27.11.2022

A15 ft high vertical wall retains an overconsolidated soil where ocr-1.5, c'-: o, ф , --33°, and 1 1 5.0 lb/ft3.

determine the magnitude and location of the thrust on the wall, assuming that the soil is at rest.

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30.05.2023, solved by verified expert
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magnitude of thrust uis  11061.65 lb/ft

location is 5 ft from bottom

Explanation:

Given data:

Height of vertical wall is 15 ft

OCR  is 1.5

A15 ft high vertical wall retains an overconsolidated, №16482173, 27.11.2022 17:20

saturated uit weightA15 ft high vertical wall retains an overconsolidated, №16482173, 27.11.2022 17:20

coeeficent of earth pressure A15 ft high vertical wall retains an overconsolidated, №16482173, 27.11.2022 17:20

A15 ft high vertical wall retains an overconsolidated, №16482173, 27.11.2022 17:20

        = 1 - sin 33 = 0.455

for over consolidate

A15 ft high vertical wall retains an overconsolidated, №16482173, 27.11.2022 17:20

            A15 ft high vertical wall retains an overconsolidated, №16482173, 27.11.2022 17:20

Pressure at bottom of wall is

A15 ft high vertical wall retains an overconsolidated, №16482173, 27.11.2022 17:20

   A15 ft high vertical wall retains an overconsolidated, №16482173, 27.11.2022 17:20

P = 1474.88 lb/ft^3

Magnitude pf thrust is

A15 ft high vertical wall retains an overconsolidated, №16482173, 27.11.2022 17:20

   A15 ft high vertical wall retains an overconsolidated, №16482173, 27.11.2022 17:20

the location must H/3 from bottom so

A15 ft high vertical wall retains an overconsolidated, №16482173, 27.11.2022 17:20

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Engineering
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magnitude of thrust uis  11061.65 lb/ft

location is 5 ft from bottom

Explanation:

Given data:

Height of vertical wall is 15 ft

OCR  is 1.5

\phi = 33^o

saturated uit weight\gamma_{sat} = 115.0 lb/ft^3

coeeficent of earth pressure K_o

K_o = 1 -sin \phi

        = 1 - sin 33 = 0.455

for over consolidate

K_{con} = K_o \times OCR

            = 0.455 \times 1.5 = 0.683

Pressure at bottom of wall is

P =K_{con} \times (\gamma_{sat} - \gamma_{w}) + \gamma_w \times H

   = 0.683 \times (115 - 62.4) \times 15 + 62.4 \times 15

P = 1474.88 lb/ft^3

Magnitude pf thrust is

F= \frac{1}{2} PH

   =\frac{1}{2} 1474.88\times 15 = 11061.65 lb/ft

the location must H/3 from bottom so

x = \frac{15}{3} = 5 ft

Engineering
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P Answered by Master

"Test Phase " is the correct choice.

Explanation:

DevSecOps seems to be a community as well as experience of corporate data science which encompasses software design, regulation, including operational activities. This same main feature of DevSecOps has always been to strengthen customer achievement as well as expedition importance by computerizing, supervising as well as implementing data protection at all stages of the development including its development tools.The testing method throughout the test phase would then help make sure that the controller is designed mostly under the responsibilities forecasted. The test focuses on either the reaction times, dependability, use of resources but instead interoperability of applications.
Engineering
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P Answered by Master

NTQ

Explanation:

The given sequence is

BHE : FLI : JPM

If is clear that, alphabets on first places are B, F, J. Difference between their place vales is 4.

B+4=F,F+4=J ; so alphabet on first place of next term of sequence is J+4=N.

Similarly, alphabets on second places are H, L, P. Difference between their place vales is 4.

H+4=L,L+4=P ; so alphabet on second place of next term of sequence is P+4=T.

Alphabets on third places are E, I, M. Difference between their place vales is 4.

E+4=I,I+4=M ; so alphabet on third place of next term of sequence is M+4=Q.

Therefore, the next term is NTQ.

Engineering
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The following code or the program will be used

Explanation:

def readFile(filename):

   dict = {}

   with open(filename, 'r') as infile:

       lines = infile.readlines()

       for index in range(0, len(lines) - 1, 2):

           if lines[index].strip()=='':continue

           count = int(lines[index].strip())

           name = lines[index + 1].strip()

           if count in dict.keys():

               name_list = dict.get(count)

               name_list.append(name)

               name_list.sort()

           else:

               dict[count] = [name]

           print(count,name)

   return dict

def output_keys(dict, filename):

   with open(filename,'w+') as outfile:

       for key in sorted(dict.keys()):

           outfile.write('{}: {}\n'.format(key,';'.join(dict.get(key

           print('{}: {}\n'.format(key,';'.join(dict.get(key  

def output_titles(dict, filename):

   titles = []

   for title in dict.values():

       titles.extend(title)

   with open(filename,'w+') as outfile:

       for title in sorted(titles):

           outfile.write('{}\n'.format(title))

           print(title)

def main():

   filename = input('Enter input file name: ')

   dict = readFile(filename)

   if dict is None:

       print('Error: Invalid file name provided: {}'.format(filename))

       return

   print(dict)

   output_filename_1 ='output_keys.txt'

   output_filename_2 ='output_titles.txt'

   output_keys(dict,output_filename_1)

   output_titles(dict,output_filename_2)  

main()

Engineering
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P Answered by PhD

Energy equation from this week’s notes, your answer from #5, and Plank’s constant (6.63E-34) to find the approximate energy of this photon

Explanation:

1.The amount of energy in those photons is calculated by this equation, E = hf, where E is the energy of the photon in Joules; h is Planck's constant, which is always 6.63 * 10^-34 Joule seconds; and f is the frequency of the light in hertz

2.The first is Planck's equation, which was proposed by Max Planck to describe how energy is transferred in quanta or packets. Planck's equation makes it possible to understand blackbody radiation and the photoelectric effect. The equation is:

E = hν

where

E = energy

h = Planck's constant = 6.626 x 10-34 J·s

ν = frequency

Engineering
Step-by-step answer
P Answered by PhD

x=float(input("Enter a number: "))

sub=(x-int(x))

print(sub)

Explanation:

Got it Right

Engineering
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P Answered by Specialist

B ). When cleaning a vessel, use regular  household detergents and cleaners.

Explanation:

Boat cleaning or vessel cleaning is a tedious job. It takes a lot of hard work to clean a vessel. We have to be careful while maintaining the vessel.

The products that we use to clean and maintain the vessels needs to chosen carefully. Many boat cleaning products have adverse effects on the marine life and aquatic animals. it also tends to pollute the water bodies.

These products may be corrosive in nature, toxic or poisonous.

So we must use the regular household detergents and cleaners to clean our vessels. The regular detergents do not cause any harm to the nature and the water bodies.

We can use phosphate free no detergent soaps which may include citrus based soap or vegetable based soaps.

Therefore, while cleaning or maintaining vessels , use the regular household detergents and cleaners.  

Thus option (B) is correct.

Learn more :

link


Which statement regarding vessel maintenance is true?

When storing a vessel for a long time, it is
Engineering
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P Answered by Specialist

understanding others first before starting to explain your own point

Explanation:

Stephen covey believes the key to effective communication is understanding others first before starting to explain your own point. Communication is an important skill in life, in which one must have a good character in order to communicate effectively.

Engineering
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P Answered by PhD
There’s 5 different types of fire extinguishers that you can differentiate by their color codes.
Red - Water based
Creme - Foam based
Blue - Powder based
Black - CO2 or carbon dioxide based
Yellow - Wet chemical based

What would determine the type of fire extinguisher used would be the class of fire it is.

Class A - Combustible materials ( i.e. paper, wood) Extinguishers to use - Red, Creme, Blue, and Yellow. (Do not use Black)

Class B - Flammable liquids ( i.e. paint, petrol, alcohol) Extinguishers to use - Creme, Blue, and Black. (Do not use Red or Yellow)

Class C - Flammable gases ( i.e. butane, methane) Extinguishers to use - Blue (Do not use Red, Creme, Black or Yellow)

Class D - Flammable metals ( i.e. lithium, potassium) Extinguishers to use - Blue (Do not use Red, Creme, Black or Yellow)

Class F - Deep fat fryers ( i.e. chip pans) Extinguishers to use - Yellow (Do not use Red, Creme, Blue or Black)

Electrical - any sort of electrical equipment
( i.e. computers, generators) Extinguishers to use - Blue and Black (Do not use Red, Creme or Yellow)

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