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solong [7]
2 years ago
13

After being purged with nitrogen, a low-pressure tank used to store flammable liquids is at a total pressure of 0.03 psig. (a) I

f the purging process is done in the morning when the tank and its contents are at 55°F, what will be the pressure in the tank when it is at 85℉ in the afternoon? (b) If the maximum design gauge pressure of the tank is 8 inches of water, has the design pressure been exceeded? (c) Speculate on the purpose of purging the tank with nitrogen

Engineering
1 answer:
g100num [7]2 years ago
8 0

Answer:

See attached picture.

Explanation:

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If a server takes precisely 15 seconds to serve a customer and customers arrive exactly every 20 seconds, what is the average wa
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The average waiting time is 10 seconds
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2 years ago
1. Add:<br>(i) 5xy, -2xy, -11xy, 8xy<br>(iv) 3a - 2b + c, 5a + 8b -70​
Cerrena [4.2K]

Answer:

(i) 0

(iv) 8a+6b+c-70

Explanation:

Hope this helps you

8 0
1 year ago
A 10-m long steel linkage is to be designed so that it can transmit 2 kN of force without stretching more than 5 mm nor having a
Fed [463]

Answer:

<em>minimum required diameter of the steel linkage is 3.57 mm</em>

<em></em>

Explanation:

original length of linkage l = 10 m

force to be transmitted  f = 2 kN = 2000 N

extension e = 5 mm= 0.005 m

maximum stress σ = 200 N/mm^2 = 2*10^{8}  N/m^{2}

maximum stress allowed on material σ = force/area

imputing values,

200 = 2000/area

area = 2000/(2*10^{8}) = 10^{-5} m^2

recall that area = \pi d^{2} /4

10^{-5} = \frac{3.142*d^{2} }{4} = 0.7855d^{2}

d^{2} = \frac{10^{-5} }{0.7855} = 1.273*10^{-5}

d = \sqrt{1.273*10^{-5}  } = 3.57*10^{-3} m = 3.57 mm

<em>maximum diameter of  the steel linkage d = 3.57 mm</em>

4 0
2 years ago
The air contained in a room loses heat to the surroundings at a rate of 50 kJ/min while work is supplied to the room by computer
Artemon [7]

Answer:

net amount of energy change of the air in the room during a 30-min period = 660KJ

Explanation:

The detailed calculation is as shown in the attached file.

4 0
2 years ago
Read 2 more answers
Write a method printShampooInstructions(), with int parameter numCycles, and void return type. If numCycles is less than 1, prin
kirill [66]

Answer:

// The method is defined with a void return type

// It takes a parameter of integer called numCycles

// It is declared static so that it can be called from a static method

public static void printShampooInstructions(int numCycles){

// if numCycles is less than 1, it display "Too few"

   if (numCycles < 1){

       System.out.println("Too few.");

   }

// else if numCycles is less than 1, it display "Too many"

    else if (numCycles > 4){

       System.out.println("Too many.");

   }

// else it uses for loop to print the number of times to display

// Lather and rinse

  else {

       for(int i = 1; i <= numCycles; i++){

           System.out.println(i + ": Lather and rinse.");

       }

       System.out.println("Done");

       

   }

}

Explanation:

The code snippet is written in Java. The method is declared static so that it can be called from another static method. It has a return type of void. It takes an integer as parameter.

It display "Too few" if the passed integer is less than 1. Or it display "Too much" if the passed integer is more than 4. Else it uses for loop to display "Lather and rinse" based on the passed integer.

8 0
2 years ago
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