Solution :
Using the data table for refrigerant-134a at P = 120 psia
∴ 
For pressure, P = 20 psia
Change in temperature, 
Now we find the quality,
The final energy,
Change in internal energy
= 38.09297-40.5485
= -2.4556
Answer:
nothing much what class r u in
Answer:
P = 80.922 KW
Explanation:
Given data;
Length of load arm is 900 mm = 0.9 m
Spring balanced read 16 N
Applied weight is 500 N
Rotational speed is 1774 rpm
we know that power is given as
T Torque = (w -s) L = (500 - 16)0.9 = 435.6 Nm
angular speed 
Therefore Power is
P = 80.922 KW
Answer:
a) the inductance of the coil is 6 mH
b) the emf generated in the coil is 18 mV
Explanation:
Given the data in the question;
N = 570 turns
diameter of tube d = 8.10 cm = 0.081 m
length of the wire-wrapped portion l = 35.0 cm = 0.35 m
a) the inductance of the coil (in mH)
inductance of solenoid
L = N²μA / l
A = πd²/4
so
L = N²μ(πd²/4) / l
L = N²μ(πd²) / 4l
we know that μ = 4π × 10⁻⁷ TmA⁻¹
we substitute
L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)
L = 0.00841549 / 1.4
L = 6 × 10⁻³ H
L = 6 × 10⁻³ × 1000 mH
L = 6 mH
Therefore, the inductance of the coil is 6 mH
b)
Emf ( ∈ ) = L di/dt
given that; di/dt = 3.00 A/sec
{∴ di = 3 - 0 = 3 and dt = 1 sec}
Emf ( ∈ ) = L di/dt
we substitute
⇒ 6 × 10⁻³ ( 3/1 )
= 18 × 10⁻³ V
= 18 × 10⁻³ × 1000
= 18 mV
Therefore, the emf generated in the coil is 18 mV
Answer:
One
For surface-mounted and pendant-hung luminaires, support rods should be placed so that they extend about _one___
<h3>what is supported mounted?</h3>
- A structure that holds up or serves as a foundation for something else. Support is a synonym for mounting.
To learn more about it, refer
to brainly.com/question/25689052
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