what is an acceptable less hazardous method of cleaning then solvents
what is an acceptable less hazardous method of cleaning then solvents

D

Explanation:

D

Explanation:

I guess that this depends on what you want to clean, but my choice is

E.g. if you are a mechanic and you need to clean auto parts (which are dirty and greasy), then a hot soap washer is very useful. A hot soap washer is basically a dishwasher that works with hot pressurized water. Generally, detergent is much less hazardous than solvents. This method applies to all the objects that can be moved into the sink in order to be washed.

If you want to clean things at home, the same applies. The least hazardous cleaning method is using hot water and detergent.

A pressurized water hose may be a solution to certain issues, e.g. clean bird poop on top of a car or a sidewalk. If you can mechanically clean something (use a broom or vacuum cleaner), then there is no reason why you would need a solvent. Using disposable wipes only increases the amount of waste that you generate.

I guess that this depends on what you want to clean, but my choice is

E.g. if you are a mechanic and you need to clean auto parts (which are dirty and greasy), then a hot soap washer is very useful. A hot soap washer is basically a dishwasher that works with hot pressurized water. Generally, detergent is much less hazardous than solvents. This method applies to all the objects that can be moved into the sink in order to be washed.

If you want to clean things at home, the same applies. The least hazardous cleaning method is using hot water and detergent.

A pressurized water hose may be a solution to certain issues, e.g. clean bird poop on top of a car or a sidewalk. If you can mechanically clean something (use a broom or vacuum cleaner), then there is no reason why you would need a solvent. Using disposable wipes only increases the amount of waste that you generate.

Code:

#include <iostream>

using namespace std;

int main()

{

  int Car_Year;

  cout<<"Please Enter the Car Model."<<endl;

  cin>>Car_Year;    

 if (Car_Year<1967)

 {

cout<<"Few safety features."<<endl;

 }

else if (Car_Year>1971 && Car_Year<=1991)

{

cout<<"Probably has head rests."<<endl;

}

else if (Car_Year>1991 && Car_Year<=2000)

{

cout<<"Probably has antilock brakes."<<endl;

}

else if (Car_Year>2000)

{

cout<<"Probably has airbags."<<endl;

  }

else

{

cout<<"Invalid Selection."<<endl;

}

  return 0;

}

Output:

Please Enter the Car Model.

1975

Probably has head rests.

Please Enter the Car Model.

1999

Probably has antilock brakes.

Please Enter the Car Model.

2005

Probably has airbags.

Please Enter the Car Model.

1955

Few safety features.

Explanation:

We were required to implement multiple If else conditions to assign car model year with the corresponding features of the car.

The code is tested with a wide range of inputs and it returned the same results as it was asked in the question.

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.

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()

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