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schepotkina [342]
3 years ago
10

Write a MATLAB program in a script file that calculate the average, standard

Engineering
2 answers:
Kryger [21]3 years ago
7 0

Answer:

Code in MATLab is given as  below:

Explanation:

grade = input('Enter the grades as elements of a vector ');

x1 = length(grade);

fprintf('There are %5.2f grades\n',x1);

x2 = mean(grade);

fprintf('The average grade is %5.2f \n',x2);

x3=std(grade);

fprintf('The standard deviation is %5.2f \n',x3);

x4 = median(grade);

fprintf('The median grade is %5.2f \n',x4);

ella [17]3 years ago
4 0

Answer:

Complete Matlab code along with explanation and output results are given below.

Matlab Code:

clear all

close all

clc

input_grades = input('Enter the grades as elements of a vector ');  

len = length(input_grades);

fprintf('There are %.2f grades\n', len);

Mean = mean(input_grades);

fprintf('The Average grade is: %.3f\n', Mean);

STD=std(input_grades);

fprintf('The Standard Deviation is: %.3f\n', STD);

Median = median(input_grades);

fprintf('The Median grade is: %.2f\n', Median);

Output:

Enter the grades as elements of a vector [92,74,53,61,100,42,80,66,71,78,91,85,79,68]

There are 14.00 grades

The Average grade is: 74.286

The Standard Deviation is: 15.760

The Median grade is: 76.00

Explanation:

The user provides input vector of grades

Using the built-in functions of Matlab the length, mean, standard deviation and median of the grades are calculated and printed.

%.2f is a text formatting command that means show up to 2 decimal digits and convert the floating point values to text.

The input vector provided in the question is tested and the program is working correctly.

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A heat engine receives heat from a heat source at 1453 C and has a thermal efficiency of 43 percent. The heat engine does maximu
xxMikexx [17]

Answer:

a) 1253 kJ

b) 714 kJ

c) 946 C

Explanation:

The thermal efficiency is given by this equation

η = L/Q1

Where

η: thermal efficiency

L: useful work

Q1: heat taken from the heat source

Rearranging:

Q1 = L/η

Replacing

Q1 = 539 / 0.43 = 1253 kJ

The first law of thermodynamics states that:

Q = L + ΔU

For a machine working in cycles ΔU is zero between homologous parts of the cycle.

Also we must remember that we count heat entering the system as positiv and heat leaving as negative.

We split the heat on the part that enters and the part that leaves.

Q1 + Q2 = L + 0

Q2 = L - Q1

Q2 = 539 - 1253 = -714 kJ

TO calculate a temperature for the heat sink we must consider this cycle as a Carnot cycle. Then we can use the thermal efficiency equation for the Carnot cycle, this one uses temperatures:

η = 1 - T2/T1

T2/T1 = 1 - η

T2 = (1 - η) * T1

The temperatures must be given in absolute scale (1453 C = 1180 K)

T2 = (1 - 0.43) * 1180 = 673 K

673 K = 946 C

8 0
3 years ago
A solid circular rod that is 600 mm long and 20 mm in diameter is subjected to an axial force of P = 50 kN The elongation of the
Kay [80]

Answer:

a) V = 0.354

b)  G = 25.34 GPA

Explanation:

Solution:

We first determine Modulus of Elasticity and Modulus of rigidity

Elongation of rod ΔL = 1.4 mm

Normal stress, δ = P/A

Where P = Force acting on the cross-section

A = Area of the cross-section

Using Area, A = π/4 · d²

= π/4 · (0.0020)²  = 3.14 × 10⁻⁴m²

δ = 50/3.14 × 10⁻⁴    = 159.155 MPA

E(long) = Δl/l  = 1.4/600 = 2.33 × 10⁻³mm/mm

Modulus of Elasticity Е = δ/ε

= 159.155 × 10⁶/2.33 × 10⁻³    = 68.306 GPA

Also final diameter d(f) = 19.9837 mm

Initial diameter d(i) = 20 mm

Poisson said that V = Е(elasticity)/Е(long)

= -  <u>( 19.9837 - 20 /20)</u>

        2.33 × 10⁻³                  

= 0.354,

∴ v = 0.354

Also G = Е/2. (1+V)

=  68.306 × 10⁹/ 2.(1+ 0.354)

= 25.34 GPA

⇒ G = 25.34 GPA

5 0
3 years ago
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