Answer:
Ammonia gas a hazardous gas to our health, when we are exposed to it for a long time. The gas is lighter than air, that means it's high concentration may not be noticed at the point of leakage, because it flows with the wind direction. Ammonia gas detector are used to determine the concentration of the gas at a particular place. We can use the dispersion modelling software program to know the exact position, where we can place the gas detector, which would be where evacuation is needed.
During evacuation, when the concentration of the gas has increased, a self-contained breathing apparatus should be used for breathing, and an encapsulated suit should be worn to prevent ammonia from reacting with our sweat or any other chemical burn. A mechanic ventilation will also be needed in the place of evacuation, so that the ammonia concentration in that area can be dispersed.
Answer:
28.6 kg
Explanation:
The final weight can be calculated from the mixing data and formulae which is given as follows:
cement content = 
Computing the parameters and checking the tables gives 28.6 kg.
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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
