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:
The frequency that the sampling system will generate in its output is 70 Hz
Explanation:
Given;
F = 190 Hz
Fs = 120 Hz
Output Frequency = F - nFs
When n = 1
Output Frequency = 190 - 120 = 70 Hz
Therefore, if a system samples a sinusoid of frequency 190 Hz at a rate of 120 Hz and writes the sampled signal to its output without further modification, the frequency that the sampling system will generate in its output is 70 Hz
True
the answer to this would be true
Answer:
Given,
Temperature;
T = 393;;K
Convert to Celcius;
T = (393-273) degrees
T = 120°C
Using Table A-4 (Saturated water - Temperature table), at T = 120 C;
vf = 0.001060 m³/kg
vg = 0.89133 m³/kg
Quality is given as;
75% = 0.75
Specific volume is given as;
v = vf + x (vg - vf) = 0.001060 + 0.75(0.89133 _ 0.001060)
v= 0.66876 m³/kg
We know;
v = V/m
0.66876 = 100/m
m = 149.53 kg
Answer:
2.8
Explanation:
The ideal mechanical advantage of the pulley IMA = D'/D where D' = diameter of output pulley = 7 inches and D = diameter of input pulley = 2.5 inches
So, IMA = D'/D
= 7/2.5
= 2.8
So, the ideal mechanical advantage of the pulley IMA = 2.8
