Answer:
John should detail his Scrum Master.
Explanation:
The Team Lead or Scrum Master coordinates the tasks of individual team members and supports the progress of the team. The Scrum Master usually receives instructions from the Product Owner and then ensures that the tasks are performed accordingly. She also coaches the Development Team and works with the Product Owner to carry out daily development activities. She also drives the Scrum Values and Principles, ensuring that the team members understand and practice them.
Answer:

Explanation:
Given data:
Diffusion constant for nitrogen is 
Diffusion flux 
concentration of nitrogen at high presuure = 2 kg/m^3
location on which nitrogen concentration is 0.5 kg/m^3 ......?
from fick's first law

Take C_A as point on which nitrogen concentration is 2 kg/m^3

Assume X_A is zero at the surface


Answer:
(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ
Explanation:
Solution
Recall that:
A 10 gr of air is compressed isentropically
The initial air is at = 27 °C, 110 kPa
After compression air is at = a450 °C
For air, R=287 J/kg.K
cv = 716.5 J/kg.K
y = 1.4
Now,
(a) W efind the pressure on [MPa]
Thus,
T₂/T₁ = (p₂/p₁)^r-1/r
=(450 + 273)/27 + 273) =
=(p₂/110) ^0.4/1.4
p₂ becomes 2390.3 kPa
So, p₂ = 2.39 MPa
(b) For the increase in total internal energy, is given below:
ΔU = mCv (T₂ - T₁)
=(10/100) (716.5) (450 -27)
ΔU =3030 J
ΔU =3.03 kJ
(c) The next step is to find the total work needed in kJ
ΔW = mR ( (T₂ - T₁) / k- 1
(10/100) (287) (450 -27)/1.4 -1
ΔW = 3035 J
Hence, the total work required is = 3.035 kJ
Answer:
The amplitude of the absorbed mass can be found
for ka:

now
![w^2=\frac{K_{a} }{m_{a} } \\m_{a} =\frac{K_{a} }{w^2} =\frac{125000}{[6000*2\pi /60]^2} =0.317kg](https://tex.z-dn.net/?f=w%5E2%3D%5Cfrac%7BK_%7Ba%7D%20%7D%7Bm_%7Ba%7D%20%7D%20%5C%5Cm_%7Ba%7D%20%3D%5Cfrac%7BK_%7Ba%7D%20%7D%7Bw%5E2%7D%20%3D%5Cfrac%7B125000%7D%7B%5B6000%2A2%5Cpi%20%2F60%5D%5E2%7D%20%3D0.317kg)
Answer:
The process from a liquid to a vapor.
Explanation:
Water is evaporated by heating up and turning into a vapor.