Amplifiers are extensively used in the baseband portion of a radio receiver system to condition the baseband signal to produce a
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Answer:
the elongation of the metal alloy is 21.998 mm
Explanation:
Given the data in the question;
K = σT/ (εT)ⁿ
given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,
strain-hardening exponent n = 0.22
we substitute
K = 345 /
K = 815.8165 Mpa
next, we determine the true strain
(εT) = (σT/ K)^1/n
given that σT = 412 MPa
we substitute
(εT) = (412 / 815.8165 )^(1/0.22)
(εT) = 0.04481 mm
Now, we calculate the instantaneous length
=
given that = 480 mm
we substitute
=
×
= 501.998 mm
Now we find the elongation;
Elongation =
we substitute
Elongation = 501.998 mm - 480 mm
Elongation = 21.998 mm
Therefore, the elongation of the metal alloy is 21.998 mm
Answer:
I am Providing Answer in C Language Program.
Explanation:
Please find attachment regarding code of taking two numbers input and adding them.
I would like to recommend you please use software which supports C language.
#include <stdio.h>
int main () {
int a, b, sum;
printf ("\ nEnter two no:");
scanf ("% d% d", & d, & e);
sum1 = d + e;
printf ("Sum:% d", sum1);
return (0);
}
Answer:
M_AD = 1359.17 lb-in
Explanation:
Given:
- T_ef = 46 lb
Find:
- Moment of that force T_ef about the line joining points A and D.
Solution:
- Find the position of point E:
mag(BC) = sqrt ( 48^2 + 36^2) = 60 in
BE / BC = 45 / 60 = 0.75
Hence, E = < 0.75*48 , 96 , 36*0.75> = < 36 , 96 , 27 > in
- Find unit vector EF:
mag(EF) = sqrt ( (21-36)^2 + (96+14)^2 + (57-27)^2 ) = 115 in
vec(EF) = < -15 , -110 , 30 >
unit(EF) = (1/115) * < -15 , -110 , 30 >
- Tension T_EF = (46/115) * < -15 , -110 , 30 > = < -6 , -44 , 12 > lb
- Find unit vector AD:
mag(AD) = sqrt ( (48)^2 + (-12)^2 + (36)^2 ) = 12*sqrt(26) in
vec(AD) = < 48 , -12 , 36 >
unit(AD) = (1/12*sqrt(26)) * < 48 , -12 , 36 >
unit (AD) = <0.7845 , -0.19612 , 0.58835 >
Next:
M_AD = unit(AD) . ( E x T_EF)
M_AD = 1835.73 + 116.49528 - 593.0568
M_AD = 1359.17 lb-in