
view there are interesting problems associated with clocks with minute (big) hands that
move at twelve times the rate of hour (little) hands. At 3:00 P.M. the hands of an analog
clock form a right angle. What is the next time that the hands of that clock will form a
If the hour hand didn’t move then the answer would be 30 minutes later at 3:30 P.M.
Since the hour hand does move, the extra time needed to once again form a right angle is
30 minutes plus the time it takes the minute hand to cover the additional distance that the
hour hand travels. Measure the distance that the hands move in “minutes on the clock”.
That is, the distance “one” is the distance on the circumference between marks that are
exactly one minute apart. Thus the circumference of the clock is 60 “minutes”. When
the hands again form a right angle, the hour hand will have traveled a distance d and the
minute hand will have traveled 30 + d . Since the minute hand travels at twelve times the
d 30 + d
rate of the hour hand and they both travel for the same amount of time, = .
Solving for d gives d€ = . This is approximately 2 minutes and 7 seconds. The next
time the hands form a right angle will be exactly 3 : 32 or approximately 3:32:07.
The diagram below shows 3 generations of the pedigree of deafness in a family. Black
circles/squares indicate deaf persons. A square indicates a male and a circle indicates a
Decide whether each of the following statements is “true” or “false”. Then choose one of
the statements and explain why it is either “true” or “false”.
a. The deafness in this pedigree could be inherited as a dominant autosomal
characteristic (allele of a gene).
b. The deafness in this pedigree could be inherited as a recessive autosomal
c. The deafness in this pedigree could be inherited as a sex-linked dominant
d. The deafness in this pedigree could be inherited as a sex-linked recessive
The explanation should include that one needs two alleles (one from mom and one from
dad for each characteristic. It should also include the fact that “autosomal” alleles are not
involved in sex determination.
a. False. If it were a dominant autosomal gene, then more offspring in the diagram
would be deaf, and for a couple to have a deaf child, one of the parents would have to
be deaf (which is not the case in generations I and II).
b. True. This is an autosomal characteristic that is recessive. This is shown in the second
family where both male and female offspring are deaf from non-deaf parents.
c. False. If this were the case then all people with the gene would be deaf.
d. False. The characteristic is carried on the X chromosome and if the male gets the
deaf gene he is deaf. If a female gets one deaf gene and one non-deaf gene they are
carriers, but would need to have both deaf genes to be deaf. In the second family
there are a male and female offspring that are deaf and who have parents who are not
Saccharin, HNC 7H 4SO 3, is used as a substitute for sugar and has a pKa = 2.32 at 25°C.
What is the pH of a 0.050 saccharin solution at 25°C?
Construct an “Initial-Change-Equilibrium” (ICE) table for the reaction.
HNC 7H 4SO 3 (aq) + H 2O ⇔ H 3O + (aq) + NC 7H 4SO− (aq)
I 0.050 - -
C −x +x +x
€ E 0.050 − x x x
Ka = 10€ = 10−2.32 € 4.8 ×10−3
This value for Ka is too close to the concentration. Therefore the quadratic equation must
be used to solve for x.
Ka = = 0.0048
0.050 − x
x 2 + 0.0048 x − 0.00024 = 0
x ≈ 0.01328 or x ≈ −0.01808
The only reasonable root is 0.03128. This represents the H+ concentration.
pH ≈ −log0.01328 ≈ 1.88
Kepler’s Laws of orbital motion can be applied to satellites in orbit around the Earth.
1. Each satellite moves in an elliptical orbit with the center of the Earth at one focus.
2. The radius vector drawn from the center of the Earth to the satellite sweeps out
equal areas in equal time intervals
3. The square of the orbital period of the satellite is proportional to the cube of the
semi-major axis of the elliptical orbit.
Suppose that a satellite is in orbit around the Earth. Suppose, further, that at its closest
approach (perigee) its altitude above Earth’s surface equals the radius of the Earth, R.
And suppose that at its most distant point (apogee) its altitude above the Earth’s surface
is 3R. What is the period of the satellite’s orbit?
(Assume that the radius of the Earth is 6.37 ×10 6 meters, its mass is 5.98 ×10 24 kgs,
and the acceleration due to gravity at the Earth’s surface is g = 9.8 m/sec . The
universal gravitational constant is 6.673 ×10−11 Nm /kg .)
Solution: € €
From Kepler’s second law, a satellite in a circular orbit would move with uniform
velocity. Therefore from Kepler’s third law, the period of a satellite in an elliptical orbit
is the same as the period of a satellite in a circular orbit that has a radius equal to the
semi-major axis of the ellipse. For the satellite described in this problem the radius of the
circular orbit would be r = 3R . For a satellite to remain in a stationary circular orbit, the
gravitational acceleration must equal the centrifugal acceleration. The period T of the
orbit is the circumference of the circle divided by the velocity.
GMm mv 2 GM GM
= ⇒ = v2 ⇒ v =
r r r r
2π r 2π r 2π r 3
T= = =
v GM GM
2π ( 3R) 3 6π R 3R
6π × 6.37 ×10 6 3 × 6.37 ×10 6 524.89 ×10 9
T= ≈ 6
≈ 2.63 ×10 4 seconds ≈ 7.31 hours
6.673 ×10−11 × 5.98 ×10 24
€ 19.98 ×10