A fiberglass composite is composed of a matrix of vinyl ester and reinforcing fibers of E-glass. The volume fraction of E-glass is 31%. The remainder is vinyl ester. The density of the vinyl ester is 0.79 g/cm3, and its modulus of elasticity is 4.04 GPa. The density of E-glass is 3.011 g/cm3, and its modulus of elasticity is 80 GPa. A section of composite 1.00 cm by 25.00 cm by 200.00 cm is fabricated with the E-glass fibers running longitudinal along the 200 cm direction. Assume there are no voids in the composite. Determine the modulus of elasticity of the composite in GPa.


Answer 1

Full Question

Calculate the modulus of elasticity of the composite

I. In the longitudinal direction of the glass fibre

2. In the perpendicular direction of glass fibre


1. 27.5876 GPa

2. 5.725 GPa



Dimensions of the fibreglass is given by: 1.00 cm x 25.00 cm x 200.00 cm

Volume = 1 * 25 * 200 = 5000cm³

P1 = The volume fraction of the E-glass is 31%

P2 = The remainder 69% is Vinyl Ester

Volume of Vinyl Ester= 69% * 5000cm³ = 3450cm³

I. In the longitudinal direction of the glass fibre


Modulus of elasticity of glass, E1 = 80 Gpa

Modulus of elasticity of vinyl ester, E2 = 4.04 Gpa

Modulus of elasticity = E1P1 + E2P2

Modulus of elasticity of vinyl esters = 31% * 80 + 69% * 4.04

Modulus = 24.8 + 2.7876

Modulus = 27.5876 GPa

ii. In the perpendicular direction of glass fibre

This is solved by

E1.E2/(P1.E2 + P2.E1)

= (80 * 4.04)/(31% * 4.04 + 69% * 80)

= 323.2/(1.2524 + 55.2)

= 5.725177317527687

= 5.725 GPa

Related Questions

An electric car is designed to run off a bank of 14.0-V batteries with total energy storage of 3.00 × 10 7 J. If the electric motor draws 7 000 W, what current will be delivered to the motor?



I =500 A    


Given that

Power ,P = 700 W

Potential difference ,ΔV = 14 V

Lets take current = I

We know that power P is given as

P = V I


ΔV =Voltage difference


Now by putting the values in above equation

7000 = 14 x I

I =500 A

Therefore the current will be 500 A.

Consider a 50-mH inductor. a. Express the voltage across the inductor and then evaluate it at t = 0.25 s if iL (t) = 5e −2t + 3te−2t −2 A. Be certain to simplify your expression.



V=L(di/dt) where i is current, V=0.208


using expression iL(t)=5e-2t+3te-2t-2 and L=0.05H(50/1000)


since there is no power of e, I'll assume the power to be 1


at t=0.25



One lap around a standard high-school running track is exactly 0.25 miles. Write the function miles_to_laps() that takes a number of miles as an argument and returns the number of laps. Complete the program to output the number of laps. Output each floating-point value with two digits after the decimal point, which can be achieved as follows: print('{:.2f}'.format(your_value)) Ex: If the input is: 1.5 the output is: 6.00 Ex: If the input is: 2.2 the output is: 8.80 Your program must define and call the following function: def miles_to_laps(user_miles)





import java.util.Scanner;


public class MyClass {

   static void miles_to_laps(float miles) {

       double laps = miles/0.25;

       System.out.printf("Number of laps: %s", String.format("%.2f", laps));



   public static void main(String args[]) {

       System.out.print("Enter miles: ");

       Scanner input = new Scanner(System.in);

       float miles = input.nextFloat();

       if (miles < 0) {

           System.out.println("Invalid input");






I have used the Scanner library for user input.


def miles_to_laps(user_miles):

return user_miles/0.25

if __name__ == '__main__':

miles = float(input(""))

lap = miles_to_laps(miles)



because it works.

The SPACE ePlanner produces ALL of the following outputs for a Service Bundle (yes or no): Strategic ICT planRFPRequirements documentSample SolutionRFPProject management plan


Answer: Yes


While handling and managing offices, means a lot of different tasks, including files,overseeing supplies, running the mailroom and worker productivity, the space planner work is to specifically focus on managing the space within your office building. When it comes to all room within each wing of each floor, the space planner's work is to layout and optimize the design in order to maximize productivity and also keep the employees happy.

Work of the space ePlanner includes all those in listed above in the question.

A 75-g projectile traveling at 600 m/s strikes and becomes embedded in a 50-kg block, which is initially stationary. Compute the kinetic energy lost during the impact. Express your answer as an absolute value |AE| and as a percentage n of the original system energy E.



Change in kinetic energy: 13479.77 J

Percentage change in kinetic energy = 99.85%


Change in momentum when the projectile becomes embedded in the block will be 0.

This means:

here subscript P means projectile, and subscript B means block.

Solving this equation we get:

(0.075 * 600) + (50 * 0) = (0.075 + 50) * V

Using this velocity we can compute the change in kinetic energy:

Initial kinetic energy = 0.5 * m * v^2

Initial kinetic energy = 0.5 * 0.075 * 600^2

Initial kinetic energy = 13500 J

Final Kinetic energy = 0.5 * m * v^2

Mass (m) = mass of block + mass of projectile = 50.075 kg

Final velocity v = 0.899 m/s

Final Kinetic Energy = 0.5 * 50.075 *  0.899^2

Final Kinetic Energy = 20.235 J

Change in kinetic energy = 13500 - 20.235 = 13479.77 J

Percentage change of kinetic energy = (13479.77 / 13500) * 100

Percentage change of kinetic energy = 99.85 %

You are asked to study the causal effect of hours spent on employee training​ (measured in hours per worker per​ week) in a manufacturing plant on the productivity of its workers​ (output per worker per​ hour). Write in the box the option letter ​(a, b, c or​ d) on the right that best describes the statement on the left. Choose a random group of employees to receive ten hours per week in additional training for a period of four weeks.​ Then, estimate the difference in productivity between workers who received the additional training and those that did not



The correct option is B.


"An ideal randomized controlled experiment " best describes the statement mentioned in the question.

A fatigue test is performed on 69 rotating specimens made of 5160H steel. The measured number of cycles to failure (L in kcycles) is given below, where f is frequency. Assuming a normal distribution, how many specimens are predicted to fail at less than 115 cycles? L 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 f 2 1 3 5 8 12 6 10 8 5 2 3 2 1 0 1 a. Estimate the mean and standard deviation of the life for the population b. Assume the distribution is normal. Predict the number of failed specimens at less than 115 k cycles. c. Assume the distribution of failures remains the same size (that is the standard deviation does not change). What should the mean value of the life expectancy be to ensure approximately 99% of the samples survive at 115 k cycles?



(a) Mean = 122.9, σ = 30.071

(b) No. of failed specimens at less than 115k cycles are 27.

(c) μ = 39.07


We are given:

L  60  70  80  90  100  110  120  130  140  150  160  170  180  190  200  210

f    2     1    3     5     8     12     6     10     8     5     2      3      2      1       0      1

(a) First we need to calculate the mean and standard deviation. The formula for calculating mean is:

Mean = ∑fx/∑f

And for standard deviation we have:

S.D. = √Var

Var = ∑fx²/∑f - (Mean)²

∑fx = (2*60) + (1*70) + (3*80) + (5*90) + (8*100) + (12*110) + (6*120) + (10*130) + (8*140) + (5*150) + (2*160) + (3*170) + (2*180) + (1*190) + (0*200) + (1*210)

         = 120 + 70 + 240 + 450 + 800 + 1320 + 720 + 1300 + 1120 + 750 + 320 + 510 + 360 + 190 + 0 + 210

∑fx = 8480

Mean = ∑fx/∑f

          = 8480/69

Mean = 122.9  

∑fx² = (2*60²) + (1*70²) + (3*80²) + (5*90²) + (8*100²) + (12*110²) + (6*120²) + (10*130²) + (8*140²) + (5*150²) + (2*160²) + (3*170²) + (2*180²) + (1*190²) + (0*200²) + (1*210²)


∑fx² = 1104600

Var = ∑fx²/∑f - (Mean)²

     = 1104600/69 - (122.9)²

     = 16008.69565 - 15104.41

Var = 904.2856

S.D = √Var

σ = √904.2856

σ = 30.071

(b) Let X be the number of failed specimen.

We will use the z-score to calculate the probability. The formula for z-score is:

z = (X-μ)/σ


When one of the data sources used for incident decision making is coming from individual or aggregated log files, the management of those sources becomes critical. What are some of the key activities associated with managing logs?





A signal suspected to be of the nominal form y(t) = 5 sin 1000 mV is measured by a first-order instrument having a time constant of 100 ms and K 1 V/V. It is then to be passed through a second-order amplifier having a K= 100 V/V, a natural frequency of 15,000 Hz, and a damping ratio of 0.8. What is the expected form of the output signal, y(t)? Estimate the dynamic error and phase lag in the output. Is this system a good choice here? If not, do you have any suggestions?


Sorry no sudgestions.



dynamic error , C(s) = K ( 1- e-T/t )

output y(t) = e-qw + [ Ae(1-q2)1/2 wt + A2e(1-q2)1/2 wt ]


Given that,

Y(t) = 5 Sin 1000t mV

t = 100 ms

K = 1v/v

Second amplifier , K = 100 V/V

Natural frequency , f = 15000 Hz

damping ratio = 0.8

Y (t) = ?

the given equation is in the form of x = X sin ( wt + \phi)

f = 1/T

  dynamic error , C(s) = K ( 1- e-T/t )

                                = 100 ( 1- e-100 / t )

    y = 5 sin 1000t mV

  Y = amplitude = 5

     f = 1/t

  t = 1/f = 1/ 5000 = 500 ms

  C(s) = 100 ( 1- e-100/500 )

         = 18.12

damping ratio ,q= 0.8

output y(t) = e-qw + [ Ae(1-q2)1/2 wt + A2e(1-q2)1/2 wt ]

                = 54

Yes it is a good choice because of having the under damped system.

A heat pump cycle operating at steady state receives energy by heat transfer from well water at 10oC and discharges energy by heat transfer to a building at the rate of 1.4x105 kJ/h. Over a period of 14 days, an electric meter records that 1620 kW·h of electricity is provided to the heat pump. These are the only energy transfers involved. Determine the total amount of energy transfer by work into the heat pump over the 14-day period, in GJ. Determine the amount of energy that the heat pump receives by heat transfer from the well water over the 14-day period, in GJ. Determine the heat pump’s coefficient of performance.


In this exercise we have to use knowledge about forces to determine the heat coefficient, so we can say that:

The total amount of energy transferred is 5.8GJ

Knowing that the potential energy formula is given by:

This energy that can be transferred within 14 days, we will have that corresponds to:

Then the formula can be written as follows:

The coefficient will be calculated as:

See more about coefficient at brainly.com/question/11808898

Answer: the total amount of energy transferred by work into the heat pump over the 14-day period is 5.8GJ

Explanation: since the meter reading for the 14 days period is 1620kW.h

Electrical power Ep is

Ep = 1620*10^3W.h

In 14 days there are 14*24 hrs = 336hrs

Ep = (1620*10^3)/336 = 4821.42W

This means 4821.42J of work is done per sec. Therefore for 14 days we have,

3600*24*14*4821.42J of enegy

= 5831989632J = 5.8GJ

Answer 2 = the amount of energy that the heat pump receives by heat transfer from the well water over the 14-day period is 41.2GJ

Explanation: rate of energy transfer as heat to building is 1.4*10^5 kJ/hr

That is 1.4*10^8 J/hr

In 14 days it becomes

24*14*1.4*10^8 = 47GJ

Energy received from well by heat pump =

Energy transfered - energy received

= 47 - 5.8 =41.2GJ

Answer 3. the heat pump’s coefficient of performance is 0.12

Explanation: COP = energy obtained/ work done

COP = 5.8/47 = 0.12

BCC lithium has a lattice parameter of 3.5089 3 10–8 cm and contains one vacancy per 200 unit cells. Calculate (a) the number of vacancies per cubic centimeter; and (b) the density of Li.


(a) The number of vacancies per cubic centimeter is 1.157 X 10²⁰

(b) ρ = n X (AM) / v X Nₐ



Lattice parameter of Li  = 3.5089 X 10⁻⁸ cm

1 vacancy per 200 unit cells

Vacancy per cell = 1/200


Number of vacancies per cubic cm = ?

Vacancies/cm³ = vacancy per cell / (lattice parameter)³

Vacancies/cm³ = 1 / 200 X (3.5089 X 10⁻⁸cm)³

Vacancies/cm³ = 1.157 X 10²⁰

Therefore, the number of vacancies per cubic centimeter is 1.157 X 10²⁰


Density is represented by ρ

ρ = n X (AM) / v X Nₐ


Nₐ = Avogadro number

AM = atomic mass

n = number of atoms

v = volume of unit cell

The purification of hydrogen gas is possible by diffusion through a thin palladium sheet. Calculate the number of kilograms of hydrogen that pass per hour through a 4.0-mm-thick sheet of palladium having an area of 0.13 m2 at 500°C. Assume a diffusion coefficient of 5.1 × 10-8 m2/s, that the concentrations at the high- and low-pressure sides of the plates are 2.9 and 0.31 kg/m3 of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.


Answer: 5.36×10-3kg/h

Where 10-3 is 10 exponential 3 or 10 raised to the power of -3.

Explanation:using the formula

M =JAt = -DAt×Dc/Dx

Where D is change in the respective variables. Insulting the values we get,

=5.1 × 10-8 × 0.13 × 3600 × 2.9 × 0.31 / 4×10-3.


2. A water pump that consumes 2 kW of electric power when operating is claimed to take in water from a lake and pump it to a pool whose free surface is 30 m above the free surface of the lake at a rate of 50 L/s. Determine if this claim is reasonable



the claim is false


Power consumed = mgh

m = ρv where volume per seconds ( 1 l = 0.001 m³) = 50 L/s × 0.001 m³ = 0.05 m³,  ρ density of water = 1000 kg/m³

P consumed = ρv gh = 0.05 m³×1000 kg/m³ × 9.8m/s  × 30 m = 14700 = 14.7 Kw

This claim is false because the power consumed is greater then the 2kW  reported.

You are riding a roller-coaster going around a vertical loop, on the inside of the loop. If the loop has a radius of 56.0 m, how fast must the cart be moving in order for you to feel 3 times as heavy at the bottom of the loop? g



46.9 m/s


When the cart is at the bottom of the loop, there are two forces acting on the person:

- The force of gravity, acting downward, of magnitude (where m is the mass and g is the acceleration due to gravity)

- The normal reaction exerted on the person, acting upward, of magnitude N

Their resultant must be equal to the centripetal force, so the equation of the forces at the bottom of the loop is:

where v is the speed of the cart at the bottom of the loop and r the radius of the loop.

N represents the apparent weight of the person in the cart: here we are told that the person must feel 3 times as heavy, this means that

Substituting into the equation,

Which means

Here the radius of the loop is

r = 56.0 m

So, the speed needed is:

In the layout of a printed circuit board for an electronic product, 12 different locations can accommodate chips. (a) If five different types of chips are to be placed on the board, how many different layouts are possible?

(b) If the five chips that are placed on the board are of the same type, how many different layouts are possible?



a) 244,140,625 different ways

b) 390,625 different ways


a) If there are 5 ways to place a chip on each location, and there are 12 locations overall, we have:

5^12 ways of placing them

This would mean a total of 244,140,625 different ways

b) If five chips are of the same type, we can first find how many ways we can place chips on the remaining 7 locations:

5^7 = 78,125

Next we can multiply this by the number of ways the next 5 chips could be the same:

78,125 * 5 = 390,625 different ways

Determine the temperature, in K, of 5 kg of air at a pressure of 0.3 MPa and a volume of 2.2 m3 . Ideal gas behavior can be assumed for air under these conditions.



The temperature is 460.45 K


From the ideal gas equation,

PV = nRT

P is the pressure of air = 0.3 MPa = 0.3×10^6 = 3×10^5 Pa

V is the volume of air = 2.2 m^3

n is the number of moles of air = mass of air/MW of air = 5/29 = 0.1724 kg mol

R is gas constant = 8314.34 m^3.Pa/kg mol.K

T is the temperature of air in Kelvin

T = PV/nR = (3×10^5 × 2.2)/(0.1724 × 8314.34) = 460.45 K

Ira, Jayla, and Kendra are all saying it would be better to move lines 22 and 24 to a single line executing print ('Code complete.') just before line 21. These three students have different reasons for their opinions. Their reasons are below. Do you think each of them is right, wrong, or somewhere in between? Explain your answer for each student. Ira: "It would be better to have a single print statement because that code is going to happen no matter what. The program will run slower by having it there twice." Jayla: "It would be better to have a single print statement because that code is going to happen no matter what. Later, if you want to change your program, you’re going to have to remember to change it in two places the way the code is now." Kendra: "It would be better to have a single print statement because it is going to happen no matter what. That program would take up less memory if you just wrote it once."


Missing Part of the Question

Here's the code segment, in question

21: if check == 2:

22: print('Code Complete')

23 else:

24 print('Code Complete')

25 print('Not all systems are ready')


Ira, Jayla and Kendra are all right



Ira thinks that wrting the same print statement in both conditions ( if and else) is not needed because the statement will be executed no matter the outcome of the conditions and would cause the program to run slower.

Ira is right and I agree; there is no need for extra code to be processed if the demand has to be executed and is not based on a condition. Basically, you don't have to write the same code twice when they serve the same purpose


Jayla thinks using a single print statement because the code would definitely be executed no matter what and one might forget the other when there's a need to modify the code e.

Jayla is also right; Jayla's reason points to the fact that there's a need to simplify codes (when needed) and do away with bulky lines of codes. Changing a line and forgetting to change the other leads to an unintended expectation of a different output from the programmer.


Kendra thinks It would be better to have a single print statement because it is going to happen no matter what. That program would take up less memory if you just wrote it once.

Kendra's reason is self explanatory; nevertheless I'll add my bit.

Repeating the same line of codes twice when not necessarily needed, takes up more memory.

A European car manufacturer reports that the fuel efficiency of the new MicroCar is 48.5 km/L highway and 42.0 km/L city. What are the equivalent fuel efficiency rates in miles per gallon?



Fuel efficiency for highway = 114.08 miles/gallon

Fuel efficiency for city = 98.79 miles/gallon


1 gallon = 3.7854 litres

1 mile = 1.6093 km

Let's first convert the efficiency to km/gallon:

48.5 km/litre = (48.5 * 3.7854) km/gallon

48.5 km/litre =  183.5919 km/gallon (highway)

42.0 km/litre = (42.0 * 3.7854) km/gallon

42.0 km/litre = 158.9868 km/gallon (city)

Next, we convert these to miles/gallon:

183.5919 km/gallon = (183.5919 / 1.6093) miles/gallon

183.5919 km/gallon = 114.08 miles/gallon (highway)

158.9868 km/gallon = (158.9868 /1.6093) miles/gallon

158.9868 km/gallon = 98.79 miles/gallon (city)

Suppose you must remove an average of 3.9×108J of thermal energy per day to keep your house cool during the summer. Part A If you upgrade from an old air conditioner with a COP of 2.3 to a new air conditioner with a COP of 6.0, by how many joules is the required mechanical work reduced each day?



The required mechanical work is required to reduce each day by 1.05×10^8 Joules.


Coefficient of Performance (COP) = Q/W

Q is thermal energy absorbed by the air conditioner

W is mechanical work done

Q = 3.9×10^8 J

COP of old air conditioner = 2.3

W = Q/COP = 3.9×10^8/2.3 = 1.70×10^8 J

COP of new air conditioner = 6

W = Q/COP = 3.9×10^8/6 = 6.5×10^7 J

Reduction in mechanical work = (1.7×10^8) - (6.5×10^7) = 1.05×10^8 J

1. You have a Co-Cr alloy with Young's Modulus: 645 MPa, Poisson ratio 0.28, and yield strength 501 MPa for that alloy when used in a medical device. From these data estimate the following for a 1 x 2 cm cross-section bar of the alloy having length 12 cm: a. The maximum tensile load that can be applied in the longitudinal direction of the bar without inducing plastic deformation. b. The length and cross-sectional area of the bar at its tensile elastic limit.



F_x = 100,200 N

x' = 21.321 cm ... Length

y' =0.7825 cm

z' = 1.565 cm

A' = ( 0.783 x 1.565 ) cm  



- The Modulus of Elasticity E = 645 MPa

- The poisson ratio v = 0.28

- The Yield Strength Y = 501 MPa

- The Length along x-direction x = 12 cm

- The length along y-direction y = 1 cm

- The length along z--direction z = 2 cm


The maximum tensile load that can be applied in the longitudinal direction of the bar without inducing plastic deformation. b. The length and cross-sectional area of the bar at its tensile elastic limit.


- The Tensile forces within the limit of proportionality is given as:

                                   F_i = б_i*A_jk

- A maximum tensile Force F_x along x direction can be given as:

                                   F_x = Y*A_yz

                                   F_x = 501*( 0.01*0.02)*10^6

                                  F_x = 100,200 N

- The corresponding strains in x, y and z direction due to F_x are:

                                    ξ_x = Y / E

                                    ξ_x = 501 / 645 = 0.7767

                                    ξ_y = ξ_z = -v*Y / E

                                    ξ_y = ξ_z = -0.28*501 / 645 = - 0.2175

- The corresponding change in lengths at tensile elastic stress are:

                                    Δx = x*ξ_x = 12*0.7767 = 9.321 cm

                                    Δy = y*ξ_y = - 1*0.2175 = -0.2175 cm

                                    Δz = z*ξ_z = - 2*0.2175 = -0.435 cm

- The final lengths are:

                                    x' = x + Δx = 12 + 9.321 = 21.321 cm

                                    y' = y + Δy = 1 - 0.2175 = 0.7825 cm

                                    z' = z + Δz = 2 - 0.435 = 1.565 cm


Other Questions
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