Answer:
D. Perform a thorough visual inspection.
Answer:
System integration can be defined as the progressive linking and testing of system components to merge their functional and technical characteristics into a comprehensive interoperable system.
Explanation:
....
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:
a)Δs = 834 mm
b)V=1122 mm/s

Explanation:
Given that

a)
When t= 2 s


s= 114 mm
At t= 4 s


s= 948 mm
So the displacement between 2 s to 4 s
Δs = 948 - 114 mm
Δs = 834 mm
b)
We know that velocity V


At t= 5 s


V=1122 mm/s
We know that acceleration a


a= 90 t
a = 90 x 5

Answer:
D. Both hosts 10.168.7.10 and 10.168.7.11 will be permitted
Explanation:
access-list 90
deny 10.168.7.0 0.0.0.255
permit 10.168.7.10
permit 10.168.7.11
permit 10.168.7.12
deny any