Answer:
1.14 * 10^-3 amperes
Explanation:
According to ohms law
V = IR ---- ( 1 )
V = voltage = 2.4 * 10^3 v
I = current = ?
R = resistance = 2.1 * 10^6 Ω
from equation 1
I = V / R
= ( 2.4 * 10^3 ) / (2.1 * 10^6 )
= 1.14 * 10^-3 amperes
Answer:
ΔL = 1.68 mm
σ = 84 MPa
Explanation:
Thermal expansion is:
ΔL = α ΔT L
Thermal stress is:
σ = α ΔT E
Given:
α = 1.2×10⁻⁵ /°C
E = 1.0×10⁵ MPa
ΔT = 80°C − 10°C = 70°C
L = 2 m
ΔL = (1.2×10⁻⁵ /°C) (70°C) (2 m)
ΔL = 0.00168 m
ΔL = 1.68 mm
σ = (1.2×10⁻⁵ /°C) (70°C) (1.0×10⁵ MPa)
σ = 84 MPa
Answer: In a clean plastic or metal container with a tightly sealed lid
Answer:
The service level for each component must be 97.91%
Explanation:
If we want a 90% confidence of starting on time, that means we need

As the probability of each component being ready is independent from the others, that means that the probability of the 5 components being ready is equal to multiply each probability:

The probability of being ready on time is equal to the service level (in fraction), and all 5 are equal so we can say:
![0.9=(\mbox{service level(in fraction)})^5\\\\\sqrt[5]{0.9} =\mbox{service level(in fraction)}=0.9791\\\mbox{In percentage}: \mbox{service level (in fraction)}*100 = 97.91\%](https://tex.z-dn.net/?f=0.9%3D%28%5Cmbox%7Bservice%20level%28in%20fraction%29%7D%29%5E5%5C%5C%5C%5C%5Csqrt%5B5%5D%7B0.9%7D%20%3D%5Cmbox%7Bservice%20level%28in%20fraction%29%7D%3D0.9791%5C%5C%5Cmbox%7BIn%20percentage%7D%3A%20%5Cmbox%7Bservice%20level%20%28in%20fraction%29%7D%2A100%20%3D%2097.91%5C%25)
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