Answer:
final volume V2 = 0.71136 m³
work done in process W = -291.24 kJ
heat transfer Q = 164 kJ
Explanation:
given data
mass = 1.5 kg
pressure p1 = 200 kPa
temperature t1 = 150°C
final pressure p2 = 600 kPa
final temperature t2 = 350°C
solution
we will use here superheated water table that is
for pressure 200 kPa and 150°C temperature
v1 = 0.95964 m³/kg
u1 = 2576.87 kJ/kg
and
for pressure 600 kPa and 350°C temperature
v2 = 0.47424 m³/kg
u2 = 2881.12 kJ/kg
so v1 is express as
V1 = v1 × m ............................1
V1 = 0.95964 × 1.5
V1 = 1.43946 m³
and
V2 = v2 × m ............................2
V2 = 0.47424 × 1.5
final volume V2 = 0.71136 m³
and
W = P(avg) × dV .............................3
P(avg) =
=
= 400 × 10³
put here value
W = 400 × 10³ × (0.71136 - 1.43946 )
work done in process W = -291.24 kJ
and
heat transfer is
Q = m × (u2 - u1) + W .............................4
Q = 1.5 × (2881.12 - 2576.87) + 292.24
heat transfer Q = 164 kJ
Answer:

Explanation:
Given that:
The direction of the applied tensile stress =[001]
direction of the slip plane = [
01]
normal to the slip plane = [111]
Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:
![cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big]](https://tex.z-dn.net/?f=cos%20%5Clambda%20%3D%20%5CBig%20%5B%5Cdfrac%7Bd_1d_2%2Be_1e_2%2Bf_1f_2%7D%7B%5Csqrt%7B%28d_1%5E2%2Be_1%5E2%2Bf_1%5E2%29%2B%28d_2%5E2%2Be_2%5E2%2Bf_2%5E2%29%20%7D%7D%20%5CBig%5D)
where;
= directional indices for tensile stress
= slip direction
replacing their values;
i.e
= 0 ,
= 0
= 1 &
= -1 ,
= 0 ,
= 1
![cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big]](https://tex.z-dn.net/?f=cos%20%5Clambda%20%3D%20%5CBig%20%5B%5Cdfrac%7B%280%5Ctimes%20-1%29%2B%280%5Ctimes%200%29%20%2B%20%281%5Ctimes%201%29%20%7D%7B%5Csqrt%7B%280%5E2%2B0%5E2%2B1%5E2%29%2B%28%28-1%29%5E2%2B0%5E2%2B1%5E2%29%20%7D%7D%20%5CBig%5D)

Also, to find the angle
between the stress [001] & normal slip plane [111]
Then;
![cos \ \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big]](https://tex.z-dn.net/?f=cos%20%5C%20%20%5Cphi%20%3D%20%5CBig%20%5B%5Cdfrac%7Bd_1d_3%2Be_1e_3%2Bf_1f_3%7D%7B%5Csqrt%7B%28d_1%5E2%2Be_1%5E2%2Bf_1%5E2%29%2B%28d_3%5E2%2Be_3%5E2%2Bf_3%5E2%29%20%7D%7D%20%5CBig%5D)
replacing their values;
i.e
= 0 ,
= 0
= 1 &
= 1 ,
= 1 ,
= 1
![cos \ \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]](https://tex.z-dn.net/?f=cos%20%5C%20%20%5Cphi%3D%20%5CBig%20%5B%20%5Cdfrac%7B%20%280%20%5Ctimes%201%29%2B%280%20%5Ctimes%201%29%2B%281%20%5Ctimes%201%29%7D%20%7B%5Csqrt%20%7B%280%5E2%2B0%5E2%2B1%5E2%29%2B%281%5E2%2B1%5E2%20%2B1%5E2%29%7D%20%7D%20%5CBig%5D)

However, the critical resolved SS(shear stress)
can be computed using the formula:

where;
applied tensile stress
13.9 MPa
∴


Answer:
c) 1.75 g/cm³
Explanation:
Given that
Radii of the A ion, r(c) = 0.137 nm
Radii of the X ion, r(a) = 0.241 nm
Atomic weight of the A ion, A(c) = 22.7 g/mol
Atomic weight of the X ion, A(a) = 91.4 g/mol
Avogadro's number, N = 6.02*10^23 per mol
Solution is attached below