Problem 1
Calculate the magnitude of acceleration of a cube that is completely submerged in oil which has a density of
500 kg/m3. The cube measures 10 cm on a side and has a mass of 10 kg.
a. 4.905 m/s2
b. 93.2 m/s2
2
c. 0 m/s
d. 9.32 m/s2
Link within Knovel
Solution
To calculate an object’s acceleration when it is subjected to forces, one must determine the net effect of all
forces, then divide this by the object’s mass (Newton’s second law). The source linked to provides, on page
273, the equation to determine the net force on a submerged object. It is reproduced here:
𝑭𝒏 = 𝐹𝑏 − 𝑊𝑂 = 𝜌𝑓 𝑉𝑠 𝑔 − 𝜌𝑜 𝑉𝑜 𝑔 = 𝜌𝑓 𝑉𝑠 𝑔 − 𝑚𝑜 𝑔
where FB is the buoyant force, WO is the weight of the object, 𝜌𝑓 is the density of the fluid, Vs is the
submerged volume, 𝜌𝑜 is the density of the object, mo is the mass of the object, and g is the acceleration due
to gravity (9.81 m/s2). Note the far right side of the equality where the product of the object’s density and
volume has been replaced by its mass is not provided in the text linked to. Substituting in values yields
𝑘𝑘
𝑚
𝑚
𝑭𝒏 = ��500 3 � (0.1𝑚)3 �9.81 2 �� − �(10 𝑘𝑘) �9.81 2 �� = −93.195 N
𝑚
𝑠
𝑠
Finally, dividing this by the mass of the object, 10 kg, yields a=-9.3195 m/s2., or approximately -9.32 m/s2.
The magnitude of the acceleration vector is approximately 9.32.
Problem 2
If the cube is not touching any edges of the container, is it sinking further down or is it rising to the
surface?
a. rising
b. sinking
c. it is not moving
d. cannot be determined
Solution
This can be answered in two equally sound ways. The simpler of the two is to note the negative value of
acceleration calculated. This means that the weight of the cube is actually greater than the buoyant force and
the cube is accelerating downwards. The second method is to calculate the density of the cube and compare
𝑚
10 𝑘𝑘
𝑘𝑘
it to that of the oil. The density of the cube is 𝜌𝑐𝑐𝑐𝑐 = 𝑐𝑐𝑐𝑐 = (0.1𝑚)3 = 10,000 3 . This is much greater than
kg/m3),
𝑉𝑐𝑐𝑐𝑐
𝑚
so again it is shown that the cube is sinking. It is stipulated in the
the density of the oil (500
question that the object is not touching any edges of the container so as to not introduce any frictional
forces, and to imply that it had not already sunk to the bottom which would make it at not moving.
Problem 3
The first law of thermodynamics is a statement of:
a. conservation of momentum
c. conservation of mass
b. increasing entropy
d. conservation of energy
Link within Knovel
On the left side of the page, just under the title of the equation is a description: the first law of
thermodynamics is the law of conservation of energy.
Problem 4
The second law of thermodynamics states that the during a real process the quality of energy of a closed
system will
a. always remain constant
b. always decrease
c. always increase
d. vary across situations
Link within Knovel
Solution
On the left side of the page, just under the title of the equation is a description: The second law of
thermodynamics states that energy has quality as well as quantity and that real processes proceed in the
direction of decreasing quality of energy.
Problem 5
The statement that if two systems are each in thermal equilibrium with a third system then they are also in
thermal equilibrium with each other is
a. the zeroth law of thermodynamics
b. the third law of thermodynamics
c. the fundamental theorem of calculus
d. Newton’s law of cooling
Link within Knovel
Solution
A Knovel search for each of these four terms will help the player to determine which is correct. Upon
searching for the first one, the zeroth law of thermodynamics, one will see a plethora of results which may
be viewed. Clicking the first one, “2.4 The Zeroth Law of Thermodynamics” from Aerothermodynamics of
Gas Turbines and Rocket Propulsion brings one to a paragraph which reads “If two bodies are separately in
thermal equilibrium with a third body, they are in thermal equilibrium with each other.” The same section
also explains the, perhaps unexpected, name. If this resource is not available in one’s subscription many
others will yield similar terminology.
The following information is included with problems 6 through 9.
An engineer has been tasked with designing a runway to be used by Boeing 747-400ERs. The maximum
takeoff mass of the planes is 412,760 kg. The planes use four Pratt & Whitney engines, each of which
produces 275 kN of thrust. The takeoff speed is approximately 250km/h. 747-400 ER Performance Summary.
Problem 6
Express the takeoff speed of the jet in m/s.
a. 0.250 m/s
c. 69.44 m/s
b. 250,000 m/s
d. 6,944 m/s
Link within Knovel
Solution
This hint brings the player to the Knovel Unit Converter. In the “input value” field, type 250. In the input
unit field, type “km/h” or “km/hr” and in the output unit field, type “m/s”. Click Convert. Knovel will return
a magnitude of 69.44.
Problem 7
Using Newton’s second law, determine the acceleration of the plane when it is fully loaded.
a. 0.666 m/s2
b. 5.17 m/s2
c. 2.66 m/s2
d. 1.5 m/s2
Link within Knovel
Solution
This hint brings the player to the populated results resulting from a search for “Newton’s second law”.
Select the resource 1.3.2.1 Newton’s Second Law from CRC Handbook of Mechanical Engineering (2nd
edition) (2004). The page opened to gives the most general form of the equation, Σ𝑭 = 𝑚𝒂. Rearranging
Σ𝑭
yields 𝒂 = . Neglecting any friction between the wheels of the plane and the tarmac, as well as air
𝑚
resistance (a simplification which is not accurate, but greatly simplifies the problem) the only force acting
on the plane in the direction of motion is the thrust created by its four engines. The total thrust is the sum
of that created by each, or 4*275,000 N. Substituting into the equation,
Σ𝑭
4∗275,000 𝑁
𝒂= =
≈ 2.66 𝑚/𝑠 2
𝑚
412,760 𝑘𝑘
Problem 8
With this acceleration, how long will it take the plane to reach its takeoff speed?
a. 26.1 s
b. 50 s
c. 13.25 s
d. 71 s
Link within Knovel
Solution
This hint brings the player to a page of a resource located within Knovel, with the pertinent equation to
answer this question. Note the “trail of breadcrumbs” in the upper part of the page that shows how this page
was found. The second breadcrumb shows “Search for ‘uniform accelerated [linear motion]’”. Try clicking
on it, then coming back to the needed page. The pertinent equation is v=v0+at, where v is the final (takeoff)
speed, v0 is the initial speed and a is the acceleration. Note that v0=0 because the plane is starting from rest.
Use the value of acceleration calculated in the previous problem, and the takeoff speed as expressed in m/s.
Rearrange the given equation, then substitute values:
𝑚
𝑣 − 𝑣0 69.44 𝑠 − 0
𝑡=
=
≈ 26.1 𝑠
𝑎
2.66 𝑚/𝑠 2
Problem 9
Using only this very simplified approach, what is the minimum length for the runway?
a. 517 m
b. 907 m
c. 257 m
d. 1.5 km
Link within Knovel
This hint brings the player to the same page as the previous problem. Again note how it was arrived at
though the “trail of breadcrumbs” in the upper part of the page. The runway can be no shorter than the
distance the plane travels in the 26.1 seconds that it takes to reach its takeoff speed. The relevant equation
for this problem is 𝑠 = 𝑣0 𝑡 +
𝑎𝑡 2
2
, where s is the distance traveled, v0 is the initial velocity, t is the time
elapsed, and a is the acceleration. Substitute the missing values, which are now known.
𝑚
�2.66 2 � (26.1𝑠)2
𝑠
𝑠 = 0+
= 906.0093 𝑚
2
In the time that it takes the plane to accelerate from rest to its takeoff speed it travels just over 906 meters.
However a runway that is 906 meters long would be too short, so the distance is rounded up to 907 meters.
The following information is included with problems 1 through 5.
Archimedes was tasked by King Hiero II with determining if his crown was made of pure gold. To
do this he placed the 5 kg crown in a tub full of water and it subsequently sank to the bottom.
Upon weighing the water that spilled over the edge, Archimedes found its mass to be 350 g.
Problem 1
What volume of water was displaced?
a. 3.5*105 m3
b. 35 m3
c. 3.5*10-4 m3
d. 1 m3
Link within Knovel
Solution
The link brings a player to a completed Knovel Data Search. View the results by clicking the
appropriate yellow button in the lower area of the screen. Table 1.8 Properties of Fluids from Food
Processing Technology-Principles and Practice (3rd Edition) (2009), shows the density of water at
0°C to be 1000 kg/m3. With the mass and density of the water, its volume can now be computed as
𝑚
0.350 𝑘𝑔
−4 3
the ratio of mass to density. 𝑉 = 𝜌 =
𝑚 . If this particular resource is not
𝑘𝑔 = 3.5 ∗ 10
1000 3
𝑚
available with one’s subscription, another title may be found in Knovel.
Problem 2
What is the volume of the crown?
a. 3.5*105 m3
b. 35 m3
c. 3.5*10-4 m3
d. 1 m3
Solution
One of Archimedes’ brilliant deductions was that a body completely submerged in a liquid will
displace a volume of that liquid equal to its own. The volume of the crown is equal to the volume
of water that flowed over the edge of the tub, which was determined in problem 1 to be
3.5 ∗ 10−4 𝑚3 .
Problem 3
What is the approximate density of the crown?
a. 19,320 kg/m3
b. 14,300 kg/m3
c. 427.35 kg/m3
c. 25,700 kg/m3
Link within Knovel
Solution
The link brings a player to Knovel generated results from a search for “definition of density”. If it
is needed the URL may be followed. Click on the resource “8.2 Units and Definitions Related to
Density” from Industrial Pressure, Level, and Density Measurement (2nd Edition) (2009). It states
“Density is defined as mass per unit volume”. So the density of the crown is 𝜌 =
𝑚
𝑉
5 𝑘𝑔
= 3.5∗10−4 𝑚3 ≈
14,286 𝑘𝑔/𝑚3 or to three significant figures 14,300 kg/m3. Again, if this particular resource is not
available with one’s subscription, another may be found which provides a similar definition—if
any such resource is needed.
Problem 4
What is the density of pure solid gold?
a. 19.35 g/cm3
b. 193.2 g/cm3
c. 1.932 g/cm3
d. 193,200 kg/m3
Link within Knovel
Solution
This link brings a player to partially completed Knovel Data Search. Find the property density by
either typing it into the find a property field in the upper right part of the page, or by locating it
under the physcial constants dropdown list. Drag and drop density into the query builder. Keep
the default option of exists, and view the results. Select “Table 3. Densities of Alloys and Metals”
from Woldman’s Engineering Alloys (9th Edition) (2000). The top row of this table shows the
density of gold to be 19.35 g/cm3. Again, any Knovel resource may be subsitituted for the one
mentioned here.
Problem 5
Is the crown pure gold?
a. Yes
b. depends on the temperature at which mass was recorded
c. not enough information to answer d. no
Link within Knovel
Solution
If the crown is made of pure gold, then its density will be the same as that of the pure substance.
The link provided brings a player to the Knovel Unit Converter main page. Convert the density of
gold, determined in problem 4, from g/cm3 to kg/m3. In the input value field type 19.35. To
designate the input unit, click New Unit and type in appropriate labels in both the numerator and
denominator, then click Enter. If ever unsure how to enter the symbol for a unit, click the Select
Input Unit button. Repeat a similar procedure on the output side, and click Convert. Knovel will
return a value of 19,350 kg/m3. The density of the crown (14,300 kg/m3) is significantly less than
that of pure gold, so some other materials must have been mixed in. Indeed this is what
Archimedes discovered.
Problem 6
The Buckingham-Pi theorem states that a system of seven variables, which are all described in
terms of density and time, can be grouped into ________ dimensionless groups.
a. 1
b. 2
c. 3
d. 4
Link within Knovel
Solution
The link provided brings the player to a section of a text about the theorem. What is significant
here is the paragraph that begins with “If a system is characterised…” (sic). The system in the
problem has seven variables, thus m=7. They are described in terms of density and time. One
might be tempted to believe that n=2. However, density is not a fundamental dimension. Density
is the ratio of mass to volume, and volume is the cube of the fundamental dimension length. Thus
the three fundamental dimensions describing these variables are mass, length, and time, and n=3.
Now it is clear:
m-n = 7-3 =4 dimensionless groups.
Problem 7
Consider an airfoil contained inside a wind tunnel, making contact with the edges. The airfoil is
characterized by its chord length, c, and the airflow around it has a velocity, v. The moving air has
a density, 𝜌 and a viscosity, 𝜇. Due to the friction between the airfoil and the air a drag force, F, is
induced.
How many dimensionless groups are needed to completely describe this scenario?
a. 5
b. 2
c. 3
d. 6
Link Within Knovel
Solution
The link brings the player to a different page of the same resource used in problem 6. The page
provides an example which is nearly identical to the question at hand. The variables describing
this system are chord length, velocity of air, density of the air, viscosity of the air, and drag force.
This means that m=5. Let L, T, and M, be the fundamental dimensions of length, time, and mass
respectively. Chord length is described by the dimension L, velocity by LT-1, density by ML-3,
viscosity by ML-1T-1, and drag force by MLT-2. The number of unique dimensions (n) is 3.
Therefore m-n = 5-3 = 2 and the number of dimensionless groups needed to completely describe
this system is 2.
Problem 1
At what thermodynamic temperature is the density of liquid acetylene equal to 0.5 g/mL?
a. -25°C
c. 352.1 K
b. 465.4 °R
d. -10°F
Link within Knovel
Solution
Following the link provided brings a player to an interactive graph in Knovel’s platform. It shows
density (of liquid acetylene) on the ordinate and temperature on the abscissa. Note that the default
units of density and temperature are g/mL and K, respectively. Under the X and Y columns click
the button Add Point, tap tab to edit the Y value and type in 0.5. The returned X value is 258.6 and
the units are Kelvin. This does not match the Kelvin value in choice c. Go to the X Unit dropdown
menu and change this to °R. In the X and Y columns, the X value has changed to 465.4. Celsius
and Fahrenheit are not thermodynamic temperature scales.
To find this interactive graph, go to the Knovel Homepage. Click the Data Search button, under
the main search bar. On this page, in the material or substance name field, type “acetylene”
(without the quotation marks). Then in the find a property field type “density”. Drag and drop the
physical constant into the query builder. Keep the default option of exists, then go to the results.
On the left side of the page, click the Interactive Graphs facet. Open “Table 14. Density of Liquid –
Organic Compounds, density…” from Yaws’ Critical Property Data for Chemical Engineers and
Chemists. Finally click the Σ under the equation plotter column, next to acetylene (located in the
second row, first column).
Problem 2
What is the density of liquid acetylene at a temperature of 50°F?
a. 4.360 g/mL
b. 4.360 kg/L
c. 5.0 lb/gal
d. 0.8460 slug/ft3
Link within Knovel
Solution
This link brings the player to the same interactive graph that the previous problem did. Using the
dropdown menu, change the X Unit to °F. Under the X and Y columns click Add Point, and enter
an X value of 50. Then experiment with the Y Unit dropdown until a combination of value and
unit is found that matches one of the four choices. This will be 0.8460 slug/ft3.
Problem 3
When Earth and Jupiter are closest together, the distance between Brooklyn, New York and the
surface of Jupiter is 5.88*1011 meters. At that time Brooklyn is on the side of Earth facing Jupiter.
Earth’s radius is 6,378.1 kilometers and its mass is 5.97*1024 kilograms. Jupiter’s radius is 70,000
kilometers and its mass is 1.898*1027 kilograms. A human baby weighing 3 kilograms is delivered
at a hospital in Brooklyn by a doctor weighing 70 kilograms. At the time that the doctor is washing
her hands she is 1 meter away from the baby.
At that moment, which object is exerting a greater gravitational force on the infant, Jupiter or the
doctor?
a. Doctor
b. Jupiter
c. Not enough information to determine
d. Equal forces
Link within Knovel
Solution
This link brings the player to a page of a book in Knovel that contains a section on Universal
Gravitation. An equation is provided to determine the gravitational force exerted by any object on
another one. The pertinent equation provided on that page is reproduced here:
𝐹=𝐺
𝑚1 𝑚2
𝑑2
where G is the universal gravitational constant, whose value is given on the
following page of that text as 6.668 * 10-11 m3/(kg s2), m1 is the mass of one object, m2 is the mass of
the other, and d is the distance between their centers of mass. When calculating the force between
the infant and Jupiter the distance to be used is the sum of the radius of Jupiter and the distance
between Earth and Jupiter (represented by d1 in the equation below). Let FJup be the force exerted
by Jupiter and Fdoc be the force exerted by the doctor. Because G and the mass of the baby will
appear in both calculations, they may each be neglected when calculating the relative magnitudes.
𝑚𝐽𝑢𝑝 𝑚𝑏𝑎𝑏𝑦
𝑚𝐽𝑢𝑝𝑖𝑡𝑒𝑟
1.898 ∗ 1027 𝑘𝑔
𝑘𝑔
𝐹𝐽𝑢𝑝 = 𝐺
∝
=
≈ 5,488 2
2
2
11
2
(70,000,000 𝑚 + 5.88 ∗ 10 𝑚)
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑚
+𝑑 )
(𝑟
𝐽𝑢𝑝𝑖𝑡𝑒𝑟
1
𝑚𝑑𝑜𝑐 𝑚𝑏𝑎𝑏𝑦 𝑚𝑑𝑜𝑐 70 𝑘𝑔
𝑘𝑔
∝ 2 =
= 70 2
2
2
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
1𝑚
𝑚
𝑑2
Note that the symbol between first and second fractions (after which G and the mass of the baby
𝐹𝑑𝑜𝑐 = 𝐺
have been eliminated) denotes “proportional to” instead of “equal to”. The following rough
sketches are provided to help illustrate the quantities.
Problem 4
The specific gravity of an oil is 0.8. What is its density?
a. 800 kg/m3
b. 8,000 kg/m3
c. 1250 kg/m3
d. 0.0008 kg/m3
Link Within Knovel
Solution
The link provided brings a player to Knovel’s Interactive Equation which defines specific gravity;
it is the ratio of an oil’s density to that of water. Thus the density of the oil is equal to the product
of its specific gravity to the density of water. To determine water’s density, either perform a
Knovel Data Search or open the interactive worksheet of this equation. To do the latter hover over
the Σ=+ 𝑜𝑝𝑒𝑛 button and select equation worksheet. If it is desired this worksheet can be
maximized in its own tab by clicking the “squares” on the right of the blue banner”. The density
of water is given towards the bottom of the worksheet, as 1000 kg/m3. Thus the density of the oil is
0.8*1000 kg/m3 = 800 kg/m3.
Problem 5
A particular alloy can sustain a maximum pressure of 50 kPa. What is the greatest depth to which
it can be submerged in the oil?
a. 10 m
b. 6.37 m
c. 15.4 m
d. 2.74 m
Link within Knovel
Solution
This link brings a player to a page which describes the relationship between depth (in water) and
the pressure experienced there. The equation provided is 𝑝 = 𝜌𝑤 x 𝑔 x 𝐷, where p is pressure, 𝜌𝑤
is the density of water, g is the acceleration due to gravity (9.81 m/s 2), and D is the depth. While
this equation is stated in terms of water, it can be generalized for any fluid by replacing the
density of water with the density of that fluid. The density of this oil was calculated in the
previous problem. Rearranging the provided equation and solving for depth yields
𝑝
50,000 𝑃𝑎
𝐷=
=
≈ 6.37 𝑚
𝜌𝑜𝑖𝑙 ∗ 𝑔 800 𝑘𝑔 ∗ 9.81 𝑚
𝑚3
𝑠2
Problem 6
What pressure would be experienced at that depth in water?
a. 62.5 kPa
b. 37.5 kPa
c. 100 kPa
d. 101.325 kPa
Link within Knovel
Solution
This link brings a player to the same page that problem 5 did. This time pressure is being asked
for. To answer this, simply use the equation provide on the page with the density of water. The
density of water is given as 1000 kg/m3 in the equation worksheet of problem 4.
𝑘𝑔
𝑚
𝑝 = 𝜌𝑤𝑎𝑡𝑒𝑟 ∗ 𝑔 ∗ 𝐷 = 1000 3 ∗ 9.81 2 ∗ 6.37 𝑚 = 62,489.7 𝑃𝑎 ≈ 62.5 𝑘𝑃𝑎
𝑚
𝑠
Problem 7
Fourier’s Law of Conduction states that the magnitude of the “heat flux” across a thick wall is
___________ than it would be if the same wall were thinner.
a. greater
b. less
c. the same as
d. exponentially less than
Link within Knovel
Solution
This link brings a player to Knovel’s Interactive Equation “Fourier’s Law of Conduction”. In case
this particular equation is not included with one’s subscription, it is reproduced here.
Δ𝑇
Δ𝑄𝑡 = −𝑘 · 𝐴 · Δ𝑋 where Δ𝑄 is “heat flux”, A is surface area, k is the material’s thermal
conductivity, Δ𝑋 is the spatial dimension, and Δ𝑇 is the change in temperature. The spatial
dimension is the distance of the material (i.e. wall) through which heat is transferred. A greater
spatial dimension (thicker wall) will result in a smaller magnitude of Δ𝑄 than a smaller spatial
dimension will. There is no exponential relationship.
Problem 8
What is the thermal conductivity of common brick, in kcal m-1 h-1 °C-1?
a. 0.34
b. 125
c. 1.31
d. 25.7
Link within Knovel
Solution
This link brings a player to a table listing common brick and its thermal conductivity. It was
obtained through a Knovel Data Search. After entering a material “common brick” only one result
is available.
Problem 9
Consider an industrial furnace constructed from 0.019-m-thick common brick. During steady
state operation the inner temperature is 11270C while the outer temperature is 876.90C. How long
would it take for 74.59 kcals to be transferred through a wall of the furnace if it measures 0.5 m by
2 m on a side?
a. 45 minutes
b. 13.25 seconds
c. 1 minute
d. 2.5 hours
Link within Knovel
Solution
This link brings one back to the same equation seen in problem 7. To work with the interactive
features, hover over the Σ=+ 𝑜𝑝𝑒𝑛 button, and select equation worksheet. To see the worksheet in a
full tab click the button on the right of the blue banner. Scroll down to the part of the worksheet
with the word “Calculation”. This is where one may input values for all of the variables. Where ΔT
is displayed, type the actual temperature difference (11270C-876.90C=250.1 0C). To insert the unit
click “Units” in the blue banner near the top of the page. In the window that appears select
temperature from the properties list, and Celsius from the units list. Click insert. ΔX is the
thickness of the wall, the distance through which the energy transfer occurs, 0.019 m. As
determined in problem 8, k=0.34
𝑘𝑐𝑎𝑙
.
𝑚 ℎ °𝐶
Area is simply 1 m2. (0.5m *2m=1m2) Knovel will return an
answer that the magnitude of the heat flux is 4475.4737 kcal/hr.
What Knovel just did is substitute the values that the player provided into Fourier’s Law of
Conduction, as follows:
(1127 − 876.9)°𝐶
Δ𝑇
𝑘𝑐𝑎𝑙
Δ𝑄𝑡 = −𝑘 · 𝐴 ·
= −0.34
∗ (2𝑚 ∗ 0.5𝑚) ∗
≈ −4475.4737 𝑘𝑐𝑎𝑙/ℎ𝑟
Δ𝑋
𝑚 ℎ𝑟 °𝐶
0.019 𝑚
The negative sign indicates the direction of the heat transfer.
To find how long it would take for 74.59 kcals to be transferred, divide this number by the rate of
4475.4737 kcal/hr to find the number of hours it will take. Then multiply by 60 to convert to
minutes
74.59 𝑘𝑐𝑎𝑙𝑠
60 𝑚𝑖𝑛
= 0.0166ℎ𝑟 ∗
≈ 1 𝑚𝑖𝑛𝑢𝑡𝑒
𝑘𝑐𝑎𝑙
ℎ𝑟
4475.4737
ℎ𝑟
It will take 1 minute for 74.59 kcals to be transferred.
The following information and diagram are included with problems 1 and 2.
Water is flowing into a circular pipe through two inlets, one with radius r1 at velocity v1, the other with
radius r2 at velocity v2. Those two inlets converge, and the flow then exits through a single outlet with
radius r3 at velocity v3. Assume the flow is incompressible (∇ • 𝒗 = 0), and
𝜕𝜌
steady ( 𝜕𝑡 = 0 𝑒𝑣𝑒𝑟𝑦𝑤ℎ𝑒𝑟𝑒), and that no water accumulates anywhere in the pipe.
v2
r3
r1
r2
v3
r2
v2
Problem 1
If r1 and r2 are equal to each other and each is half of r3, express v3 as a function of the inlet velocities
𝟏
𝟒
𝜌2
𝑣
𝜌1 2
𝟏
𝟒
𝜌
= 𝜌3 𝑣3
1
1
2
1
2
a. 𝒗𝟑 = 𝒗𝟏 + 𝒗𝟐
b. 𝑣3 = 𝜌1 𝑣1 + 𝜌2 𝑣2
c. 𝑣1 =
d. 𝑣3 = 4𝑣1 + 4𝑣2
Link within Knovel
Solution
Note: the v2 that appears next to the upper inlet should be v1.
The link included brings a player to Knovel’s Interactive Equation “Continuity Equation”. As a
consequence of conservation of mass and no accumulation of water in the pipe the total mass of water
flowing into the pipe through all inlets must be equal to the total mass of water flowing out through all
exits, at any instant. Knovel provides the equation for the specific case of a pipe with two inlets
(denoted by subscripts 1 and 2) and one outlet (denoted by subscript 3):
𝑣3 =
𝜌2 𝐴2 𝑣2 + 𝜌1 𝐴1 𝑣1
𝜌3 𝐴3
where 𝜌 is the density of the water, A is the cross sectional area of the pipe, v is the velocity of the
flowing water, and the subscripts denote the opening. It is stated in the introductory information that
the flow is steady, thus density is constant. This means that 𝜌1 = 𝜌2 = 𝜌3 and all three may be dropped
from the equation. Eliminating density, then rearranging slightly yields
𝑣3 =
𝜌2 𝐴2 𝑣2 + 𝜌1 𝐴1 𝑣1 𝐴2 𝑣2 + 𝐴1 𝑣1 𝐴2
𝐴1
=
=
𝑣2 + 𝑣1
𝜌3 𝐴3
𝐴3
𝐴3
𝐴3
𝐴
𝐴
(1)
At this point the ratios 𝐴2 and 𝐴1 ought to be computed. (From the problem statement r3=2r1=2r2)
3
3
𝐴2 𝜋(𝑟2 )2
𝑟22
1
=
=
=
2
2
(2𝑟2 )
𝐴3 𝜋(𝑟3 )
4
𝐴1 𝜋(𝑟1 )2
𝑟12
1
=
=
=
2
2
(2𝑟1 )
𝐴3 𝜋(𝑟3 )
4
Finally substitute these values into (1)
1
1
𝑣3 = 𝑣2 + 𝑣1
4
4
Problem 2
If r1=1m, r2=0.5 m, r3=10 m, and v1=10 m/s, and v2=1160 m/s what is v3?
a. 1000 m/s
b. 110 m/s
c. 50 m/s
d. 3 m/s
Link within Knovel
Solution
The link included with this problem is the same as the previous one. This time one may make use of the
interactive features to quickly answer the problem. To open the interactive worksheet hover over the
Σ=+ 𝑜𝑝𝑒𝑛 button and select “equation worksheet”. To view it in its on tab, click the icon in on the right
side of the blue banner. Under Legend with variables and units go ahead and enter the appropriate
values. Again, the flow in the pipe is said to be incompressible and steady, so the values entered for the
densities will not affect v3, so long as all three values are the same, finite, and nonzero. The default value
used in this worksheet is that of water. One may manually enter approximate values for the areas, or
use the Lowercase Greek pallet on the right to select 𝜋, then type in the square of the radius. Knovel will
then calculate the values of the area. Using the former approach
𝐴1 = 𝜋(1𝑚)2 ≈ 3.14159 𝑚2
𝐴2 = 𝜋(0.5𝑚)2 ≈ 0.785𝑚2
𝐴3 = 𝜋(10𝑚)2 ≈ 314.159𝑚2
Finally, enter v1=10 m/s and v2=1160 m/s. After typing the value for each variable, hit enter/return on
the keyboard to allow Knovel to reflect the new information in the value of v3. All of the default units in
this interactive worksheet are consistent with those stated in the problem, so there is no need to
change them. Knovel will substitute all of these values into the Continuity Equation and return v3=3 m/s.
Problem 3
An engineer working for an eyewear company wants to design a new line of sunglasses for specialized
use. She wants them to utilize two very thin polarizing lenses which allow 25% of the incident sunlight
through. At what angle should the two be oriented to one another?
a. 75.52°
b. 60°
c. 45°
d. 80°
Link within Knovel
Solution
The link provided brings the player to a section on Law of Malus, the relationship which governs
polarizing lenses, named for French engineer Étienne Louis Malus. The Law states
𝐼𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑
1 + cos(2𝜃)
= cos 2(θ) =
𝐼𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡
2
where I transmitted is the intensity of the light transmitted through a lens, I incident is the intensity of the light
incident upon the same lens, and θ is the angle between.
The law will be applied twice, once for each of the lenses. Let I1 be the sunlight incident upon the first
lens, I2 be the now polarized light transmitted through it and thus incident upon the second lens, and I3
be the polarized light transmitted through the second lens. The first paragraph of the source states a
polarizer “will transmit half of the illumination of an incident beam of unpolarized light”. Sunlight is
unpolarized, thus
𝐼2 = 0.5 𝐼1
Applying the Law of Malus to the second lens:
𝐼3
1 + cos(2𝜃)
= cos 2(𝜃) =
(1)
𝐼2
2
It is stated in the problem that the light which has passed through both lenses is equal to 25% of the
initial incident light, thus 𝐼3 = 0.25 𝐼1
From (1)
𝐼3
0.25 𝐼1
𝜃 = acos (√ ) = acos (√
) = acos(√0.5) = 45°
𝐼2
0.5 𝐼1
Or
𝜃=
acos (2 ·
2
𝐼3
− 1)
𝐼2
=
acos (2 ·
0.25 𝐼1
− 1)
0.5 𝐼1
= 45°
2
Problem 4
A calorie is defined as the amount of energy required to raise one gram of water by
a. 10 °C
b. 1 °F
c. 1° N
d. 𝟏 °C
Link within Knovel
Solution
This link brings a player to the Knovel homepage. A search for “calorie” will yield many dictionary
results, almost all of which will include the sentence of the problem completed with 1°C. It is
alternatively stated as 1/100 of the heat required to raise 1 gram of water from 0°C to 100°C at constant
atmospheric pressure.
Problem 5
The specific heat of water is
a. 10 cal/ (g*°C)
c. 15 J/ (kg*°C)
b. 100 cal/ (g*°C)
d. 1 BTU/ (lb*°F)
Link within Knovel
This link brings the player to the Knovel Data Search homepage. In the Material or Substance Name type
water. In the Find a property field type specific heat, then drag and drop it into the query builder. Go to
the results. Open “Table 6-17 Properties of Liquids” from Pressure Vessel Design Manual (3rd Edition)
(2004). This shows the specific heat as 1 BTU/ (lb °F).
Alternatively, one may use the Knovel Unit Converter. From the previous problem, the specific heat of
water is 1 cal/ (g* °C). This immediately eliminates choices a and b from this problem. In the Input value
field of the Unit Converter type 1. Under the Input Unit field click the button new unit. Enter “cal” in the
numerator and “g*degC” in the denominator. Click Enter. Repeat the procedure on the output side. Click
New Unit, then enter each of the units in choice c of this problem. In the numerator type “J”, in the
denominator “kg*degC”. Click Enter. Knovel returns an output value of 4187, not 15. Finally try the units
of choice d on the output side, keeping the input the same. In the output numerator type “BTU”, in the
denominator type “lb*degF”. Click Enter. Click Convert. Knovel returns 1.000. Hence choice d is again
shown to be correct. If ever unsure how to enter a particular unit, click “Select Output Unit”, select
“Property” and find it under the alphabetical listing of properties.
Problem 6
1 BTU/ (lb*°F) is equal to:
𝐽
b. 1.725*10-7
a. 5 𝑘𝑔∗𝐾
𝑒𝑉
c. 6*1010 𝑙𝑏∗𝑑𝑒𝑔𝐹
𝒆𝒒𝒖𝒊𝒗 𝒃𝒃𝒍, 𝒐𝒊𝒍
𝒍𝒃∗𝒅𝒆𝒈𝑭
𝑐𝑎𝑙
d. 10 𝑘𝑔∗𝑑𝑒𝑔𝐶
Link within Knovel
Solution
This problem provides the player an opportunity to use the Knovel Unit Converter. In the Input value
field type 1. Under Input Unit, click the New Unit button. In the numerator type “BTU”, in the
denominator type “lb*degF”. Click Enter. On the output side, click the button New Unit. Try each of the
units that appear in choices a through d of this problem, entering them exactly as they appear, then click
Enter. Click Convert after entering the Units. Only choice b will match the numerical value returned by
Knovel.
Problem 7
All of the following are SI base units, except:
a. ampere
c. second
b. mole
d. ohm
Link within Knovel
Solution
This link brings a player to a completed Knovel search for “SI base units”. Open the resource “SI Base
Units” from Oxford Dictionary of Biochemistry and Molecular Biology (2nd Edition) (2006). The page
opened to lists all seven SI base units, just above SI prefixes. The only unit from this problem which does
not appear on that list is ohm. The table just to the right of that shows that ohm is the SI derived unit of
electrical resistance, defined as a volt per ampere, or in terms of SI base units m2 kg s-3 A-2.
Problem 8
What is the pressure of air at an altitude of 10.5 km?
a. 101.325 kPa
c. 16.12 kPa
b. 57.625 kPa
d. 24.475 kPa
Link within Knovel
Solution
This link brings a player to the Knovel Data Search homepage. In the Material or Substance Name field
type “air” (exclude quotation marks). In the Find a property field type “altitude”, then drag and drop the
property into the query builder. Notice how a specific search can greatly reduce the number of results
one must sort through. Go to the results. Open Table 19-1 Ambient Conditions Versus Pressure Altitude.
See that at an altitude of 10500 meters, the air pressure is 24.475 kPa.
The following diagram is included with problems 9 through 11.
Problem 9
What is the equivalent resistance of this parallel circuit?
a. 150 Ω
b. 25 Ω
c. 6 Ω
d. 0.1666 Ω
Link within Knovel
Solution
This link brings a player to a completed Knovel search for “equivalent resistance in parallel” Open the
resource 3.2.1.3 Equivalent Parallel Resistance. The page opened to shows that the equivalent
resistance, Req, of n resistors set up in parallel is given by
1
𝑅𝑒𝑞 =
1
1
1
𝑅1 + 𝑅2 + ⋯ + 𝑅𝑛
For two resistors in parallel this simplifies to:
𝑅𝑒𝑞 =
1
𝑅1 ∗ 𝑅2
=
1
1
𝑅1 + 𝑅2
𝑅1 + 𝑅2
Substituting in the values given for these two parallel resistors:
𝑅1 ∗ 𝑅2
10Ω ∗ 15Ω
=
= 6Ω
𝑅1 + 𝑅2 10Ω + 15Ω
Problem 10
Using the principle of current dividers, determine the current flowing through R1.
a. 10 A
b. 6 A
c. 1.6666 A
d. 4 A
Link within Knovel
Solution
This hint brings a player to another completed Knovel search, this time for “current dividers”. Open the
resource 7.4.2.1 Current Dividers from Exploring Engineering – An Introduction to Engineering and
Design (2nd Edition) (2010). Pages 129 through 131 of this resource were used to write problems 9
through 15 of this week’s Challenge. As per the resource, according to Ohm’s Law, the current flowing
through the circuit before the division is
𝑉
1
𝑅1 + 𝑅2
10 Ω + 15 Ω
𝐼=
=𝑉∗
=𝑉∗
= 60𝑉 ∗
= 10 𝐴
𝑅𝑒𝑞
𝑅𝑒𝑞
𝑅1 ∗ 𝑅2
10 Ω ∗ 15 Ω
The current through R1 is given by Ohm’s law to be:
𝑉
𝐼1 = 𝑅
1
Verbatim from the resource: dividing this equation by the previous one and solving for I1 gives
𝑅2
15 Ω
𝐼1 = 𝐼 ∗
= 10 𝐴 ∗
=6𝐴
𝑅1 + 𝑅2
10 Ω + 15 Ω
The preceding approach utilized the principle of current dividers. Alternatively, one may simply apply
Ohm’s Law to R1, as is also done in the resource:
𝑉
60𝑉
𝐼1 =
=
=6𝐴
𝑅1 10 Ω
Problem 11
Again use the principle of current dividers to determine the current flowing through R2.
a. 10 A
b. 6 A
c. 1.6666 A
d. 4 A
Link within Knovel
Solution
This hint brings the player to the same search results as the previous one; open the same resource if it is
not already open in one’s browser. Using the principle of current dividers, as in the previous problem
𝑅1
10 Ω
𝐼2 = 𝐼 ∗
= 10𝐴 ∗
=4𝐴
𝑅1 + 𝑅2
10 Ω + 15 Ω
The value of I used here is the same value calculated in the solution to problem 10.
A simpler approach is to apply Ohm’s Law to R2. About a quarter of the way down page 130 the resource
includes the equation 𝐼1 = 𝑉/𝑅1 . Replace the 1 subscripts with 2, then substitute in the values included
with the diagram
𝑉
60𝑉
𝐼2 =
=
=4𝐴
𝑅2 15Ω
The following diagram is included with problems 12 through 15
Problem 12
Using Kirchhoff’s Current Law express I1 as a function of I2 and I3, with the directions indicated by the
arrows in the diagram. Remember this relationship, as it will be utilized in later problems.
a. I1 = I2 + I3
b. 𝐈𝟏 = 𝐈𝟑 − 𝐈𝟐
c. I1 = I2 /I3
d. I1 = I2 − I3
Link within Knovel
This link brings a player to a completed Knovel search for Kirchhoff’s Current Law (KCL). Open 7.4.2
Kirchhoff’s Current Law from Exploring Engineering – An introduction to Engineering and Design (2nd
Edition) (2010), if it is not already open. The Law is defined as the statement that the sum of the
currents into a given node equals the sum of the currents out of that node. A node in an electrical circuit
is defined simply as any point where two or more elements meet. A new diagram of the circuit is
included on the next page, this time with each of its nodes labeled, A, B, C, and D. Applying KCL to node
A, using the directions indicated, yields:
Sum of currents entering=sum of currents exiting
I1 + I2 = I3
I1 = I3 − I2
(1)
𝑅1 = 10 Ω, 𝑅2 = 15 Ω, and R 3 = 80 Ω
I1
I2
A
C
D
R2
R1
I3
V1=20 V
1
R3
B
2
V2=50 V
Problem 13
Using Kirchhoff's Voltage Law and Ohm's Law determine the value of I3. This value will be used in
subsequent calculations.
a. 0.372 A
b. 10.5 A
c. 0.796 A
d. 5.12 A
Link within Knovel
Solution
This link brings a player to a completed search in Knovel for “Kirchhoff’s voltage law”. Note how these
search terms can lead to the same resource that the previous few problems have, but a different page.
Kirchhoff’s Voltage Law (KVL) states that the sum of the voltage drops in a closed circuit is equal to the
algebraic sum of the voltage sources. Combine this with Ohm’s Law which states that voltage is equal to
the product of resistance and current. Scroll to the bottom of page 130 to see how it is done in the text.
Applying KVL to Loop 1 yields
𝑅1 𝐼1 + 𝑅3 𝐼3 = 20 𝑉
(2)
Applying it to Loop 2 yields
𝑅2 𝐼2 + 𝑅3 𝐼3 = 50 V
(3)
To proceed substitute (1), the relationship derived in problem 12, and the values provided for the
resistances, into (2).
10(𝐼3 − 𝐼2 ) + 80𝐼3 = 20
𝐼2 = 9𝐼3 − 2
(4)
Substitute (4), along with the resistances, into (3)
15(9𝐼3 − 2) + 80𝐼3 = 50
𝐼3 =
80
𝐴
215
≈ 0.372𝐴
(5)
Problem 14
Use Ohm's Law and the value of I3 to determine the value of V3, the voltage across the 80 ohm resistor.
This voltage will be used in the next problem.
a. 50.76 V
b. 76.50 V
c. 29.76 V
d. 10.67 V
Link within Knovel
Solution
This link brings a player to a completed Knovel search for “ohm’s law”. The dictionary result states that
Ohm’s Law is the relationship between voltage, E, current, I, and resistance, I=E/R. This relationship is
also shown on page 128 of the previous question’s resource. Returning to the example on page 130 of
that resource:
Apply Ohm’s law to R3.
𝑉3 = 𝐼3 𝑅3
Now substitute the value for I3 obtained in problem 13 and the value of R3 provided in the problem:
𝑉3 =
80
𝐴
215
∗ 80𝛺 ≈ 29.76 𝑉
Problem 15
Use the relationship between I1, I2, and I3 to determine the magnitude of I1.
a. 1.764 A
b. 25 A
c. 0.05 A
d. 0.98 A
Link within Knovel
Solution
In case this pages that these problems were taken from has not been found by the player by this point,
this hint links directly to them.
The relationship between I1, I2, and I3 was determined in problem 12.
I1 = I3 − I2
Substitute (4) from the calculations in problem 13’s solution into this relationship, then simplify
𝐼1 = 𝐼3 − (9𝐼3 − 2)
𝐼1 = 2 − 8𝐼3
Finally substitute the value obtained for I3 in problem 13:
𝐼1 = 2 − 8 ∗
80
𝐴
215
≈ −0.98 𝐴
The negative indicates that the direction of I1 assumed in the diagram and in calculations is incorrect.
Thus I1 is actually flowing from right to left.
Problem 1
Convert 4.5 kg to ounces (oz)
a. 0.1276 oz
c. 57.62 oz
b. 158.7 oz
d. 0.25 oz
Link within Knovel
Solution
This link brings the player to the Knovel homepage. To navigate to the Unit Converter, select Tools on
the left side of the upper banner, then click Unit Converter. A new tab will open with the tool. In the
Input value field type 4.5; in the Input Unit field type “kg”. In the Ouput Unit field type “oz”. Click
Convert. Knovel will return a quantity of 158.7.
Problem 2
Convert -40°F to °C
a. 40°C
c. -40°C
b. -80°C
d. 20°C
Link within Knovel
Solution
This hint brings the player directly to the Knovel Unit Converter. In the Input value field type -40, in the
Input Unit field type degF, in the Output Unit field type degC. Click Convert. Knovel will return -40.00
Problem 3
A change of 1°F is equal to a change in Celsius of, very roughly,
a. -17.22°
c. 0.25°
b. 0.5°
d. 5°
Link within Knovel
Solution
Again the player is brought to the Knovel Unit Converter. Converter two adjacent inters in °F to °C and
compute the difference between the converted values. For example,
20°F= -6.667°C, 21°F= -6.111°C. A rise of 1°F here is equal to a 0.556°C rise.
200°F= 93.33°C, 201°F= 93.89°C. A rise of 1°F here is equal to a 0.56°C rise.
More experimentation will yield similar results, so of the choices 0.5° is the most reasonable. Of course,
this is not suitable proof, but it yields a reasonable result. 1 °F is actually equal to 5/9 °C.
Problem 4
Convert 3*108 m/s to mi/h.
a. 186,000 mph
c. 134,100,000 mph
b. 671,100,000 mph
d. 500,000 mph
Link within Knovel
Solution
Once again the player is brought to the Unit Converter. In the Input value field type “3*10^8”, in the
Input Unit field type “m/s”, and in the Output Unit Field type “mi/h” or “mi/hr”. Click Convert. Knovel
will return the value in choice d, without the commas.
The instructions provided with problems 5 through 8 are “evaluate the following derivatives or limits:”
Problem 5
a. 𝑒 𝑥 sin(𝑒 𝑥 )
c. 𝑒 −𝑥 sin(𝑒 𝑥 )
Link within Knovel
𝑑
[cos(𝑒 −𝑥 )]
𝑑𝑑
b. 𝒆−𝒙 𝐬𝐬𝐬(𝒆−𝒙 )
d. −𝑒 −𝑥 sin(𝑒 −𝑥 )
Solution
The hint provided brings the player to Knovel’s Interactive Equation “The Chain Rule for the Derivative of
a Composite Function”. Open the interactive worksheet by hovering the cursor over Σ=+ open then
selecting “equation worksheet”. To view the worksheet larger open it in its own tab by clicking the
squares on the right side of the blue banner. In the second green box, replace the existing derivative
with the one from the problem. (If the text below the green box is covering the function that Knovel
returns, click in white space outside the green box and hit enter/return on the keyboard a few times).
Knovel will produce this:
1
sin �𝑒 𝑥 �
𝑑
−𝑥 )
(cos(𝑒
=
𝑑𝑑
𝑒𝑥
The equality
1
𝑒𝑥
= 𝑒 −𝑥 is applied, and the expression in choice b is obtained.
Problem 6
2𝑠𝑠𝑠(𝑥) − 𝑠𝑠𝑠(2𝑥)
𝑥→0
𝑥 − 𝑠𝑠𝑠(𝑥)
lim
a. 0
c. 6
Link within Knovel
b. does not exist
d. 2
Solution
The hint provided with this problem brings the player to Knovel’s Interactive Equation “L’Hopital’s Rule
for the Limit of the Ratio of Two Functions”. Open the equation worksheet as was done in the previous
problem. The worksheet states that the L’Hopital’s Rule (L’H) is “for calculating the limit of the ratio of
two functions whose individual limits evaluate simultaneously to zero or infinity”. Indeed the expression
in the problem evaluates to 0/0. Apply L’H by computing the ratio of the derivative of the numerator to
the derivative of the denominator. To compute any of these derivatives, one may use the worksheet
provided in the previous problem’s hint.
2𝑠𝑠𝑠(𝑥) − 𝑠𝑠𝑠(2𝑥) 0 − 0
=
𝑥→0
0−0
𝑥 − 𝑠𝑠𝑠(𝑥)
lim
𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖, 𝑎𝑎𝑎𝑎𝑎 𝐿′𝐻
𝑑
[2𝑠𝑠𝑠(𝑥) − 𝑠𝑠𝑠(2𝑥)] 2 cos(𝑥) − 2cos(2𝑥)
𝑑𝑑
=
𝑑
1 − cos(𝑥)
[𝑥 − 𝑠𝑠𝑠(𝑥)]
𝑑𝑑
2 cos(𝑥) − 2cos(2𝑥) 2 − 2 0
=
=
𝑥→0
1−1 0
1 − cos(𝑥)
lim
𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖, 𝑎𝑎𝑎𝑎𝑎 𝐿′ 𝐻 𝑎𝑎𝑎𝑎𝑎
𝑑
[2 cos(𝑥) − 2 cos(2𝑥)] −2 sin(x) + 4 sin(2x)
𝑑𝑑
=
𝑑
sin(𝑥)
[1 − cos(𝑥)]
𝑑𝑑
lim
−2 sin(x) + 4 sin(2x) 0 + 0
=
sin(𝑥)
0
lim
−2 cos(𝑥) + 8cos(2𝑥) −2 + 8
=
=6
1
cos(𝑥)
𝑥→0
𝑥→0
𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖, 𝑎𝑎𝑎𝑎𝑎 𝐿′ 𝐻 𝑜𝑜𝑜 𝑓𝑓𝑓𝑓𝑓 𝑡𝑡𝑡𝑒
𝑑
[−2 sin(x) + 4 sin(2x)] −2 cos(𝑥) + 8cos(2𝑥)
𝑑𝑑
=
𝑑
cos(𝑥)
[sin(𝑥)]
𝑑𝑑
This means that
2 sin(𝑥) − sin(2𝑥)
=6
lim
𝑥→0
𝑥 − sin(𝑥)
a. 𝑐𝑐𝑐 2(𝑥)
c. 𝑙𝑙(𝑥)
Problem 7
𝑑
tan(𝑥)
𝑑𝑑
b. 𝒔𝒔𝒄𝟐 (𝒙)
d. 1
Link within Knovel
Solution
This hint brings the player to Knovel’s Interactive Equations on Calculus and Differential Equations. In
the filter by keyword search, type “trig” then select “Derivative of Trigonometric Functions”. Open the
equation worksheet. The third identity provided by Knovel is:
𝑑
1
tan(𝑥) =
= (sec(𝑥))2
𝑑𝑑
cos(𝑥)2
The formatting in choice c is slightly different, but it is the same function.
Problem 8
a. −𝟐𝟐 𝐬𝐬𝐬�𝒙𝟐 � 𝐜𝐜 𝐬�𝒄𝒄𝒄�𝒙𝟐 ��
c −𝑠𝑠𝑠(𝑥)𝑐𝑐𝑐(𝑐𝑐𝑐(𝑥))
Link within Knovel
𝑑
sin(cos(x 2 ))
𝑑𝑑
b. −2𝑥 𝑠𝑠𝑠(𝑥 2 )𝑐𝑐𝑐(𝑥 2 )
d. 𝑠𝑠𝑠(𝑐𝑐𝑐(2𝑥))
Solution
This hint brings the player to the same Interactive Equation that the hint in problem 5 did. Open the
equation worksheet. Again, in the second green box, replace the existing expression with the one in the
problem. Knovel will return
𝑑
(sin(cos(𝑥 2 ))) = −2 · 𝑥 · sin(𝑥 2 ) · cos(cos(𝑥 2 ))
𝑑𝑑
The expression on the right side of this equality is the same as the one given in choice a.
Problem 9
What is the electron configuration of silicon?
a. [Ne]3s23p2
c. [Ne]3s23d10
b. [Ne]3s23d2
d. [Ne]3s1
Link within Knovel
Solution
This hint brings the player to Knovel’s Periodic Table. Click the element silicon (atomic symbol Si, atomic
number 14). The new tab which opens contains some key information about the element, among which
is its electron configuration. A screenshot is provided here:
Problem 10
Which element, in its ground state, has the electron configuration [Ar]4s23d104p1?
a. germanium
b. gallium
c. zinc
d. aluminum
Link within Knovel
Solution
This link also brings the player to Knovel’s Periodic Table. The element with the given configuration is in
the period (row) after argon, thus the fourth period. It has a full 4s subshell (2 electrons), a full 3d
subshell (10 electrons), and 1 electron in its 4p subshell. Thus it is the first element in the p-block of the
fourth period. That element is gallium (atomic symbol Ga, atomic number 31). Click it to view key
information and verify.
Note: the 4s subshell is listed (and fills) before the 3d subshell because the former is at a slightly lower
energy level. The 4p subshell is higher in energy than either of those two, so it is listed (and fills) after
they are both filled.
Problem 11
What is the electron configuration of Ga+?
a. [Ar] 4s23d10
c. [Ar]4s2
b. [Ar] 4s23d104p1
d. [Ar]4s23d94p1
Link within Knovel
Solution
This hint is identical to the last two. In problem 10 the electron configuration of a neutral gallium atom
in its ground state was found. The cation in this problem has one electron less than the stable atom, and
that one electron is necessarily the one that was at the highest energy level, the one in the 4p subshell.
Thus the configuration of this ion is identical to the neutral atom, without the last electron.
Problem 12
A Ni2+ ion has the same number of electrons as a neutral atom of which element?
a. Zn
b. Ba
c. Kr
d. Fe
Link within Knovel
Solution
This hint also brings the player to Knovel’s Periodic Table. The cation in this problem has two electrons
less than a neutral nickel atom. A neutral nickel atom (atomic number 28) has 28 protons and therefore
28 electrons. This ion has 26 electrons. A neutral element with 26 electrons also has 26 protons. That
element is iron (atomic symbol Fe, atomic number 26).
Problem 13
What is the unique temperature and pressure combination at which water exists as solid, liquid, and gas
in equilibrium?
a. 273.16 K, 101.325 kPa
b. 273.16 K, 611.73 Pa
c. 298.15 K, 611.73 Pa
d. 298.15 K, 611.73 Pa
Link within Knovel
Solution
This hint tells the player “This is called the triple point of a material”. The link brings the player to the
Knovel Data Search page. In the Material or Substance Name field type “water”. In the Find a property
field type “triple point”. Drag and drop “triple point pressure” into the query builder. Only one result
exists. Go to the result, then open it. The table contains two rows, but the first is for the material
deuterium oxide, a form of water with an uncommon isotope of hydrogen. The second row clearly
states the material is water. Scroll to the right to find the triple point temperature and triple point
pressure. The values are 273.16 K and 611.73 Pa, respectively.
Problem 1
What is the molar mass of glucose, C6H12O6?
a. 180 g/mol
c. 160 g/mol
b. 144 g/mol
d. 360 g/mol
Link within Knovel
Solution
Included with this link is the hint “Try ‘Tools’ at the top pf the page”. The link brings the player to the
Knovel homepage and in the Tools dropdown at the top, one may find a link to the Periodic Table. It
shows that the atomic mass of a carbon atom is 12.01 atomic mass units (u), so the molar mass is 12.01
grams per mole. There are 6 carbon atoms in 1 molecule of glucose, so carbon contributes
6*12.01=72.06 u, or 72.06 grams per mole. Performing this for all of the elements that comprise the
molecule:
Carbon:
6*12.01 g/mol
Hydrogen:
12*1.008 g/mol
Oxygen:
6*16.00 g/mol
180.156 g/mol
Problem 2
What is the molar mass of water, H2O?
a. 33 g/mol
b. 44 /mol
b. 18 g/mol
d. 50 g/mol
Link within Knovel
Solution
This link brings the player directly to the Knovel Periodic Table. Using the atomic masses of hydrogen
and oxygen provided, the molar mass of water is found to be just greater than 18 g/mol:
Hydrogen:
2*1.008 g/mol
Oxygen:
1*16.00 g/mol
18.016 g/mol
Problem 3
6CO2 + 6H2O (+light)  C6H12O6 + 6O2
Given this balanced chemical equation describing photosynthesis, and an excess of CO2, how much
water is needed to produce 100 grams of glucose?
a. 10 g
b. 60 g
c. 3.33 g
d. 10.8 kg
Link within Knovel
Solution
This link brings the player to a page of a resource with an example that is almost identical to the
problem at hand. The pertinent pieces of information to extract from the provided chemical equation
are the coefficients of water and glucose, 6 and 1 respectively. These coefficients tell the reader that to
produce 1 mole of glucose 6 moles of water are needed (provided there is enough CO2 to complete the
reaction). Now following the example provided in the text and using the molar masses calculated in
problems 1 and 2:
𝑚𝑜𝑙 𝑜𝑓 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 6 𝑚𝑜𝑙 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
18 𝑔
100 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 ∗
∗
∗
= 60 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
180 𝑔
1 𝑚𝑜𝑙 𝑜𝑓 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 1 𝑚𝑜𝑙 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
Problem 4
The Maclaurin series of sin(x2) expanded to four terms is
𝑥6
𝑥 10
𝑥 14
−
+
3!
5!
7!
𝑥3
𝑥5
𝑥7
− 3! + 5! − 7!
𝒙𝟔
𝒙𝟏𝟎
𝒙𝟏𝟒
−
𝟓!
𝟕!
𝑥5
𝑥7
𝑥9
2
𝑥 − 3! + 5! − 7!
a. −𝑥 2 +
b. 𝒙𝟐 − 𝟑! +
c. 𝑥
d.
Link within Knovel
Solution
This hint brings the player to a page of a resource that provides Maclaurin series of some key functions
expanded to a few terms. The expansion of sin(x) is provided to four terms as
sin(𝑥) = 𝑥 −
𝑥3 𝑥5 𝑥7
+ −
3! 5! 7!
and the domain is all x. Replacing x above with x2, and making use of the equality (xa)b=xab, one obtains
the expression provided in choice b.
Problem 5
The sum of the first five terms of the Maclaurin expansion of e2 is
a. 7.389
b. 7
c. 5
d. 6.333
Link within Knovel
Solution
This hint brings the player to the same page as the previous one. It provides the Maclaurin expansion of
ex to five terms as:
𝑥2 𝑥3 𝑥4
𝑒𝑥 = 1 + 𝑥 + + + + ⋯
2! 3! 4!
for all x. In this particular problem, x=2. Substituting in the value of x and evaluating the factorials in the
denominators, one obtains:
4 8 16
𝑒2 = 1 + 2 + + +
+⋯
2 6 24
𝑒2 = 7 + ⋯
The sum of the first five terms is simply 7.
Problem 6
The difference between the true value of e3, and the sum of the first five terms of the Maclaurin
expansion is approximately
a. 0
b. 3.71
c. 2.718
d. 𝜋
Link within Knovel
Solution
The true value of e3 is approximately 20.0855. The link provided brings the player to the same page that
the hint in the previous two problems did. It shows that the expansion of ex, to five terms, is
1+𝑥+
𝑥2 𝑥3 𝑥4
+ + +⋯
2! 3! 4!
Substituting in x=3 and evaluating, one obtains:
9 27 81
1+3+ +
+
+⋯
2 6 24
16.375 + ⋯
So the difference between the true value and the sum of the first five terms of the expansion is
approximately 3.71
𝑒 3 − 16.375 ≈ 3.71
Included with problems 7 through 9 are the instructions: “Consider the function f(x) = -2x6+x5-x4+x3+5x2 ”
Problem 7
What are the roots?
a. -1, 0, 1.3711
c. 1, 0, 1.3711, 2.7422, 5, 7.142
b. 0, 1.3711, 2.7422
d. -1, 1, 1.3711, 5, 7.142, 8
Watch the Functions Palette (Part II) How-to video here:
Link within Knovel
The link included with this problem brings the player to a video that will walk one through a process to
answer questions 7 through 9. A problem similar to this one is considered at approximately 3:51 in the
video. In the Knovel Interactive Equations homepage, click the blue button on the right to open a new
interactive worksheet. Open it in its own tab if desired. Click anywhere in the whitespace then begin
typing what is italicized here.
f(x [space] [colon] -2x^6 [space] + x^5 [space] –x^4 [space] + x^3 [space] + 5x^2 [enter]
This will define the function so that it may be called. Notice that a second parenthesis is not typed.
In another whitespace area type
p [colon] solve (f(x [space] [comma] x [enter]
p=
Knovel will return a vector containing the three roots,
−1
p=( 0 )
1.3711
Problem 8
What is the area below the curve, for which f(x)>0?
a. 1.1173
c. 9.81
b. 6.67
d. 3.4793
Link within Knovel
Solution
This hint is identical to the last one, and the problem is referring to the same function used in problem 7,
which is also defined in the instructions provided with this problem. The solution to a problem very
similar to this one is addressed in the video immediately following the portion that addressed the
subject of the previous problem. Continue working in the same interactive worksheet.
With p defined as the vector with the three roots of the function, in any whitespace type
a [colon] int(f(x [space] [comma] x [comma] p [left bracket] 1 [space] [comma] p [left bracket] 3 [space]
[space] =
Notice how Knovel interprets the keystrokes, as indicated by the emerging integrand and movement of
the cursor. Knovel will return 3.4793.
To see what this looks like click any whitespace and type @. In the area under the graph that appears
type f(x [space] [enter]. Knovel will produce a graph that shows the function. In more familiar symbols,
Knovel just evaluated the following definite integral:
1.3711
𝑎= ∫
𝑓(𝑥) 𝑑𝑥
−1
Alternatively, one could have used Knovel to compute the sum of each of the areas on either side of the
y axis:
0
1.3711
𝑎 = ∫ 𝑓(𝑥)𝑑𝑥 + ∫
−1
𝑓(𝑥) 𝑑𝑥
0
but the former approach is quicker and just as accurate.
Problem 9
What is the slope of the tangent at x=6?
a. 0
c. -87,528
b. 87,528
d. 14
Link within Knovel
Solution
This hint is identical to the previous two. The video begins a problem nearly identical to this one at
approximately 0:33. Continue working with the same interactive worksheet that has already been
created. Click in any whitespace and type
s(x [space] [colon] [derivative] f(x [right arrow] [right arrow] x [right arrow] [enter]
In place of [derivative] click the derivative symbol in the functions palette on the right.
s(6 [space] =
Knovel will return -87,528.
Included with problems 10 through 13 is the following introduction:
An uninsulated pipe, with an outer diameter of 6 inches, and a surface temperature of 300°F passes
through a room in which the air and the walls are at 77°F. The emissivity of the pipe is 0.7, and the heat
𝐵𝑇𝑈
transfer coefficient associated with this flow is 3ℎ𝑟 𝑓𝑡 2 °𝑅𝑎 .
Problem 10
Does the pipe behave as a blackbody?
a. yes
c. impossible to determine
b. no
d. only in this environment
Link within Knovel
Solution
This hint brings the player to a list of Interactive Equations on thermodynamics. From this list, one can
search for “black”. This will return two equations. Upon examining the second, Non-Blackbody Emissive
Power one will read in the description “If a surface is not black, the emissive power of the surface
depends on its emissivity which is a surface characteristic whose value ranges from 0 to 1, with the
emissivity of a blackbody being 1.” A value of 0.8 implies not a blackbody, and such is called a gray body.
A blackbody is an idealized body that radiates energy at the maximum possible rate for a given
temperature and absorbs all radiation that falls upon it. It is used an ideal approximation for many real
scenarios.
Problem 11
What is the emissive power of the surface?
a. 99.4 BTU/(hr*ft2)
c. 571 BTU/(hr*ft2)
b. 400 BTU/(hr*ft2)
d. 2.71828 BTU/(hr*ft2)
Link within Knovel
This hint brings the player directly to the Knovel Interactive Equation Non-Blackbody Emissive Power,
the same one that was used to answer the previous problem. Open the equation worksheet. Enter an
emissivity (𝜀) of 0.7 and press enter. The Stefan-Boltzmann constant is constant, so do not change its
value or units. Enter a surface temperature of 300°F and allow the interactive worksheet to convert to
°Ra(not shown, only reflected in the answer), or use the Knovel Unit Converter to convert. (To convert
from °F to °Ra, simply add 459.7, or 460) To enter the temperature in °F, dele the F and type Ra. Tap
enter on the keyboard. Knovel will return
𝐵𝑇𝑈 1
𝐸 = 399.12
·
ℎ𝑟 𝑓𝑡 2
The most reasonable answer of those provided is choice c.
Problem 12
What is the rate of energy flow, per unit length, from the surface of the pipe to the air?
a. 131.36 BTU/hr
b. 334.5 BTU/hr
c. 1050.86 BTU/hr
d. 78.05 BTU/hr
Link within Knovel
Solution
This hint brings the player to Knovel’s Interactive Equation Convection. From the description, convection
is the method of energy transfer “between a solid surface and a moving fluid”. Open the interactive
worksheet. Replace the default value of the heat transfer coefficient, h, with 3 as given in the
introduction; the units are the same. Tap enter or return on the keyboard. The surface area of the pipe
with circular cross section and open on both ends is equal to the circumference of the cross section
multiplied by the length of the pipe. Because the problem asks for the rate of energy flow per unit
length, only the circumference needs to be considered.
𝐶 = 2𝜋𝑟 = 𝜋𝐷 = 0.5 𝑓𝑡 ∗ 𝜋. The area may be entered as 0.5 𝜋 with units of ft2.
(If the length of the pipe were given, one would multiply the circumference of the pipe by it, complete
the process to obtain a value for the energy transfer, then divide this by the length of the pipe, to find
the value per unit length. In the end, the length will cancel out, so it is not included in the calculations)
The surface temperature, Ts , is 300°F. Delete the Ra and type F, or keep the units of °Ra and convert the
300°F to 759.7°Ra (simply add 459.7, or 460). The fluid temperature, 𝑇∞ , is 77°F or 536.7°Ra. Knovel
returns
qconv = 1050.86
BTU
.
hr
Problem 13
What is the rate of energy transfer, per unit length, from the surface of the pipe to the walls of the
room?
a. 15.18 BTU/hr
b. 20.71 BTU/hr
c. 470 BTU/hr
d. 1.48 BTU/hr
Link within Knovel
Solution
This hint brings the player to the Knovel Interactive Equation Net Radiant Heat Transfer. Because the
problem asks for the rate of energy transfer between two surfaces, radiation is the method to consider.
Open the interactive worksheet. The Stefan-Boltzmann constant is a fundamental physical constant, so
it is unchanged. The gray surface area is that of the pipe (per unit length), or 0.5 ∗ 𝜋 ft2. (To enter 𝜋, use
the Lowercase Greek palette on the right.) ε is the emittance, or emissivity, of the gray surface and is
given in the introductory statement as 0.7. T1 is the temperature of the pipe, 300°F or 759.7°F. T2 is the
temperature of the walls, 77°F or 536.7°F. Knovel will return
qr = 470.84
BTU
.
hr
As mentioned in the solution to problem 10 a true blackbody does not exist, and Knovel’s reference to
T2 as the “surrounding black surface temperature” is an approximation. This simply means that the wall
does not absorb all of the energy radiated to it by the pipe; it does not affect the rate at which the pipe
radiates it.
Included with problems 1 and 2 is the following information and accompanying diagram.
A rod measuring 20 mm in diameter is subjected to a tension force of 150 kN. This increases its original
length of 100 mm by 0.2387 mm.
L0=100 mm
𝜹L=0.2387 mm
Problem 1
What is Young’s modulus of the unknown material? This value will be used in the next problem.
a. 2.00*1011 Pa
c. 500 Pa
6
b. 10.00*10 Pa
d. 5.00*109 Pa
Link within Knovel
Solution
This link brings the player to Knovel’s list of Interactive Equations which fall under the heading “Stress
and Strain”. In the filter by keyword field type “young” and open the equation Modulus of Elasticity or
Young’s Modulus. Open the equation worksheet. In the area under calculation, type in the appropriate
values for each variable, and change units by deleting the US customary ones and typing the SI ones
given above.
P:=150 kN
l:=100 mm
A:=100·𝜋 mm² (To enter pi use the Lowercase Greek palette on the right)
𝛿l:=0.2387 mm
Be aware when typing units that Knovel is case sensitive. After all of this information is entered Knovel
will return
E = 2.9011 · 107 psi
Delete the units in the answer and type Pa for pascals. Knovel will then change the value to
E = 2.0003 · 1011 Pa
The values and units were converted to SI units by Knovel and substituted into the equation. This is now
shown
P=150kN=150,000N
l=100mm=0.1 m
r=10mm=0.01 m
𝛿𝛿=0.2387mm=2.387*10-4 m
(150,000 𝑁) · (0.1 𝑚)
𝑃·𝑙
=
≈ 2.00 · 1011 𝑃𝑃
𝐸=
𝐴 · 𝛿𝛿 𝜋 · (0.01𝑚)2 · (2.387 · 10−4 𝑚)
Problem 2
If the stretching that occurs is entirely in the elastic regime of the material and a digital gauge reports a
strain of 0.2387%, what is the stress at that point?
a. 477 MPa
b. 4.77*1010 Pa
c. 875.3 kPa
d. 1.2 kPa
Link within Knovel
Solution
This link brings the player to a page on the subject mechanics of materials. Scroll up one page to the
beginning of the section to find the sentence “Within the elastic limits of the material, i.e. within the
limits in which Hooke's law applies, it has been shown that
stress
= constant
strain
This constant is given the symbol E and termed the modulus of elasticity or Young’s modulus.” Thus
E=
stress σ
=
strain ε
Rearranging for stress yields
stress=Young’s modulus * strain, or σ = Eε.
Substituting the value of E obtained in the previous problem, and the value of ε given in this problem:
0.2387
� ≈ 477.47 · 108 𝑃𝑃 ≈ 477 𝑀𝑀𝑀
𝜎 = (2.0003 · 1011 𝑃𝑃) �
100
Problem 3
Light with a wavelength of 200 nm strikes cesium metal. How much energy is contained in one photon
of this light? This will be used in a subsequent problem.
a. 1.33*10-40 J
b. 9.94*10-37 J
c. 1500 J
d. 9.94*10-19 J
Link within Knovel
Solution
The text provides the relationship between light’s wavelength and its energy.
𝐸 = ℎ𝜈
The paragraph immediately above that states 𝜈 = 𝑐/𝜆. E is the energy, h is Planck’s constant
(0.66252*10-34 J·s), 𝜈 is the frequency of the light, c is the speed of light in a vacuum (3·108 m/s), and 𝜆
is the wavelength. Substituting in the wavelength given (in meters) in the question one obtains an
energy of 9.938*10-19 J.
𝑚
3 · 108
𝑐
𝑠 ≈ 9.9378 · 10−19 𝐽
−33
𝐸 = ℎ = (0.66252 · 10 𝐽 · 𝑠)
200 · 10−9 𝑚
𝜆
Problem 4
If the energy needed to remove an electron from the surface of cesium is 2.1 eV, what is it in joules? Use
this in the next problem.
a. 3.14159 J
b. 2.71828 J
c. 3.365 *10-19 J
d. 6.022*1023 J
Link within Knovel
Solution
Use the Knovel Unit Converter. Be aware when entering units, that Knovel is case sensitive. In the Input
value field type 2.1, in the Input Unit field type “eV” (electron-volts). In the Output Unit field type “J”.
Click Convert. Knovel will return 3.365*10-19.
Problem 5
What is the kinetic energy of an electron leaving the surface of the cesium? Use this in the next
problem.
a. 6.58*10-19 J
b. 4.099*10-14 J
c. 4.68*10-25 J
d. 6.626*10-23 J
Link within Knovel
Solution
This link brings a player to a different page of the same resource that is linked to in problem 3. The page
contains the Einstein photoelectric equation:
1
ℎ𝜈 = 𝐸1 + 𝑚𝑣 2
2
where h𝜈 is the energy contained in one photon of the light, E1 is the minimum amount of energy a
photon must have to eject an electron from the surface of the cesium, and
1
𝑚𝑣 2 ,
2
is the familiar
expression from Newtonian mechanics for kinetic energy (K). That minimum energy, 𝐸1 , is called the
work function and it is a material property unique to each metal. Replacing
rearranging yields
1
𝑚𝑣 2
2
with K and
𝐾 = ℎ𝜈 − 𝐸1
In words this equation states that the kinetic energy of an electron ejected from the metal is equal to
the difference between the energy of the photon that strikes it and the work function of the metal. The
energy of the incident photon was found in problem 3 and the work function of cesium was converted
to joules in problem 4 therefore
𝐾 = 9.94 ∗ 10−19 𝐽 − 3.365 ∗ 10−19 𝐽 ≈ 6.58 ∗ 10−19 𝐽
Problem 6
With what velocity will the electrons leave the surface of the cesium?
a. 1.2 *106 m/s
c. 343 m/s
b. 2.998*106 m/s
d. 6.022*1023 m/s
Link within Knovel
Solution
This link brings the player to a list of fundamental physical constants, found within the Knovel Sampler.
The constant that will be used in this problem is the electron mass, which is provided to a large number
of significant figures. It is often approximated as 9.109*10-31 kg. Now using the familiar equation for
1
2
kinetic energy, 𝐾 = 𝑚𝑣 2 , and the value obtained in the previous problem, one can solve for the
velocity of the electron.
1
𝐾 = 𝑚𝑣 2
2
2𝐾
(2) ∗ (6.58 ∗ 10−19 𝐽)
𝑚
𝑣=�
=�
≈ 1.2 ∗ 106
−31
𝑚
9.109 ∗ 10 𝑘𝑘
𝑠
That’s more than 2,600,000 mph!
Included with problems 7 and 8 is the following sentence: “A beam supported only on its left end is
subjected to a uniformly distributed load of 2000 lb/ft from above, across its entire 10 foot length”.
Problem 7
Determine the magnitude of the bending moment at the midpoint of the beam.
a. 500 lb*ft
b. 25,000 lb*ft
c. 1000 lb*ft
d. 2000 lb*ft
Link within Knovel
This hint brings the player to Knovel’s list of Interactive Equations on Strength of Materials. A beam that
is supported only on one end is called a cantilever, so in the filter by keyword field type cantilever. Select
the equation Bending Moment at Any Section in a Cantilever Beam Carrying a Uniformly Distributed
Load. One may use the equation and description of variables provided to obtain the answer by hand, or
use the interactive features to answer the question. To do the latter, open the equation worksheet, and
in the area under Calculation replace all of the default values for the variables with the ones given in the
problem. Change the units as well and tap enter or return on the keyboard after each entry. Pounds are
entered as lbf to differentiate pounds-force from pounds-mass, lbm. Because z is measured from the left
end, and it is the point at which the moment is being calculated, for this problem it is 5 feet.
w:=2000 lbf/ft
L:=10 ft
z:= 5 ft
Knovel will return
M = −3 ∗ 105 lbf in
Change the inches to feet (in to ft) and Knovel will now show
M = −25000 lbf ft
The question only asks for the magnitude of the bending moment; the negative sign indicates hogging
moment (curving upwards in the middle).
Problem 8
Where is the moment equal to -4000lb*ft. Use the convention that clockwise rotation is negative, as
suggested in the appropriate equation.
a. 8 ft from right end
b. 2 ft from right end
c. 6 ft from right end
d. 4 ft from right end
Link within Knovel
Solution
This link brings the player to Knovel’s interactive equation Bending Moment at Any Section in a
Cantilever Beam Carrying a Uniformly Distributed Load. The interactive features of this equation will
only solve for M, the actual bending moment. To solve this problem one must manually rearrange the
provided equation to solve for z, the distance of the section (from fixed end) at which moment is to be
determined.
1
If 𝑀 = − · 𝑤 · (𝐿 − 𝑧)2 , then
2
−2 · 𝑀
𝑧 =𝐿−�
𝑤
From the introduction L=10 ft, w=2000 lb/ft and from this problem M=-4000lb*ft. Substituting all of this
information into the equation:
𝑧 = 10𝑓𝑓 − �
(−2) · (−4000 𝑙𝑙 ∗ 𝑓𝑓)
= 8 𝑓𝑓
𝑙𝑙
2000
𝑓𝑓
However z is measured from the fixed end of the beam, in this case the left end. 8 feet from the left end
of a 10 foot beam is 2 feet from the right end.
The following diagram is provided with problems 9 and 10
a
30°
100°
c
67
A
Problem 9
What are the missing side lengths? Use these in the next problem.
a. a=34.017, c=52.12
b. a=52.12, c=34.02
c. a=72.16, c=32.45
d. a=105.2, c=37.62
Link within Knovel
Solution
This hint brings the player to a basic Knovel search for “triangles”, with the Equations facet on the left
chosen. Select Law of Sines for Triangles. Open the interactive worksheet. Scroll to example 2 which is
an example for solving a triangle given one side and two angles. The first thing to deduce is that the
measure of angle A is 50° (the interior angles of any planar triangle must sum to 180°). Now notice that
in the Knovel diagram the angle labeled 𝛾, is opposite the side labeled c, so for this particular triangle
the measure of angle gamma is 30°, ∡𝛾=30°. In the diagram, angle 𝛼 (lowercase Greek letter ‘alpha’) is
opposite side a, so ∡𝛼=50°. Finally b is the only known side length, 67. Substitute these values into
example 2:
𝑏 ≔ 67
α ≔ 50 deg
γ ≔ 30 deg
Knovel will return
β = 100 deg (this was already known)
a = 52.1167 deg
c = 34.0168
Problem 10
What is the area of the triangle?
a. 1000 square units
c. 873.12 square units
b. 947.76 square unis
d. 1570.6 square units
Link within Knovel
Solution
This hint brings the player to Knovel’s Interactive Worksheet Area of a Triangle Given Three Sides. Now
that the lengths of all three sides are known from the previous problem, one may use this equation. The
interactive features, or a calculator, may be used
𝑎 + 𝑏 + 𝑐 52.12 + 67 + 34.02
𝑠=
=
= 76.57
2
2
A
= �𝑠 · (𝑠 − 𝑎) · (𝑠 − 𝑏) · (𝑠 − 𝑐)
= �76.57 · (76.57 − 52.12) · (76.57 − 67) · (76.57 − 34.02)
≈ 873.12 square units
Problem 11
Suppose that a new planet in our solar system in discovered. Its orbital distance is 35 times greater than
Earth’s. Approximately how many Earth years will it take for the planet to make one orbit around the
sun?
a. 0.02
b. 207
c. 6
c. 42,875
Link within Knovel
Solution
This hint brings the player to a page that contains Kepler’s three laws of planetary motion. Notice
through the “trail of breadcrumbs” that it was found from a search for “kepler’s laws”. The third law will
be used to answer this question. It is stated as “The squares of the periods of revolution of the planets
about the Sun are proportional to the cubes of their mean distances from it”. This means, for every
planet, that the square of the time to complete one orbit around the sun divided by the cube of the
planet’s average distance from the sun is equal to a constant. That constant is the same for every planet
in our solar system.
Let TE be Earth’s period, or 1 year, and let dE be the Earth’s mean distance from the sun. The new
planet’s mean distance is equal to 35dE, and its period will be represented by Tnew.
2
𝑇𝑛𝑛𝑛
𝑇𝐸2
=
𝑑𝐸3 (35𝑑𝐸 )3
Rearranging:
𝑇𝐸2 · (35𝑑𝐸 )3
= �1 · 353 ≈ 207.06 Earth years
𝑇𝑛𝑛𝑛 = �
𝑑𝐸3
Problem 12
A planet takes 1.63 Earth years to travel from one point in its orbit to another, and the area swept
during that movement is 71% of the total area covered during its orbit. What is this planet’s period in
Earth years?
a. 2.3
b. 20.1
c. 3.2
d. 10.5
Link within Knovel
Solution
This hint sends the player to the same page that the previous one did. However, this time the pertinent
law is Kepler’s second. This law states “Each planet moves in such a way that the (imaginary) line joining
it to the sun sweeps out equal areas in equal times”.
A
71%
29%
B
The diagram above is provided to help illustrate this. The planet takes 1.63 Earth years to travel
counterclockwise from point A to point B. The area swept during that time by an imaginary line
connecting it to the sun covers 71% of the total area of the ellipse. Kepler’s second law can be used to
determine the amount of time the planet will take to travel counterclockwise from point B back to point
A, and cover the remaining 29% of the area. The ratio of the time elapsed sweeping an area to the area
swept is constant, so
1.63 𝑦𝑦𝑦𝑦𝑦 𝑥 𝑦𝑦𝑦𝑦𝑦
=
0.29
0.71
0.71
≈ 0.66577 𝐸𝐸𝐸𝐸ℎ 𝑦𝑦𝑦𝑦𝑦
0.29
This is the time required to complete the remaining portion of the orbit. The total period of the planet is
therefore the sum of this time and the time to complete the first portion
𝑥 = 1.63 ·
1.63 𝑦𝑦𝑦𝑦𝑦 + 0.66577 𝑦𝑒𝑒𝑒𝑒 ≈ 2.3 𝑦𝑦𝑦𝑦𝑦
Those are Earth years, of course
Problem 13
How many valence electrons does a neutral sodium atom have in its ground state?
a. 10
b. 14
c. 1
d. 8
Link within Knovel
Solution
This link brings the player to Knovel’s Periodic Table. Click on sodium (atomic symbol Na, atomic number
11). Its electron configuration is given as [Ne]3s1. The highest energy shell is the third, and there is one
electron in this shell (in the s subshell).
Problem 14
The number of protons in a potassium atom is:
a. 4
c. 20
b. 19
d. 37
Link within Knovel
This link also brings the player to the Periodic Table. Potassium is represented by the symbol K (first
element in the fourth row). Its atomic number is shown to be 19, and this number is dictated by the
number of protons in the atom.
Problem 15
What is the lightest noble gas?
a. neon
c. xenon
b. oxygen
d. Helium
Link within Knovel
Again the player is brought to the Periodic Table. The legend below the table shows that noble gasses
are in the rigthmost column. The atomic weight of each element is shown below its symbol, and the
lightest one is He (helium) with a weight of 4.003 u.
Problem 16
What is the only common metal that is a liquid at normal room temperature?
a. K
b. Na
c. Hg
d. Co
Link within Knovel
Solution
Click the symbol for each of the elements presented in the multiple choice options. Upon reading the
description for Hg (mercury), one will encouner the sentence “Mercury is the only common metal liquid
at ordinary temperatures”.
Problem 17
The most abundant element in the universe is
a. He
c. O
b. H
d. N
Link within Knovel
Solution
Once again, click on the symbol of each of these elements and review the descriptions. Hydrogen’s
includes the sentence “Named by Lavoisier, hydrogen is the most abundant of all elements in the
universe”.
Problem 1
What is the rest mass of a photon?
a. 0 kg
c. 9.109*10-31 kg
Link within Knovel
b. 1.9*10-30 kg
d. 5.17*10-27 kg
Solution
This link brings the player to a page of a dictionary with an entry for photon. It begins “a particle with
zero rest mass consisting of…” This is to say to that a photon is never at rest, or equivalently light is
always moving.
Problem 2
The Nobel Prize in Physics 2014 was awarded “for the invention of efficient blue light-emitting diodes
which has enabled bright and energy-saving white light sources”. Which of the following is a reasonable
value for the wavelength of the light from the LED?
a. 500 nm
b. 475 nm
c. 390 nm
d. 700 nm
Link within Knovel
Solution
The page linked to includes a table which shows approximate ranges of wavelengths for light of each
color within the visible spectrum. Light with wavelengths between 450 and 490 nanometers is called
blue. The only choice which falls within this range is 475 nm.
Problem 3
With a wavelength of 475 nm, what is the frequency of the light?
a. 0.00063 Hz
b. 6.316*1011 kHz
c. 6.316*1017 Hz
d. 142.5 Hz
Link within Knovel
Solution
The page linked to contains a section titled “Frequency and Wavelength”. This section provides the
equation relating the frequency of light waves to their frequency: 𝑐 = 𝑓𝜆. c is the speed of light in free
space, f is the frequency, and 𝜆 is the wavelength. Rearranging and solving:
𝑐
3 · 108 𝑚/𝑠
𝑓= =
≈ 6.316 · 1014 𝐻𝑧 = 6.316 · 1011 𝑘𝐻𝑧
𝜆 475 · 10−9 𝑚
Problem 4
What is the rest mass of an electron? You will need this value for subsequent problems.
a. 9.109*10-31 kg
b. 11*10-31 kg
c. 8.23*10-10 kg
c. 1.67*10-27 kg
Link within Knovel
Solution
This page of fundamental physical constants includes electron mass. It is provided to several significant
figures, but is commonly rounded to 9.109·10-31 kg.
Problem 5
According to Newtonian mechanics the kinetic energy of an electron, traveling at 95% the speed that
light does in a vacuum, is:
a. 7.4*10-14 J
b. 3.699*10-14 J
c. 1.298*10-22 J
d. 4.57*10-8 J
Link within Knovel
Solution
This link brings the player to Knovel’s Interactive Equation Kinetic Energy. One may use the embedded
math engine to solve this. Open the equation solver and under the section “Legend with variables and
units”, enter the information provided in the problem. Alternatively, the solution may be obtained with
a calculator. As a reminder the speed of light in a vacuum is 3·108 m/s. Therefore:
𝑚 2
𝐾𝐸 = 0.5𝑚𝑣 2 = (0.5)(9.109 · 10−31 𝑘𝑔) [(0.95) (3 · 108 )] ≈ 3.699 · 10−14 𝐽
𝑠
Problem 6
The true kinetic energy of an electron traveling at 95% the speed that light does in a vacuum is
a. 7.588*10-13 J
b. 2.625*10-13 J
c. 8.76*10-14 J
d. 1.806*10-13 J
Link within Knovel
Solution
The page linked to in this hint provides information on the relativistic correction. Newton’s equation,
used in the previous problem, predicted that if a natural speed limit is imposed then there is also a
maximum kinetic energy that an object of a given mass can have. Einstein famously showed that this is
not the case, that in fact kinetic energy approaches infinity as an object’s velocity approaches c. He
introduced a new equation to determine kinetic energy of objects that are traveling at significant
fractions of c. That equation is provided on the page in Knovel, and reproduced here:
1
𝐾 = 𝑚0 𝑐 2 (
− 1)
√1 − 𝛽 2
where K is the kinetic energy, m0 is the rest mass of an object, and 𝛽 is the ratio of an object’s speed, v,
to c (v/c). For the electron in this problem 𝛽 =
0.95 𝑐
,
𝑐
or simply 0.95. Now substitute the rest mass of the
electron from problem 4, along with 𝛽 and c:
𝑚 2
1
) (
− 1) ≈ 1.806 · 10−13 𝐽
2
𝑠
√1 − 0.95
This is almost five times greater than the kinetic energy that Newton’s equation gave.
𝐾 = (9.109 · 10−31 𝑘𝑔) (3 · 108
Problem 7
What is the effective inertial mass of an electron traveling at 95% the speed that light does in a vacuum?
a. 9.109*10-31 kg
b. 2.917*10-30 kg
c. 6.67*10-30 kg
d. 9.34*10-30 kg
Link within Knovel
Solution
This link brings the player to the same page that the previous one did. Einstein also showed that if an
object is traveling at a significant fraction of c, then its mass must also be reconsidered. His equation for
effective inertial mass, m, is
𝑚0
𝑚=
√1 − 𝛽 2
where again, m0 is the rest mass of the object, and 𝛽 is the ratio of the object’s speed to c. Substituting
the information pertaining to the electron at hand:
9.109 · 10−31 𝑘𝑔
𝑚=
≈ 2.917 · 10−30 𝑘𝑔
√1 − 0.952
This is more than three times greater than the electron’s rest mass.
Problem 8
The difference between the color of light emitted from a display and the color observed after the light
has passed through a thin touch screen is called
a. interference
b. chromaticity shift
c. relativity
d. modulation
Link within Knovel
Solution
This link brings the player to an article titled Ways to Avoid Chromaticity Shift on Touch Screens. Taken
verbatim from the second paragraph under the heading Compatibility Issues and Chromaticity Shift
“Color or chromaticity shifts can occur when the output light wavelengths of the display are interfered
with by the reflection of the front and back edges within the thin films of the touchscreen. As a result,
the output color from the combined system has changed or shifted from the original color in the
display.”
Problem 9
A radar system detects a target as echoes of radio waves reflected off of it reach a receiver. The strength
of the echo is proportional to the target’s range to the power of
a. 4
b. -4
c. 2
d. -2
Link within Knovel
Solution
The hint brings the player to a pertinent section of a book in Knovel. At the bottom of the first column is
the sentence “The strength of a target’s echoes is inversely proportional to the target’s range to the
fourth power (I/R4)”. Being inversely proportional to the fourth power is equivalent to being
proportional to the negative fourth power.
Problem 10
What is the molar mass of acenaphthene, C12H10? Use this in the next problem.
a. 154 g/mol
b. 132 g/mol
c. 178 g/mol
d. 144 g/mol
Link within Knovel
Solution
The link provided brings the player to the Knovel homepage. However along with the link is the hint “Try
Tools!” From the Tools menu on the homepage, one can go to the Periodic Table. Using the atomic
masses of both carbon and hydrogen, one can obtain the molar mass of acenaphthene.
Carbon:
12*12.01 g/mol
Hydrogen:
10*1.008 g/mol
154.2 g/mol
Problem 11
How much energy must be delivered to 308 grams of acenaphthene to raise its temperature from 900K
to 901K, at a constant pressure?
a. 732.5 J
b. 775.12 J
c. 810.74 J
d. 840.72 J
Link within Knovel
Solution
The link provided in the hint brings the player to an interactive table in Knovel. The table lists different
substances and their heat capacities, cp, at constant pressure, at a range of temperatures. From it one
may glean that the heat capacity of acenaphthene at 900K is 387.56
𝐽
.
𝑚𝑜𝑙·𝐾
A familiar equation for
calculating energy requirement in such scenarios is
𝑄 = 𝑚𝑐Δ𝑇
where Q is the energy that flows into the system, m is the mass of the substance, c is the specific heat of
the substance, and ΔT is the difference between the final and initial temperatures. However, because
specific heat is provided on a molar basis (and hence called molar heat capacity), the mass of the
substance ought to be converted to moles, using the value obtained in problem 10:
308 𝑔
= 2 𝑚𝑜𝑙
154 𝑔/𝑚𝑜𝑙
Now substitute into the first equation provided:
𝐽
𝑄 = (2 𝑚𝑜𝑙) (387.56
) (901𝐾 − 900𝐾) = 775.12 𝐽
𝑚𝑜𝑙 · 𝐾
Problem 12
The statement that an object’s momentum and position cannot be simultaneously determined to
arbitrary accuracy is known as the
a. Riemann Hypothesis
b. Pauli Exclusion Principle
c. Heisenberg Uncertainty Principle
d. Dirac Delta Function
Link within Knovel
Solution
This hint brings the player to the Knovel homepage. Upon searching for “Heisenberg uncertainty
principle” one will find a slew of results, the first of which is from Laser Fundamentals (2nd edition)
(2004). After opening the resource one will read that, according to the principle certain variables like
distance and momentum, have an uncertainty in their product below which the value of the product
could never be determined.
Problem 13
According to the uncertainty principle, as the measured value of an object’s momentum gets sufficiently
closer and closer to the true value, the accuracy of the measured value of its original position,
necessarily:
a. can increase or decrease
b. decreases
c. increases
d. is unaffected
Link within Knovel
Solution
The link in this hint brings the player to the same page that was referenced in the previous solution. The
section states that the product of the uncertainty of an object’s momentum and the uncertainty of the
distance (or position) cannot be below a certain value. Therefore if the uncertainty in the momentum is
less than that minimum, as it gets smaller the uncertainty in the position becomes as great as necessary
such that the product of the two is equal to the minimum value. This increase in uncertainty in the
position is due to the position changing, and thus the accuracy of a measurement decreases.
Problem 14
What is the de Broglie wavelength associated with a 1600 kg automobile moving at 65 mph?
a. 6.37*10-39 m
b. 1.425*10-38 m
c. 1.014*10-39 m
d. 2.27*10-39 m
Link within Knovel
Solution
This link opens to a book section on de Broglie’s hypothesis. It states “French physicist Louis de Broglie
prosed that just as light exhibits both wavelike and particle like properties, so material particles might
also exhibit a wavelike behavior.” The equation that he gave for determining these wavelengths, 𝜆 is:
ℎ
𝜆=
𝑚·𝑢
where h is Planck’s constant, m is the mass of the particle and u is its velocity. Using the Knovel unit
converter, one finds that 65 miles per hour is equal to 29.06 m/s. Planck’s constant is equal to
0.66252 · 10−33 𝐽 · 𝑠. Substituting this information, along with the mass into the equation yields:
0.66252 · 10−33 𝐽 · 𝑠
𝜆=
≈ 1.425 · 10−38 𝑚
1600 𝑘𝑔 · 29.06 𝑚/𝑠
Problem 15
As bodies collapse under their own gravity their mass is concentrated in a rapidly decreasing volume,
and the gravitational pull on outside objects increases. If light is to escape from that body, then it may
not collapse beyond a certain point. This critical parameter is called the
a. Schwarzschild radius
b. Kretschmann scalar
c. de Broglie wavelength
d. vanishing point
Link within Knovel
Solution
This hint brings the player to the Knovel homepage. Search for each of the terms in the order presented.
The first result returned for “Schwarzschild radius” is from Dictionary of Science (6th edition) (2010), and
the entry reads: “a critical radius of a body of given mass that must be exceeded if light is to escape from
that body”.
Problem 16
The mass of the Earth is approximately 5.972·1027 g. If the planet begins to collapse, what will be its
radius the moment that it becomes a black hole?
a. 8.849 m
b. 8.849 mm
c. 884.9 km
d. 8.849 cm
Link within Knovel
Solution
This link brings the player to a dictionary entry for Schwarzschild radius, the same one that the previous
solution used. The entry provides an expression for the radius and explains that if an object collapses to
such an extent that it is less than this, it will become a “black hole”. The Schwarzschild radius is 2𝐺𝑀/𝑐 2 ,
where G is the universal gravitational constant, M is the mass of the object, and c is the speed of light in
free space.
3
−11 𝑚
(2)
(6.668
·
10
) (5.972 · 1024 𝑘𝑔)
2𝐺𝑀
𝑘𝑔 · 𝑠 2
=
≈ 0.008849 𝑚 = 8.849 𝑚𝑚
𝑚 2
𝑐2
8
(3 · 10 )
𝑠