Use a 150-amp breaker and No. 1/0 copper conductors
Explanation:
An installation consists of a 10-kVA, single-phase transformer with a 440-volt primary and a 110-volt, 2-wire secondary using insulation type THWN copper conductors. No secondary OCPD is installed. The secondary loads are considered to be noncontinuous loads. Calculate the minimum size secondary conductors required to handle the secondary full load current for this transformer
Calculate primary full-load amps:
10 kVA ÷ (440 volts ÷ 1000) = 22.72 amps
Calculate required feeder breaker size and conductor ampacity:
22.72 amps x 1.5 = 34.09 amps
Use an 80-amp breaker and No. 3 AWG copper conductors 1
Calculate secondary full-load amps:
10 kVA ÷ (110 volts ÷ 1000) = 90.9 amps
Calculate required secondary breaker size and conductor ampacity:
90.9 amps x 1.25 = 113.6 amps
Use a 150-amp breaker and No. 1/0 THWN copper conductors.
Use a 150-amp breaker and No. 1/0 copper conductors
Explanation:
An installation consists of a 10-kVA, single-phase transformer with a 440-volt primary and a 110-volt, 2-wire secondary using insulation type THWN copper conductors. No secondary OCPD is installed. The secondary loads are considered to be noncontinuous loads. Calculate the minimum size secondary conductors required to handle the secondary full load current for this transformer
Calculate primary full-load amps:
10 kVA ÷ (440 volts ÷ 1000) = 22.72 amps
Calculate required feeder breaker size and conductor ampacity:
22.72 amps x 1.5 = 34.09 amps
Use an 80-amp breaker and No. 3 AWG copper conductors 1
Calculate secondary full-load amps:
10 kVA ÷ (110 volts ÷ 1000) = 90.9 amps
Calculate required secondary breaker size and conductor ampacity:
90.9 amps x 1.25 = 113.6 amps
Use a 150-amp breaker and No. 1/0 THWN copper conductors.
True, a retractable service pit cover can help other shop employees from falling into the pit while another technician is in the pit servicing a car.
Retractable service pit cover is a special cover made to cover service pit which aid inspection and repair of vehicle in an auto workshop.
Retractable service pit ensure safety of workshop technicians around service pits.Retractable service pits aid proper inspection of the beneath of the vehicleRetractable service pits enable the technician to have a close view of the vehicle in repair.The pit also serves as guide to short-eyed people or people to are not aware of the structure.In essence, the pit prevent against injury or death of workers and occupier.In conclusion, the retractable service pit cover help other shop employees from falling into the pit even if they are aware of the structure.
Learn more about retractable service pit in picture attached
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.
"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.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
x=float(input("Enter a number: "))
sub=(x-int(x))
print(sub)
Explanation:
Got it Right
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