An environmental engineer is considering three methods for disposing of a nonhazardous chemical sludge: land application, fluidized-bed incineration, and private disposal contract. The estimates for each method are below. (a) Determine which has the least cost on the basis of an annual worth comparison at 10% per year. (b) Determine the equivalent present worth value of each alternative using its AW value

Answers

Answer 1
Answer:

Answer:

Explanation:

The annual worth of land application = $ 110,000 ( A/P , 10% , 5 years ) + $ 95,000 - $ 15,000 (A/F , 10% , 5 years )

The annual worth of land application = $ 110,000 X 0.263797 + $ 95,000 - $ 15,000 X 0.163797

The annual worth of land application = $ 29,017.54 + $ 95,000 - $ 2,456.95

The annual worth of land application = $ 121,560.59

The annual worth of land Incineration = $ 800,000 ( A/P , 10% , 6 years ) + $ 60,000 - $ 250,000 X (A/F , 10% , 6 years )

The annual worth of land Incineration = $ 800,000 X 0.229607 + $ 60,000 - $ 250,000 X 0.129607

The annual worth of land Incineration = $ 211,283.85

The annual worth of contract = $ 190,000

The annual worth of contract = $ 190,000

The land application has the least cost , hence it is preferred .


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If you answer the whole question and show your work/coding I will rate 5 stars/brainliest!!! Walnut Orchard has two farms that grow wheat and corn. Because of different soil conditions, there are differences in the yields and costs of growing crops on the two farms. The yields and costs are shown in the following table. Each farm has 100 acres available for cultivation. 11,000 bushels of wheat and 7,000 bushels of corn must be grown. Please have an LP model to minimize the total cost while meeting the demand and solve it with Lindo or Excel. You need to have all parts of a model: notation, objective function, constraints, and sign restrictions.

The sports car has a weight of 4500-lb and a center of gravity at G. If it starts from rest it causes the rear wheels to slip as it accelerates. Determine how long it takes for it to reach a speed of 10 ft/s. Also, what are the normal reactions at each of the four wheels on the road? The coefficients of static and kinetic friction at the road are us=0.5 and uk=0.3, respectively. Neglect the mass of the wheels.

Answers

Answer:

Time=2.72 seconds

Front wheel reactions= 1393 lb

Rear wheel reactions= 857 lb

Explanation:

The free body diagram is assumed to be the one attached here

The mass, m of the car is  

M=\frac {W}{g} where W is weight and g is acceleration due to gravity

Taking g as 32.2 ft/s^(2) then  

M=\frac {4500}{32.2}=139.75 lbm

Considering equilibrium in x-axis

Ma_G-f=0

Ma_G-(\mu_g* 2N_B)=0

139.75* a_G-(0.3* 2* N_B)=0

0.6N_B=139.75a_g

N_B=232.92a_g

At point A using the law of equilibrium, the sum of moments is 0 hence

-2N_B(6)+4500(2)=-Ma_G(2.5)

-12N_B+9000=-139.75a_G* 2.5

-12(232.92a_G)+900=-349.375a_G

a_g\approx 3.68 ft/s^(2)

The normal reaction at B is therefore

N_B=232.92a_G=232.92* 3.68\approx 857 lb

Consider equilibrium in y-axis

4500-2N_A-2N_B=0

N_A+N_B=2250

N_A+857=2250

N_A=1393 lb

To find time that the car takes to a speed of 10 ft/s

Using kinematic equation

V=u+at

10=0+3.68t

t=\frac {10}{3.68}\approx 2.72 s

Calculate how large a mass would be necessary to obtain a mechanical noise limit of [Equation] = 1 nG, 1 µG, and 1 mG if the mechanical resonance frequency is [Equation] = 100 Hz. If the mass is to be made of cube of `silicon, what would its physical dimensions be?

Answers

Answer:

Mechanical resonance frequency is the frequency of a system to react sharply when the frequency of oscillation is equal to its resonant frequency (natural frequency).

The physical dimension of the silicon is 10kg

Explanation:

Using the formular, Force, F = 1/2π√k/m

At resonance, spring constant, k = mw² ( where w = 2πf), when spring constant, k = centripetal force ( F = mw²r).

Hence, F = 1/2π√mw²/m = f ( f = frequency)

∴ f = F = mg, taking g = 9.8 m/s²

100 Hz = 9.8 m/s² X m

m = 100/9.8 = 10.2kg

In contrasting the read-evaluation loop and the notification-based paradigm for inter- active programs, construction of a pre-emptive dialog was discussed. How would a programmer describe a pre-emptive dialog by purely graphical means? (Hint: Refer to the discussion in Sec- tion 8.5 concerning the shift from external and independent dialog management to presentation control of the dialog)

Answers

The way a programmer describe a pre-emptive dialog by purely graphical means is; by producing a window that covers the entire screen to make it the currently selected window.

What is Pre - emptive Dialogue?

In a graphics - based interaction, it is supposed that the user can only interact with parts of the system that are visible. However, In a windowing system, the user can only direct input to a single window that was currently selected and the way to change that selected window is to indicate with some gesture within that window.

Finally, to create a pre-emptive dialog, the system would do so through the production of a window that covers the entire screen to make it the currently selected window. Thereafter, all user input would be directed to that window and the user would have no means of selecting any other window. Then the covering window will now pre-empt any other user action with the exception of that which it is defined to support.

Read more about dialogue at; brainly.com/question/5962406

Answer:

In an illustrations based communication, it is expected that the client can just associate with parts of the framework that are obvious. In a windowing framework, for instance, the client can just direct contribution to a solitary, at present chosen window, and the main methods for changing the chose window would be by demonstrating with some signal inside that window. To make a preemptive exchange, the framework can create a window that covers the whole screen and make it the right now chosen window. All client information would then be coordinated to that window and the client would have no methods for choosing another window. The 'covering' window in this way preempts some other client activity with the exception of that which it is characterized to help

You are analyzing an open-return wind tunnel that intakes air at 20 m/s and 320K. When the flow exits the wind tunnel it is moving at a speed of 250 m/s. What is the temperature of the flow exiting that wind tunnel

Answers

The solution is in the attachment

Answer:

please find attached.

Explanation:

A rocket is launched vertically from rest with a constant thrust until the rocket reaches an altitude of 25 m and the thrust ends. The rocket has mass 2 kg and thrust force 35 N. Neglecting air resistance, determine (a) the speed of the rocket when the thrust ends, (b) the maximum height reached by the rocket, and (c) the speed of the rocket when it returns to the ground.

Answers

Answer:

a) v=19.6 m/s

b) H=19.58 m

c) v_(f)=29.57 m/s  

Explanation:

a) Let's calculate the work done by the rocket until the thrust ends.

W=F_(tot)h=(F_(thrust)-mg)h=(35-(2*9.81))*25=384.5 J

But we know the work is equal to change of kinetic energy, so:

W=\Delta K=(1)/(2)mv^(2)

v=\sqrt{(2W)/(m)}=19.6 m/s

b) Here we have a free fall motion, because there is not external forces acting, that is way we can use the free-fall equations.

v_(f)^(2)=v_(i)^(2)-2gh

At the maximum height the velocity is 0, so v(f) = 0.

0=v_(i)^(2)-2gH

H=(19.6^(2))/(2*9.81)=19.58 m  

c) Here we can evaluate the motion equation between the rocket at 25 m from the ground and the instant before the rocket touch the ground.

Using the same equation of part b)

v_(f)^(2)=v_(i)^(2)-2gh

v_(f)=\sqrt{19.6^(2)-(2*9.81*(-25))}=29.57 m/s

The minus sign of 25 means the zero of the reference system is at the pint when the thrust ends.

I hope it helps you!

Given asphalt content test data: a. Calculate the overall mean and standard deviation for the entire test period.
b. The contract specifications require an average asphalt content of 5.5% +/- 0.5% every day. Plot the daily average asphalt content. Show upper and lower control limits.
c. Do all of these samples meet the contract specifications? Explain your answer.
d. What trend do you observe based on the data? What could cause this trend?"

Answers

Answer:

hello your question is incomplete attached below is the complete question

A) overall mean = 5.535,  standard deviation ≈ 0.3239

B ) upper limit = 5.85, lower limit = 5.0

C) Not all the samples meet the contract specifications

D) fluctuation ( unstable Asphalt content )

Explanation:

B) The daily average asphalt content has to obtained in order to determine the upper and lower control limits using an average asphalt content of 5.5% +/- 0.5% everyday

The upper limit : 14 may = ( 5.8 + 5.1 ) / 2 = 5.85

The lower limit : 16 may = ( 5.2 + 4.8 ) / 2 = 5.0

attached below is the required plot

C ) Not all the samples meet the contract specifications and the samples that do not meet up are samples from :

15 may and 16 may . this is because their Asphalt contents are 6.2 and 4.8 respectively and sample number 18 and 20

D ) what can be observed is that the ASPHALT content fluctuates between the dates while the contract specification is fixed