Answer:
The work furnished by the compressor is 
The minimum work required for the state to change is 
Explanation:
The explanation to these solution is on the first, second , third and fourth uploaded image respectively
Answer:
a) V = 0.354
b) G = 25.34 GPA
Explanation:
Solution:
We first determine Modulus of Elasticity and Modulus of rigidity
Elongation of rod ΔL = 1.4 mm
Normal stress, δ = P/A
Where P = Force acting on the cross-section
A = Area of the cross-section
Using Area, A = π/4 · d²
= π/4 · (0.0020)² = 3.14 × 10⁻⁴m²
δ = 50/3.14 × 10⁻⁴ = 159.155 MPA
E(long) = Δl/l = 1.4/600 = 2.33 × 10⁻³mm/mm
Modulus of Elasticity Е = δ/ε
= 159.155 × 10⁶/2.33 × 10⁻³ = 68.306 GPA
Also final diameter d(f) = 19.9837 mm
Initial diameter d(i) = 20 mm
Poisson said that V = Е(elasticity)/Е(long)
= - <u>( 19.9837 - 20 /20)</u>
2.33 × 10⁻³
= 0.354,
∴ v = 0.354
Also G = Е/2. (1+V)
= 68.306 × 10⁹/ 2.(1+ 0.354)
= 25.34 GPA
⇒ G = 25.34 GPA
Answer:
Technician B only is correct
Explanation:
Voltage drop testing is a method used to find the amount of electrical resistance available in an high amperage circuit that involves connecting the leads of the meter in parallel to the circuits being tested such that disassembly is not required
In voltage drop test, the red voltmeter lead and black voltmeter lead are placed at two points on the same side of the circuit connection such that the leads are in between two positive connection or two negative connection (rather than connecting the red to the positive and the black to the negative sides of the circuit) and digital voltmeter is used for the voltage drop measurement across the lead while the connection is under load.
Therefore, Technician B only is correct
Answer:
def theRoundTrip(movement):
x=0
y=0
for i in movement:
if i not in ["U","L","D","R"]:
print("bad input")
return
if i=="U":
y+=1
if i=="L":
x-=1
if i=="D":
y-=1
if i=="R":
x+=1
return x==0 and y==0
