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Nitella [24]
3 years ago
15

A) For Well A, provide a cross-section sketch that shows (i) ground elevation, (ii) casing height, (iii) depth to

Engineering
1 answer:
Ad libitum [116K]3 years ago
7 0
Don’t go on that file will give a virus! Sorry just looking out and I don’t know how to comment!
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How can I solve this sequence problem?​
Vinvika [58]

Answer:

Podes  hablar en español?

Explanation:

8 0
3 years ago
Read 2 more answers
A designer chooses a font size and color for a web page. What language was most likely to have been used?
Katarina [22]

The language was most likely to have been used is HTML as it help the designer to choose a font size and color for a web page.

<h3>What is HTML a programming language? </h3>

HTML  is known as HyperText Markup Language. This is known to be a computer code that is often employed to put together or structure a web page and all of its content.

Note that The language was most likely to have been used is HTML as it help the designer to choose a font size and color for a web page Because HTML is regarded as the markup language that is often used to edit web pages.

Learn more about HTML from

brainly.com/question/4056554

#SPJ1

5 0
2 years ago
Power cords can be damaged by Which of the following?
Elena L [17]
What are the options??
3 0
3 years ago
g (b) (4 pt) Write a function unique that identifies the repeated elements of a list and returns a list with unique elements. $
Taya2010 [7]

Answer:

I am writing a Python function unique()    

def unique(list):  # function unique that takes a list as parameter

 unique_list = []  #list to store unique elements

 for elements in list:  # loop that checks every element of the list

   if elements not in unique_list:  # takes unique elements from list

     unique_list.append(elements)  

#appends unique elements  from list to unique_list

 return unique_list      #outputs unique_list elements

         

Explanation:

The unique() function takes a list as argument which is named as list.

unique_list is a new list which stores unique element from the list.

The loop moves through the elements of the list one by one.

if condition checks if the element in list is not present in the unique_list which means that element is unique to the unique_list.

If this condition is true this means that the element is not repeated and is not already present in unique_list. Then that element is included to the unique_list using append() function which appends an element into the unique_list from the list.

If you want to check if this function works you pass a list with repeated elements to this function so that it can print the unique elements as follows:

print(unique([1,2,2,2,2,3,4,4,4,4,4,5]))

Output:

[1, 2, 3, 4, 5]

The screen shot of the function along with its output is attached.

6 0
4 years ago
Assume that a person skiing high in the mountains at an altitude of h = 14200 ft takes in the same volume of air with each breat
goldfiish [28.3K]

Answer:

The ratio of the mass of oxygen inhaled for each breath at this high altitude compared to that at sea level is 64.7 %

Explanation:

Mass of air at sea level is given by;

m_o = \rho_o V_o

Mass of air at 14,200 ft altitude is given by;

m_{14.2} = \rho _{14.2} V_{14.2}

The ratio of the mass of oxygen at high altitude to that at sea level is given by;

\frac{m_{14.2}}{m_o} = \frac{\rho _{14.2} V_{14.2}}{\rho _oV_{o}}\\\\ Assume \ that \ the \ air \ composition \ (V_{14.2} = V_o)\\\\\frac{m_{14.2}}{m_o} = \frac{\rho _{14.2} }{\rho _o}\\\\

density of air at sea level, \rho _o = 0.002378 \ slug/ft^3

density of air at 10,000 ft = 0.001756 slug/ft³

density of air at 14,200 ft = x

density of air at 15,000 ft, = 0.001496 slug/ft³

Interpolate between 10,000 ft and 15,000 ft

\frac{14,200 - 10,000}{15,000-10,000} = \frac{X - 0.001756}{0.001496 -0.001756}\\\\ 0.84(-0.00026) = X - 0.001756\\\\-0.0002184 = X - 0.001756\\\\X = 0.001756  - 0.0002184\\\\ X = 0.001538 \ slug/ft^3

The ratio of the mass of oxygen at 14,200 ft to that at sea level is given by;

\frac{m_{14.2}}{m_o} = \frac{\rho _{14.2}}{\rho_o} \\\\\frac{m_{14.2}}{m_o} =\frac{0.001538}{0.002378} = 0.647 = 64.7 \%

Therefore, the ratio of the mass of oxygen inhaled for each breath at this high altitude compared to that at sea level is 64.7 %

6 0
3 years ago
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