By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.
<h3>How to determine the differential of a one-variable function</h3>
Differentials represent the <em>instantaneous</em> change of a variable. As the given function has only one variable, the differential can be found by using <em>ordinary</em> derivatives. It follows:
dy = y'(x) · dx (1)
If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:
By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.
To learn more on differentials: brainly.com/question/24062595
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Answer:
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The Emergency Stop Button icon on the Inputs toolbar can be used to press or release the Emergency Stop button on the CNC machine - False
<h3><u>
Explanation:</u></h3>
The Emergency Stop button is practiced to stop machine operation. When actuated, machine operation terminates instantly. The execution of the part program can be hindered by touching the Control and Spacebar buttons on the computer keyboard. Unlike applying the Emergency Stop button, this way of stopping the milling center does not create the software to drop the path of the tool position.
The Inputs Toolbar is idle. It contributes information only on the state of the Emergency Stop, the safety shield, and the limit switches. An input is active (on) when the button is depressed.
Answer:
Yes, fracture will occur
Explanation:
Half length of internal crack will be 4mm/2=2mm=0.002m
To find the dimensionless parameter, we use critical stress crack propagation equation
and making Y the subject
Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness, is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain
When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m
and making K the subject
and substituting 260 MPa for while a is taken as 0.003m and Y is already known
Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material
Answer:
See explaination and attachment for theprogram code and output.
Explanation:
matlab program:
x1(1)=0;
y1(1)=0;
x2(1)=0;
y2(1)=0;
x3(1)=0;
y3(1)=0;
for i=1:10
x1(i+1)=0.5*x1(i);
y1(i+1)=0.5*y1(i);
x2(i+1)=0.5*x2(i)+0.25;
y2(i+1)=0.5*y2(i)+(sqrt(3)/4);
x3(i+1)=0.5*x3(i)+0.5;
y3(i+1)=0.5*y3(i);
end
figure
hold on
plot(x1,y1,'*')
plot(x2,y2,'*')
plot(x3,y3,'*')
hold off
Please kindly check attachment for output of 10, 100 and 1000 respectively.