Answer:
Technician A and Technician B both are correct.
Explanation:
Technician A accurately notes that perhaps the forms of combustion process most widely used are hemispherical and cross.
Technician B also correctly notes that in several cylinder heads, cooling system and greases gaps and pathways are found.
Answer:
// Program is written in C++ Programming Language
// Comments are used for explanatory purpose
#include<iostream>
using namespace std;
int main ()
{
// Variable declaration
string name;
int numQuestions;
int numCorrect;
double percentage;
//Prompt to enter student's first and last name
cout<<"Enter student's first and last name";
cin>>name; // this line accepts input for variable name
cout<<"Number of question on test"; //Prompt to enter number of questions on test
cin>> numQuestions; //This line accepts Input for Variable numQuestions
cout<<"Number of answers student got correct: "; // Prompt to enter number of correct answers
cin>>numCorrect; //Enter number of correct answers
percentage = numCorrect * 100 / numQuestions; // calculate percentage
cout<<name<<" "<<percentage<<"%"; // print
return 0;
}
Explanation:
The code above calculates the percentage of a student's score in a certain test.
The code is extracted from the Question and completed after extraction.
It's written in C++ programming language
Answer:
a) the power consumption of the LEDs is 0.25 watt
b) the LEDs drew 0.0555 Amp current
Explanation:
Given the data in the question;
Three AAA Batteries;
<---- 1000mAh [ + -] 1.5 v ------1000mAh [ + -] 1.5 v --------1000mAh [ + -] 1.5 v------
so V_total = 3 × 1.5 = 4.5V
a) the power consumption of the LEDs
I_battery = 1000 mAh / 18hrs { for 18 hrs}
I_battery = 1/18 Amp { delivery by battery}
so consumption by led = I × V_total
we substitute
⇒ 1/18 × 4.5
P = 0.25 watt
Therefore the power consumption of the LEDs is 0.25 watt
b) How much current do the LEDs draw
I_Draw = I_battery = 1/18 Amp = 0.0555 Amp
Therefore the LEDs drew 0.0555 Amp current
Answer:
Consider A is square matrix of order 4 x 4 generated using magic function. Augmented matrix can be generated using:
Aug=[A eye(size(A))]
Above command is tested in MATLAB command window and is attached in figure below
Answer:
Time taken for the capacitor to charge to 0.75 of its maximum capacity = 2 × (Time take for the capacitor to charge to half of its capacity)
Explanation:
The charging of a capacitor/the build up of its voltage follows an exponential progression and is given by
V(t) = V₀ [1 - e⁻ᵏᵗ]
where k = (1/time constant)
when V(t) = V₀/2
(1/2) = 1 - e⁻ᵏᵗ
e⁻ᵏᵗ = 0.5
In e⁻ᵏᵗ = In 0.5 = - 0.693
-kt = - 0.693
kt = 0.693
t = (0.693/k)
Recall that k = (1/time constant)
Time to charge to half of max voltage = T(1/2)
T(1/2) = 0.693 (Time constant)
when V(t) = 0.75
0.75 = 1 - e⁻ᵏᵗ
e⁻ᵏᵗ = 0.25
In e⁻ᵏᵗ = In 0.25 = -1.386
-kt = - 1.386
kt = 1.386
t = 1.386(time constant) = 2 × 0.693(time constant)
Recall, T(1/2) = 0.693 (Time constant)
t = 2 × T(1/2)
Hope this Helps!!!
