Answer:
the pressure reading when connected a pressure gauge is 543.44 kPa
Explanation:
Given data
tank volume (V) = 400 L i.e 0.4 m³
temperature (T) = 25°C i.e. 25°C + 273 = 298 K
air mass (m) = 3 kg
atmospheric pressure = 98 kPa
To find out
pressure reading
Solution
we have find out pressure reading by gauge pressure
i.e. gauge pressure = absolute pressure - atmospheric pressure
first we find absolute pressure (p) by the ideal gas condition
i.e pV = mRT
p = mRT / V
p = ( 3 × 0.287 × 298 ) / 0.4
p = 641.44 kPa
so
gauge pressure = absolute pressure - atmospheric pressure
gauge pressure = 641.44 - 98
gauge pressure = 543.44 kPa
Answer:
Please see the attached file for the complete answer.
Explanation:
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
