Answer:
![Ma=\frac{260}{330} \\Ma=0.7878](https://tex.z-dn.net/?f=Ma%3D%5Cfrac%7B260%7D%7B330%7D%20%5C%5CMa%3D0.7878)
So, Ma < 1 Flow is Subsonic
Explanation:
Mach Number:
Mach Number is the ratio of speed of the object to the speed of the sound. It is used to categorize the speed of the object on the basis of mach number as sonic, supersonic and hyper sonic. (It is a unit less quantity)
Mach < 1 Subsonic
Mach > 1 Supersonic
Ma= Speed of the object/Speed of the sound
![Ma=\frac{260}{330} \\Ma=0.7878](https://tex.z-dn.net/?f=Ma%3D%5Cfrac%7B260%7D%7B330%7D%20%5C%5CMa%3D0.7878)
So, Ma < 1 Flow is Subsonic
Answer:
B only
Explanation:
Squeeze-type resistance spot welding (STRSW)is a type of electric resistance welding that brings about the weld on interfacing sheet metal pieces through which heat generated from electric resistance bring about fusion and welding of the two pieces together
Therefore, it is not meant for opening but joints but it can be used for making replacement spot welds adjacent to the original spot weld due to the smaller heat affected zone (HAZ) created by the STRSW process.
The modulus of elasticity is 28.6 X 10³ ksi
<u>Explanation:</u>
Given -
Length, l = 5in
Force, P = 8000lb
Area, A = 0.7in²
δ = 0.002in
Modulus of elasticity, E = ?
We know,
Modulus of elasticity, E = σ / ε
Where,
σ is normal stress
ε is normal strain
Normal stress can be calculated as:
σ = P/A
Where,
P is the force applied
A is the area of cross-section
By plugging in the values, we get
σ = ![\frac{8000 X 10^-^3}{0.7}](https://tex.z-dn.net/?f=%5Cfrac%7B8000%20X%2010%5E-%5E3%7D%7B0.7%7D)
σ = 11.43ksi
To calculate the normal strain we use the formula,
ε = δ / L
By plugging in the values we get,
ε = ![\frac{0.002}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B0.002%7D%7B5%7D)
ε = 0.0004 in/in
Therefore, modulus of elasticity would be:
![E = \frac{11.43}{0.004} \\\\E = 28.6 X 10^3 ksi](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B11.43%7D%7B0.004%7D%20%5C%5C%5C%5CE%20%3D%2028.6%20X%2010%5E3%20ksi)
Thus, modulus of elasticity is 28.6 X 10³ ksi
Answer:
M_c = 61.6 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two casesshown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *88
M_c = 2.5*( 42 / 150 ) *88
M_c = 61.6 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.