Answer:
0.5°c
Explanation:
Humidity ratio by mass can be expressed as
the ratio between the actual mass of water vapor present in moist air - to the mass of the dry air
Humidity ratio is normally expressed in kilograms (or pounds) of water vapor per kilogram (or pound) of dry air.
Humidity ratio expressed by mass:
x = mw / ma (1)
where
x = humidity ratio (kgwater/kgdry_air, lbwater/lbdry_air)
mw = mass of water vapor (kg, lb)
ma = mass of dry air (kg, lb)
It can be as:
x = 0.005 (100) / [(100 - 100)]
x = 0.005 x 100 / (100 - 100)
x = 0.005 x 100 / 0
x = 0.5°c
So the temperature to which atmospheric air must be cooled in order to have humidity ratio of 0.005 lb/lb is 0.5°c
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:
b. 10A
Explanation:
Using the formula, E= k × r×I
200= 0.5 ×2000×0.02×I
200=20×I
Dividing with 20
I = 200/20= 10A