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3 years ago

Working with which of these systems requires a technician that has been certified in an EPA-approved course?

1 answer:

6 0

EPA Regulations provides a certified course for the technicians involved in the Air-conditioning system.

Answer: Option (b)

<u>Explanation:</u>

The EPA regulation has implemented an act called the "Clean Air Act" under the "section of 609".

This act provides some basic requirements for EPA Regulation such as follows;

Refrigerant: This unit must be approved by EPA Regulations before being implemented into the atmosphere.
Servicing: This system provides a certified course for technicians in service and also approve them with proper refrigerant equipment.
Reuse Refrigerants: The use of recycled refrigerants must be properly monitored before it comes in to serve.

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When removing a diesel engine from a truck, Technician A says it is OK to disconnect an air conditioning hose, but the refriger

agasfer [191]

Answer:

Technician B is right.

Explanation:

Air conditioning refrigerant contains Freon R22 and R410a, which have been linked to environmental damages, including ozone depletion, global warming, and energy-inefficiency. For environmentally-savvy entities and individuals, there is the modern move to a more environment-friendly refrigerant, known as R-32. Technician A's advice to vent the refrigerant outside the shop is in bad taste. He does not seem to be aware of the environmental footprint of such an action. Venting gas outside, in addition to the environmental damages, is also a waste of resources, and therefore, costly. This is why Technician B's advice should be preferred.

5 0

3 years ago

A box-shaped aquarium has horizontal dimensions 0.5 m by 1 m, and depth 0.5 m, and is filled two-thirds of the way to the surfac

antoniya [11.8K]

Answer:

3270 N/m^2

Explanation:

we can calculate the pressure difference between the bottom and surface of the tank by applying the equation for the net vertical pressure

Py = - Ph ( g ± a )

for a downward movement

Py = - Ph ( g - a ) ------ ( 1 )

From the above data given will be

p = 1000 kg/m^3, h = 2/3 * 0.5 = 0.33 m , a =2g , g = 9.81

input values into equation 1 becomes

Py = -Ph ( g - 2g ) = Phg ------ ( 3 )

Py = 1000 * 0.33 * 9.81

= 3270 N/m^2

6 0

3 years ago

What is the minimum recommended safe distance from an X-ray source?

lara31 [8.8K]

Hi! I believe the answer is 2 meters(:

3 0

2 years ago

Read 2 more answers

Write a iterative function that finds the n-th integer of the Fibonacci sequence. Then build a minimal program (main function) t

Natasha2012 [34]

Answer:

Codes for each of the problems are explained below

Explanation:

PROBLEM 1 IN C++:

#include<iostream>

using namespace std;

//fib function that calculate nth integer of the fibonacci sequence.

void fib(int n){

// l and r inital fibonacci values for n=1 and n=2;

int l=1,r=1,c;

//if n==1 or n==2 then print 1.

if(n==1 || n==2){

cout << 1;

return;

}

//for loop runs n-2 times and calculates nth integer of fibonacci sequence.

for(int i=0;i<n-2;i++){

c=l+r;

l=r;

r=c;

cout << "(" << i << "," << c << ") ";

}

//prints nth integer of the fibonacci sequence stored in c.

cout << "\n" << c;

}

int main(){

int n; //declared variable n

cin >> n; //inputs n to find nth integer of the fibonacci sequence.

fib(n);//calls function fib to calculate and print fibonacci number.

}

PROBLEM 2 IN PYTHON:

def fib(n):

print("fib({})".format(n), end=' ')

if n <= 1:

return n

else:

return fib(n - 1) + fib(n - 2)

if __name__ == '__main__':

n = int(input())

result = fib(n)

print()

print(result)

7 0

2 years ago

Read 2 more answers

-Mn has a cubic structure with a0 = 0.8931 nm and a density of 7.47 g/cm3. -Mn has a different cubic structure, with a0 = 0.63

Fudgin [204]

Answer:

The percentage volume change is -3.0%

Explanation: We are to determine the percentage change that will occurs is alpha-Mn is transformed to beta-Mn

Value are defined as;

Cubic structure (a0) for alpha-Mn = 0.8931nm = 0.8931e-9m = 7.1236e-28cm3

Cubic structure (a0) for beta-Mn = 0.6326nm = 0.6326e-9m = 2.5316e-28cm3

Density of alpha-Mn = 7.47g/cm3

Density of beta-Mn = 7.26g/cm3

Atomic weight of Mn = 54.938g/mol

Atomic radius of Mn = 0.112nm

STEP1: CALCULATE THE ATOM NUMBER PER CELL IN THE ALPHA-Mn;

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight ) × 100000

(7.47× 7.1236e-28 × 6.02e23) ÷ 54.938 = 58.31

Therefore the number of Atom in alpha-Mn is 58.31 atom per cell

STEP2: CALCULATE THE NUMBER OF ATOM PER CELL IN THE BETA-Mn

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight) × 1000000

(7.26 × 2.5316e-28 × 6.02e23) ÷ 54.938 = 20.14

Therefore the number of Atom in beta-Mn is 20.14 atom per cell

STEP3: CALCULATE THE PERCENTAGE VOLUME OF ALPHA-Mn AND BETA-Mn

V% = [(volume of atom × number of atom per cell) ÷ volume of unit cell] × 1000

For Alpha-Mn:

[(1.4049e-30 × 58.31) ÷ 7.1236e-28] × 1000 = 114.998%

For Beta-Mn:

[(1.4049e-30 × 20.14) ÷ 2.5316e-28] × 1000 = 111.766%

STEP4: CALCULATE THE CHANGE IN PERCENTAGE VOLUME FOR ALPHA TO TRANSFORM TO BETA

change = final state - initial state

Therefore;

Change = 111.766 - 114.998 = -3.23%

Therefore for a transformation of Alpha-Mn to Beta-Mn they will be a decrease in volume

3 0

2 years ago