The magnitude of applied stress in the direction of 101 is 12.25 MPA and in the direction of 011, it is not defined.
<u>Explanation</u>:
<u>Given</u>:
tensile stress is applied parallel to the [100] direction
Shear stress is 0.5 MPA.
<u>To calculate</u>:
The magnitude of applied stress in the direction of [101] and [011].
<u>Formula</u>:
zcr=σ cosФ cosλ
<u>Solution</u>:
For in the direction of 101
cosλ = (1)(1)+(0)(0)+(0)(1)/√(1)(2)
cos λ = 1/√2
The magnitude of stress in the direction of 101 is 12.25 MPA
In the direction of 011
We have an angle between 100 and 011
cosλ = (1)(0)+(0)(-1)+(0)(1)/√(1)(2)
cosλ = 0
Therefore the magnitude of stress to cause a slip in the direction of 011 is not defined.
Answer:
a)Wt =25.68 lbf
b)Wt = 150 lbf
F= 899.59 N
Explanation:
Given that
m= 150 lbm
a)
Weight on the spring scale(Wt) = m g
We know that
Wt = 150 x 5.48/32 lbf
Wt =25.68 lbf
b)
On the beam scale
This is scale which does not affects by gravitational acceleration.So the wight on the beam scale will be 150 lbf.
Wt = 150 lbf
If the plane is moving upward with acceleration 6 g's then the for F
F = m a
We know that
a=6 g's
So
F = 90 x 9.99 N
F= 899.59 N
Answer:
During film condensation on a vertical plate, heat flux at the top will be higher since the thickness of the film at the top, and thus its thermal resistance, is lower.
Explanation:
https://www.docsity.com/pt/cengel-solution-heat-and-mass-transfer-2th-ed-heat-chap10-034/4868218/
https://arc.aiaa.org/doi/pdf/10.2514/1.43136
https://arxiv.org/ftp/arxiv/papers/1402/1402.5018.pdf
Answer:
Qin = 1857 kJ
Explanation:
Given
m = 0.5 Kg
T₁ = 25°C = (25 + 273) K = 298 K
P₁ = 100 kPa
P₂ = 500 kPa
First, the temperature when the piston starts rising is determined from the ideal gas equations at the initial state and at that state:
T₂ = T₁*P₂/P₁
⇒ T₂ = 298 K*(500 kPa/100 kPa) = 1490 K
Until the piston starts rising no work is done so the heat transfer is the change in internal energy
Qin = ΔU = m*cv*(T₂-T₁)
⇒ Qin = 0.5*3.1156*(1490 - 298) kJ = 1857 kJ
