Answer:
Explanation gives the answer
Explanation:
% Using MATLAB,
% Matlab file : fieldtovar.m
function varargout = fieldtovar(S)
% function that accepts single structure as input, assigning each
% of the field values to user-defined variables
fields = fieldnames(S); % get the field names of the input structure
% check if number of user-defined variables and number of fields in
% structure are equal
if nargout == length(fields)
% if equal assign each value of structure to user-defined varable
for i=1:nargout
varargout{i} = getfield(S,fields{i});
end
else
% if not equal display an error message
error('The number of output variables does not equal the number of fields');
end
end
%This brings an end to the program
uhh , not all girls are like that. many of em these days are more career oriented and thats good i guess , so if anything ur just being judgy a f here.
Answer:
a. 2.9218x10^(-29) m^3
b. 7.1230x10^(28) atoms/m^3
c. 2.0812 BM/atom
Explanation:
Atomic radius r = 0.154 nm
saturation flux density Bs = 0.83 tesla
- the formula for the volume of the simple cubic crystal (Vc) = a^3 = (2r)^3
= (2 x 0.154x10^-9)^3
= 2.9218x10^-29 m^3
- The formula for the Bohr magneton per atom with respect to VC, Bs, permeability of the vacuum Uo and magnetic moment per Bohr magneton Ub is;
Bohr magneton per atom nb = (Bs x VC) / (Ub x Uo)
= (0.83 x 2.9218x10^-29) / (9.27x10^-24) x(1.257x10^-6)
= 2.4251x10^-29 / 1.1652x10^-29
= 2.0812 BM/atom
- Number of atoms per Vc, N = nb / Vc
= 2.0812 / 2.9218x10^-29 = 7.1230x10^28 atoms
Question:
The following tabulated data were gathered from a series of Charpy impact tests on a commercial low-carbon steel alloy.
Temperature(∘C)50403020100-10-20-30-40Impact energy (J)76767158382314951.5
(a) Plot the data as impact energy versus temperature.
(b) Determine a ductile-to-brittle transition temperature as the temperature corresponding to the average of the maximum and minimum impact energies.
(c) Determine a ductile-to-brittle transition temperature as the temperature at which the impact energy is 20 J.
Answer:
a) see the attached graph
b) max E = 76 and Min E =1.53 average = 77.5/2 = 38.75 J
this corresponds to about 10° C
c ) at E = 20 J temperature is about -2°C