Answer:
4140 steel contains 0.4% C having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C
Explanation:
we have given 4140 steel contains 0.4% C
we know here that 4140 steel is low steel alloy , and it have low amount of chromium , manganese etc alloying element
and these elements which are present in 4140 steel they increase yield strength and ultimate strength of steel
while in 1045 steel contains 0.45 % c is plain carbon steel
and it do not contain any alloying element
so that 4140 steel contains 0.4% C having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C
Answer: both mm and inches on each dimension in a sketch (with the main dimension in one format and the other in brackets below it), in the way you can have dual dimensions shown when detailing an idw view.
personally think it would look a mess/cluttered with even more text all over the sketch environment, but everyone's differenent.
If it's any help - you know you can enter dimensions in either format? If you're working in mm you can still dimension a line and type "2in" and vice-versa. Probably know this already, but no harm saying it, just in case.
You can enter the units directly in or mm and Inventor will convert to current document settings (which you can change - maybe someone can come up with a simple toggle icon to toggle the document settings). Tools>Document Settings>Units
Unlike SolidWorks when you edit the dimension the original entry shows in the dialog box so it makes it easy to keep track of different units even if they aren't always displayed. (SWx does the conversion or equation and then that is what you get.)
I work quite a bit in inch and metric and combination (ex metric frame motor on inch machine) and it doesn't seem to be a real difficulty to me.
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
Answer:
a) 70.29 %
b) 37%
Explanation:
percent reduction can be found from:
PR = 100*(π(do/2)^2-π(df/2)^2)/π(do/2)^2
= 100*(π(11.34/2)^2-π(6.21/2)^2)/π(11.34/2)^2
=70.29 %
percent elongation can be found from:
EL =L_f - Lo/Lo*100
= (73.17 -53.3/53.3)*100
= 37%
Answer:
A wheelbarrow, a bottle opener, and an oar are examples of second class levers
