Serves as a protective barrier to prevent contact with engergized ("hot") parts within the unit
1 answer:
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Answer:
Attached below are the sketches
answer :
c) G(s) = 100 / ( s + 100 )
d) y'(t) + 100Y(s) = 100 X(s)
e) g(t) = e^-100t u(t)
Explanation:
a) Sketch the bode plot
The filter here is a low pass filter
b) Sketch the s-plane
attached below. pole ( s ) is at 100
c) write the transfer function of the filter
Transfer function ; G(s) = 100 / ( s + 100 )
d) write the differential equation
Y(s) / X(s) = 100 / s + 100
Y(s) [ s + 100 ] = 100 X(s)
= sY(s) + 100Y = 100 X(s)
∴ differential equation = y'(t) + 100Y(s) = 100 X(s)
e) write out the unforced transient response
g(t) = e^-100t u(t)
f) write out the frequency response
attached below
Answer:
Both Brass and 1040 Steel maintain the required ductility of 20%EL.
Explanation:
Solution:-
- This questions implies the use of empirical results for each metal alloy plotted as function of CW% and Yield Strength.
- So for each metal alloy use the attached figures as reference and determine the amount of CW% required for a metal alloy to maintain a Yield Strength Y = 345 MPa.
- Left Figure (first) at Y = 345 MPa ( y -axis ) and read on (x-axis):
1040 Steel --------> 0% CW
Brass ---------------> 22% CW
Copper ------------> 66% CW
The corresponding ductility (%EL) for cold Worked metal alloys can be determined from the right figure. Using the %CW for each metal alloy determined in first step and right figure to determine the resulting ductility.
- Right Figure (second) at respective %CW (x-axis) read on (y-axis)
1040 Steel (0% CW) --------> 25% EL
Brass (22% CW) -------------> 21% EL
Copper (66% CW) ----------> 4% EL
We see that both 1040 Steel and Brass maintain ductilities greater than 20% EL at their required CW% for Yield Strength = 345 MPa.
