Answer:
Explanation:
% Clears variables and screen
clear; clc
% Asks user for input
n = input('Total number of objects: ');
r = input('Size of subgroup: ');
% Computes and displays permutation according to basic formulas
p = 1;
for i = n - r + 1 : n
p = p*i;
end
str1 = [num2str(p) ' permutations'];
disp(str1)
% Computes and displays combinations according to basic formulas
str2 = [num2str(p/factorial(r)) ' combinations'];
disp(str2)
=================================================================================
Example: check
How many permutations and combinations can be made of the 15 alphabets, taking four at a time?
The answer is:
32760 permutations
1365 combinations
==================================================================================
Question
Determine the average water exit velocity
Answer:
53.05 m/s
Explanation:
Given information
Volume flow rate, 
Diameter d= 8cm= 0.08 m
Assumptions
- The flow is jet flow hence momentum-flux correction factor is unity
- Gravitational force is not considered
- The flow is steady, frictionless and incompressible
- Water is discharged to the atmosphere hence pressure is ignored
We know that Q=AV and making v the subject then
where V is the exit velocity and A is area
Area, 
where d is the diameter
By substitution
To convert v to m/s from m/s, we simply divide it by 60 hence
Answer:
hello your question is incomplete attached below is the missing equation related to the question
answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°
Explanation:
<u>Determine the friction angle at each depth</u>
attached below is the detailed solution
To calculate the vertical stress = depth * unit weight of sand
also inverse of Tan = Tan^-1
also qc is in Mpa while σ0 is in kPa
Friction angle at each depth
2 meters = 40.389°
3.5 meters = 38.987°
5 meters = 38.022°
6.5 meters = 39.869°
8 meters = 40.265°
Answer:
A selective surface with large absorption for solar radiation and high reflectance for thermal infrared radiation was produced by use of surface oxidation of stainless steel. The surfaces were studied for use with concentrated light in a solar power plant at temperatures of 400°C and higher.
In order to investigate the relation between surface treatment and optical properties, stainless steels (AISI 304 and 430) which were submitted to different chemical and mechanical surface treatments, were used. To increase the spectral selectivity, these surfaces were treated in air and in vacuum at different temperatures and times. The optical properties of these films were investigated. Visual and infrared spectral absorptances were measured at room temperature. The thermal hemispherical emittance and absorptance were obtained by a calorimetric method at 200°C. It was noticed that these chemically and mechanically treated stainless steel surfaces have good spectral properties without further oxidations. This is very important for high temperature uses. The best values are found for samples 7 and 8 under vacuum and air. These two samples with mechanically ground surfaces retained their selectivity and specularity after several hours oxidation. One can conclude that the surface ground treatment confers good selectivity on the steel surfaces for use in concentrating solar collectors with a working temperature of 500°C.
Sample surfaces were subjected to long temperature ageing tests in order to gain some idea of the thermal stability of the surfaces. The results promise better-performing surface and the production of durable selective finishes at, possibly, lower cost than competing processes.
Explanation:
Answer:
The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm
Explanation:
For steel bolt
Stress = 210 MPa or 210 N/mm2
Pressure = Stress* Area
Pbolt = 210 N/mm2 * 16^2 *(pi)/4
Pbolt = 210 N/mm2 * 200.96 mm^2 = 42201.6 N
For Brass spacer
Pressure = 42201.6 N
Area of Brass spacer = Pressure/Stress
Area of Brass spacer = 42201.6 N/145 N/mm^2 = 291.044 mm^2
Area of Brass spacer = (pi) (d^2 - 16^2)/4 = 291.044 mm^2
d^2 - 16^2 = 291.044 mm^2* 4/(pi) = 370.758
d^2 = 370.758 + 16^2
d^2 = 626.758
d = 25.03 mm
The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm
